Notes on the t Distribution Critical Points Pictorially

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```					       8. Inference for Population Means                                  8.1.2 Student’s t Distribution

8.1 One Sample Inference                                     Estimation
¯ is                                          ¯
X √ an estimator of µ, with standard error SE(X) =
8.1.1 Introduction
S/ n.
Sampling Distribution of X¯
• Let X1, X2, . . . , Xn be a random sample from a
population with mean µ and variance σ 2. In this                 ¯          2
Since X ∼ N (µ, σ /n), we know that
section we shall use the data to make inferences
X −µ
Z=           √ ∼ N (0, 1)
σ/ n
• We will assume that the sampling distribution of
¯
X is N (µ, σ 2/n). This assumption is valid if
either (or both) of the following are true:                If we replace the (population standard deviation) σ
by the (sample standard deviation) S then we get
A1: X1, X2, . . . , Xn are normally distributed;
A2: n is suﬃciently large for the Central Limit                                                 ¯
X −µ   ¯
X −µ
Theorem to apply.                                                               T =           √ =    ¯
S/ n  SE(X)

which has Student’s t distribution with ν = (n − 1)
degrees of freedom.

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Notes on the t Distribution                                     Critical Points Pictorially

(i) Distribution of T depends upon n, but not upon
µ or σ.

(ii) When ν = (n − 1) is large, the t distribution is
virtually identical to the N (0, 1) distribution.

(iii) ‘t-tables’ give critical points of the t distribution.
That is, values tα/2(ν) such that
Area=α/2                                               Area=α/2
P (T > tα/2(ν)) = α/2

where T has a t distribution with ν degrees of
freedom. Note that                                                     − t α/2(ν)                             t α/2(ν)

P (T < −tα/2(ν)) = α/2
Figure 35: Critical point ±tα/2(ν) for a t distribution
by symmetry.                                             with ν degrees of freedom.

Some t-tables are available at the S160 web page.

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8.1.3 Conﬁdence Intervals for a                                                            Example 8.1
Population Mean

A car is tested for fuel economy. In 7 road trials the
¯            ¯
Since T = (X − µ)/SE(X) has a t distribution with                     number of km per litre was recorded, giving
ν = (n − 1) degrees of freedom,
¯
x = 15.2              s = 2.2
¯         ¯
1 − α = P −tα/2(ν) < (X − µ)/SE(X) < tα/2(ν)
¯    ¯                ¯                        Let µ be the population mean km per litre. Estimate
= P −tα/2(ν)SE(X) < X − µ < tα/2(ν)SE(X)
µ and ﬁnd a 90% conﬁdence interval for it.
¯              ¯        ¯            ¯
= P X − tα/2(ν)SE(X) < µ < X + tα/2(ν)SE(X)

Conﬁdence Interval for µ
From observed data x1, x2, . . . , xn, a 100(1 − α)%
conﬁdence interval for µ is given by

¯             x ¯               x
x − tα/2(ν)SE(¯), x + tα/2(ν)SE(¯)
√
where SE(¯) = s/ n.
x

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Example 8.2: Conﬁdence Interval for                                Computation by Hand
Mean Pulmonary Compliance                                                                ¯
The mean of the data is x = 209.75 and the sample
standard deviation is s = 24.16. the standard error
√
for x is SE(¯) = 24.16/ 16 = 6.04. To ﬁnd a 95%
¯       x
conﬁdence interval for µ, we need to look up critical
Chronic exposure to asbestos ﬁbre is a well known                     value tα/2 in t-tables with ν = n − 1=15, where
health hazard. The table below gives measurements                     α = 0.05 (since 95% = 100(1 − 0.05)%). From
of pulmonary compliance (a measure of the how                         tables we ﬁnd that t0.025 = 2.131. Hence, a 95%
eﬀectively the lung can inhale and exhale) for                        conﬁdence interval for µ is given by
16 construction workers, 8 months after they had
left a site on which they had suﬀered prolonged                        ¯          x ¯            x
x − tα/2SE(¯), x + tα/2SE(¯)
exposure to asbestos. (Data source: ‘The Acute                               = (209.75 − 2.131 × 6.04, 209.75 + 2.131 × 6.04)
Eﬀects of Chrysotile Asbestos Exposure on Lung                               = (196.9, 222.6)             (1 dp)
Function’, Environmental Research, 1978, pp. 360-
372.) In this example we provide an estimate, and a
95% conﬁdence interval, for the population mean                       Using MINITAB
pulmonary compliance for people with prolonged                        First enter the data into a column of a MINITAB
asbestos ﬁbre exposure.                                               worksheet; this column is named PulCom in this
example. Then go through the menu sequence Stat
167.9         180.8         184.8   189.8   194.8   200.2         → Basic Statistics → 1-Sample t... which
201.9         206.9         207.2   208.4   226.3   227.7         produces the 1-Sample t dialogue box. Choose the
228.5         232.4         239.8   258.6                         conﬁdence level (95% in this case) and click on OK.

Table 16: Pulmonary compliance for 16 workers                         MTB > TInterval 95.0 ’PulCom’.
subjected to prolonged exposure to asbestos ﬁbre
Confidence Intervals
Deﬁne µ to be the population mean pulmonary
compliance for people with prolonged asbestos ﬁbre                    Variabl N   Mean                    StDev SE Mean    95.0 % C.I.
exposure.                                                             PulCom 16 209.75                    24.16   6.04 (196.87, 222.63)
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8.1.4 One Sample t-Test                                                     P-Values for T-test

Aim: to investigate the credibility of some claim                     P-Value: Suppose that the observed value of the
regarding µ.                                                          t-statistic is t. The way of calculating the P-value
depends on H1:
Hypotheses: H1 is a statement of the claim about
µ; H0 is a statement negating the claim.                                                         H1             P-value
µ > µ0          P (T ≥ t)
Test Statistic: Suppose that we are testing
µ < µ0          P (T ≤ t)
µ = µ0         2P (T ≥ |t|)
H0 : µ = µ 0
where T has a t distribution with ν = n − 1 degrees
against any one of the alternatives                                   of freedom.
P-Value Interpretation and Signiﬁcance Levels:
H1 : µ > µ 0               or H1 : µ < µ0            or H1 : µ = µ0   The interpretation of P-values, and the choice and
application of signiﬁcance levels, is exactly the
The appropriate test statistic is the t-statistic                     same as described in section 7.2.1 (for testing a
proportion).
¯      ¯
X − µ0 X − µ0
T =       √ =      ¯
S/ n   SE(X)

• If H0 is true then T has a t sampling distribution
with ν = n − 1 degrees of freedom.

• T is likely to take extreme values when H1 is
correct.

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Example 8.3                                  Example 8.4: Shoshoni Rectangles
The Greeks called a rectangle ‘golden’ if the ratio
√
of its width to its length was 1 ( 5 − 1) = 0.618.
2
12 king-size chocolate bars weighed (in grams) giving                 Such rectangles are aesthetically pleasing to most
people, being not ‘too square’ or ‘too oblong’. The
¯
x = 149.10                 s = 1.73               Shoshoni Indians used beaded rectangles to decorate
their goods. The data tabulated below are the
The producer wants to know whether the                                width-to-length ratios for eighteen such rectangles.
(population) mean weight diﬀers from the advertised
value of 150.0 grams. Write down appropriate                              0.693         0.662         0.690     0.606   0.570   0.749
hypotheses and calculate the t-statistic to test this.                    0.672         0.628         0.609     0.844   0.654   0.615
0.688         0.601         0.576     0.670   0.606   0.611
Table 17: Width-to-length ratios

See also M&M Examples 7.4 and 7.5, Exercise 7.11,                     It is of interest to anthropologists to know whether
7.23, 7.29 for t-tests and conﬁdence intervals. (Use                  the Shoshoni Indians aimed to produce rectangles
MINITAB for tests.)                                                   with the golden ratio. To investigate this, we test

H0 : µ = 0.618 versus H1 : µ = 0.618,

where µ is the population mean width-to-length ratio
of Shoshoni rectangles. To do so, we enter data into
the data into a column of a MINITAB worksheet. In
this example, the column has been named Ratio.
To perform a t-test of H0 against H1, go through
the menu sequence Stat → Basic Statistics →
1-Sample t.... This will produce a dialogue box

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in which you must enter the variable to be tested                                           8.1.5 Paired t-Test
(Ratio in this case). You also need to click on the
circle by Test mean, and to ﬁll in the H0 value for
µ (0.618 in this case). Finally, you must choose
the type of alternative hypothesis (in our case not                  Consider an experiment in which there is one
equal). When all that’s done, click on OK. This                      treatment and one control group.      Diﬀerences
should produce the following output in the MINITAB                   between the proﬁle of subjects in each group can
command window:                                                      cause problems.

T-Test of the Mean

Test of mu = 0.6180 vs mu not = 0.6180

Var            N         Mean         StDev    SE Mean   T P-Value
Ratio         18       0.6513        0.0665     0.0157 2.13  0.049
This tells us (amongst other things) that

• the value of the test statistic is 2.13 (which is
computed from (0.6513 − 0.6180)/0.0157);

• the P-value for this test is 0.049.

Because the P-value is less that 0.05 (if only just!)
we reject H0. However, the evidence in favour of H1
is not very strong.
Conclusion: the data provides some evidence that
the population mean width-to-length ratio for the
Shoshoni rectangles is not the golden ratio.

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Example 8.5: A Small Clinical Trial                                  outcomes than the control group. It will be
impossible to say whether this is due to the
treatment, or due to the fact that the disease
tends to be worse in women (or in the young).
Four patients with a rare disease are recruited to
a clinical trial. The patient’s characteristics are as
follows.
Patient    Age     Sex
A        71       M
B        68       M
C        28       F
D        34       F
Two patients are to be assigned to the treatment
group, and two to the control group. If we did this
completely at random, we could get patients A and
B in the treatment group, and C and D in the control
group. This could cause problems.

• Suppose that the disease tends to be worse in
men than women. Then the outcomes for the
treatment and control group may be very similar
even though the treatment is beneﬁcial, since
the beneﬁts are counterbalanced by the increased
acuteness of the disease in men. (The same
argument could also be made in terms of age
rather than sex.)

• Suppose that the treatment group do have better

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Matched Pairs                                       More Matched Pairs

• A powerful way of designing an experiment to
It is often possible to ‘match’ an individual to
avoid such problems is matching.
him/herself, and look at results under two diﬀerent
experimental conditions (e.g. before and after some
• The idea is to select experimental units in
treatment).
pairs that are ‘matched’ on potentially important
characteristics, such as subject age and sex in a
clinical trial.
Example 8.7
Twenty students studying European languages went
• We then randomly assign one member of each pair
on a two week intensive course in spoken French. To
to the treatment group, and the other member of
assess the eﬀectiveness of the course, each student
the pair to the control group.
took a spoken French test before the course, and an
equivalent test after the course. The before and after
• The result is a treatment group and a control         test scores on each student are paired observations.
group that are very similar in terms of important
characteristics of the experimental units (i.e. the
variability between subjects is controlled).

Example 8.6
In Example 8.5, patients A and B form a matched
pair, as do patients C and D. We should randomly
assign A and B to diﬀerent groups, and C and D to
diﬀerent groups.

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Analysis of Matched Pairs Data                     Example 8.8: Analysis of Shoe Wear

This example is concerned with data from an
experiment on shoe wear. A manufacturing company
• Matched pairs data should be analysed by looking      wanted to compare two synthetic materials (A and
at the diﬀerences between paired observations.        B) for use in the soles of boy’s shoes. Data on the
reduction in sole thickness were taken from a number
• These diﬀerences can be analysed using the            of shoes after boys had worn them for a period of
methods described earlier (e.g. one sample t-test).   three months. The results are displayed in a dot
plot in Figure 36 below (data source: Box, Hunter
& Hunter (1978) Statistics for Experimenters).
See M&M Example 7.7, 7.8 and Exercises 7.17, 7.19
(use MINITAB).                                                 Material B
Material A

6                9     12       15

Wear

Figure 36: Dot plot of shoe wear data.

Looking at the dot plot there appears to be little
evidence of diﬀerence in wear between the two
materials. However, that Figure does not tell the
whole story. The data were collected via a matched
shoe soled with material A, and one shoe soled with
material B. Figure 37 redisplays the data on a dot
plot, but with paired observations joined by lines.

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in order to assess whether the materials have diﬀerent
Material B
Material A                                                           levels of durability. The test proceeds as a standard
one sample t-test using the diﬀerences as data. We
6                9             12           15
conduct this test in MINITAB – see output below.
Wear
MTB > Name c4 = ’x’
Figure 37: Dot plot with paired observations joined                         MTB > Let ’x’ = ’A’ - ’B’
by lines.                                                                   MTB > TTest 0.0 ’x’;
SUBC>   Alternative 0.
The data are displayed in Table 18.
T-Test of the Mean
Subject                  1            2      3       4      5
A                   13.2          8.2   10.9    14.3   10.7         Test of mu = 0.000 vs mu not = 0.000
B                   14.0          8.8   11.2    14.2   11.8
x=A−B                  -0.8         -0.6   -0.3     0.1   -1.1         Var        N   Mean                 StDev    SE Mean    T          P-Value
Subject                  6            7      8       9     10         x         10 -0.410                 0.387     0.122 -3.35           0.0086
A                    6.6          9.5   10.8     8.8   13.3
B                    6.4          9.8   11.3     9.3   13.6         • The worksheet shoewear.mtw contains the shoe
x=A−B                   0.2         -0.3   -0.5    -0.5   -0.3           wear data. The ﬁrst command gives the name
x to the fourth column of this worksheet, while
Table 18: Shoe wear data: A indicates wear on shoe                            the second deﬁnes x to be the required diﬀerence
soled with material A, B indicates wear on shoe sold                          (between the A and B wear data which are stored
with material B, and x is the paired diﬀerence A-B.                           in columns C2 and c3 respectively).
Let µ be the (population) mean diﬀerence, and let
• The t-test was conducted via the menu sequence
σ be the (population) standard deviation of the
Stat → Basic Statistics → 1-Sample t....
diﬀerences. We wish to test
• The observed test statistic is t = −3.35 which
H0 : µ = 0                      against             H1 : µ = 0            gives a P-value of 0.0086. The data therefore

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provide strong evidence against H0, and hence in                                  8.2 Two Independent Samples
favour of a diﬀerence in durability between the two
materials. Inspection of the dot plot (or simply
comparing the mean wear for A and B) indicates
that material A is the more durable (i.e. is less                       8.2.1 Introduction
subject to wear).                                                       Suppose we want to compare the means of two
populations:
It is generally a wise move to accompany a test
result with a conﬁdence interval for the parameter                                         population 1                   population 2
under study. The reason for this is that statistical                                         mean µ1                        mean µ2
2                              2
signiﬁcance, as a concept, is only concerned with the                                       variance σ1                    variance σ2
detection of a departure from H0. It does not tell                          We take random samples from the populations:
you whether the magnitude of the departure from
H0 is of practical importance. We therefore ﬁnish
this example by calculating a conﬁdence interval for                                                                        sample 1
the (population) mean diﬀerence between the wear                              data                                X11, X12, . . . , X1n1 , size n1
in materials A and B.
sample mean                                        ¯
X1
√¯
From the data above, x = −0.410 and sx = 0.387.
√                                                                                               n1
sample std. dev.                  S1 =       1                  ¯
− X 1 )2
Hence SE(¯) = s/ n = 0.387/ 10 = 0.122. A
x                                                                                                              n1 −1    i=1(X1i
95% conﬁdence interval for µ is given by
sample 2
x
(¯ − t0.025(9)SE(¯), x + t0.025(9)SE(¯))
x ¯                 x                                data                               X21, X22, . . . , X2n2 , size n2
sample mean                                       ¯
X2
where the critical point t0.025(9) = 2.262. Hence the                                                                   1         n2           ¯
sample std. dev.                  S2 =     n2 −1    i=1(X2i − X2)2
95% conﬁdence interval is given by

(−0.410 − 2.262 × 0.122, −0.410 + 2.262 × 0.122)
= (−0.69, −0.13)

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Estimation                                          8.2.2 Diﬀerent Population Variances

¯   ¯
We can estimate µ1 −µ2 using the estimator X1 −X2.                               Assumptions (at least one assumed true)
¯     ¯
Properties of X1 − X2 are:
(A3): Both populations are normal;
¯     ¯     ¯      ¯
(i) E[X1 − X2] = E[X1] − E[X2] = µ1 − µ2. Hence
it is unbiased.                                                            (A4): n1 and n2 are suﬃciently large for the Central
2
Limit theorem to apply.
¯     ¯          ¯          ¯                       σ1     σ2
(ii) Var[X1 − X2] = Var[X1] + Var[X2] =                      n1   + n2 .
2
Hence the standard deviation is

2
σ1 σ22
+                                         Result for Normal Random Variables
n1 n2                                                        2                 2
If W ∼ N (µW , σW ), Y ∼ N (µY , σY ) and W and
Y are independent, then
¯    ¯
The standard error of X1 − X2 is therefore
2    2
W ± Y ∼ N (µW ± µY , σW + σY ).
2
S1 S22
¯    ¯
SE X1 − X2 =                   + .
n1 n2

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Two Sample T-Statistic                                                               Conﬁdence Intervals

Using normal r.vs result,                                                        An (approximate) 100(1 − α)% conﬁdence interval
for µ1 − µ2 is
2   2
¯    ¯                                σ1 σ2
X1 − X2 ∼ N            µ1 − µ2 ,         +                                       ¯    ¯                ¯   ¯
n1 n2                                      X1 − X2 − tα/2(ν)SE(X1 − X2),
¯     ¯              ¯   ¯
X1 − X2 + tα/2(ν)SE(X1 − X2)
Standardizing and replacing (unknown) population
variances by sample variances gives us the statistic                             where

¯    ¯
X1 − X2 − (µ1 − µ2)                                                ¯    ¯               2        2
T =                                                                      SE(X1 − X2) =          S1 /n1 + S2 /n2.
2
S1        2
S2
n1   +   n2

• This statistic has an approximate t distribution,
whose degrees of freedom, ν, depend on n1, n2
and the data in a complex way.

• Using exactly the same type of arguments as were
presented in section 8.1, we can derive conﬁdence
intervals and tests for µ1 − µ2.

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Two-Sample t-Test                                                                         Example 8.9
Hypotheses: H1 is a statement of some claim about
µ1 and µ2; H0 is the negation of this statement. Null
hypothesis of the form                                                            To monitor educational standards, a particular type
of test was given to a random sample of year 10
H0 : µ 1 − µ 2 = δ0                                   students in 1995, and to a random sample of year
10 students in 2000. Results:
for some ﬁxed number δ0 (often zero).                                The                                                    1995    2000
alternative will be one of                                                                         Sample size                50      60
Sample mean              63.1    60.9
H1 : µ1−µ2 > δ0,                     H1 : µ1−µ2 < δ0,           H1 : µ1−µ2 = δ0                    Sample std. dev.         17.8     8.9

Write down hypotheses and the observed value of
Test Statistic: Use the t-statistic                                               the two-sample t-test statistic for assessing whether
¯    ¯
X1 − X2 − δ0                                there is a diﬀerence in standards between 1995 and
T =              2
2000.
S1     S2
n1   + n2
2

P-Value: Suppose observed value of the t-statistic
is t.
H1                    P-value
µ 1 − µ 2 > δ0            P (T ≥ t)
µ 1 − µ 2 < δ0            P (T ≤ t)
µ 1 − µ 2 = δ0           2P (T ≥ |t|)
where T has a t distribution with ν degrees of
freedom.

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Example 8.10: Cavendish’s Experiment                                              (population) mean experimental measurement after
on the Density of the Earth                                                   these changes. We wish to test

H0 : µ 1 − µ 2 = 0                against   H 1 : µ 1 − µ2 = 0
In Example 6.15 we looked at data collected by                                    The analysis can be conducted in MINITAB. The
Cavendish from an experiment to determine the                                     data are stored in the worksheet cavendish2.mtw,
density of the earth. We reproduce the data in                                    in which the ﬁrst column of data (named x)
Table 19. As it happened, Cavendish made some                                     contains the experimental results, and the second
changes to his experimental apparatus after the ﬁrst                              column (named app) indexes these results with a
six measurements. The observations made before                                    1 for observations before the change and a 2 for
this change are marked with a dagger in the Table.                                observations after the change. To perform a two-
In this example we will assess the evidence that                                  sample t-test on the data, we select the 2-sample
the changes to the apparatus made a systematic                                    t dialogue box via the menu sequence Stat →
diﬀerence to Cavendish’s experimental results.                                    Basic Statistics → 2-sample t.... Note that
5.50†        5.61†            4.88†    5.07†      5.26†    5.55†            we use the Samples in One Column section of the
5.36         5.29             5.58     5.56       5.57     5.53            resulting dialogue box to enter the variable names,
5.62         6.29             5.44     5.34       5.79     5.10            since this is the way in which our worksheet has been
5.27         5.39             5.42     5.47       5.63     5.34            set up. Having entered x as the Samples and app as
5.46         5.30             5.75     5.68       5.85                     the Subscripts we select the appropriate alternative
hypothesis (and conﬁdence level if required), then
Table 19: Cavendish’s results for the (mean)                                      click on OK. The resulting output is given below.
density of the Earth (in grams per cubic
centimetre). Observations taken before the change                                 Two Sample T-Test and Confidence Interval
in experimental apparatus are marked with a dagger.
Twosample T for x
Let µ1 be the (population) mean experimental                                      app   N      Mean                       StDev     SE Mean
measurement of the density of the earth before                                    1     6     5.312                       0.293        0.12
the changes to the apparatus, and let µ2 be the                                   2    23     5.483                       0.190       0.040

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8.2.3 Common Population Variance
95% C.I. for mu 1 - mu 2: ( -0.48, 0.136)
T-Test mu 1 = mu 2 (vs not =): T= -1.36 P=0.22               DF= 6
Assume that A3 and/or A4 hold. Sometimes it will
• The observed t-statistic is t = −1.36. The                         be reasonable to make a further assumption:
degrees of freedom for the t-test have been
evaluated as ν = 6 by the software. The P-                    (A5) : Both populations have the same variance; i.e.
value for the test is P = 0.22. We therefore                         2    2
σ1 = σ 2 = σ 2 .
conclude that there is no statistically signiﬁcance
evidence for a systematic change in the mean level
We can judge the appropriateness of this assumption
of experimental results after the alterations to the
by comparing the the sample standard deviations, S1
experimental apparatus.
and S2.
• A 95% conﬁdence interval for µ1 − µ2 is given by
(−0.48, 0.136).

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Pooled Estimation of Variance                                      Two Sample T-Statistic (Pooled
Variance)

If A5 assumed, then can estimate the common
population variance, σ 2, by the pooled estimator                    Adapting the working in section 8.2.1 to take into
of variance,                                                         account the common population variance:
2            2
2        (n1 − 1)S1 + (n2 − 1)S2                                          ¯     ¯
X1 − X2 − (µ1 − µ2)
Sp =                              .                          Z      =
n1 + n 2 − 2
σ 2/n1 + σ 2/n2
¯     ¯
X1 − X2 − (µ1 − µ2)
Example 8.11                                                                         =                                 ∼ N (0, 1)
In a medical study, 12 men with high blood pressure                                             σ        1/n1 + 1/n2
were given calcium supplements, and 16 men with
high blood pressure were given placebo.            The
2
Replacing the unknown σ 2 by the estimator Sp gives
reduction in blood pressure (in mm Hg) over a
ten week period was measured for both groups. The                                                  ¯    ¯
X1 − X2 − (µ1 − µ2)
summary statistics for the ﬁrst (i.e. treatment) group                                    T =
Sp 1/n1 + 1/n2
were
¯
x1 = 4.10               s1 = 6.42                          which has a t-distribution with ν = n1 + n2 − 2
and for the second (i.e. control group) were                         degrees of freedom.

¯
x2 = 0.03                s2 = 5.25

Assuming that treatment and control population
variances are equal, calculate a pooled estimate of
the common population standard deviation.

Statistics 160 – Semester 2, 2003                      239           Statistics 160 – Semester 2, 2003                              240
Conﬁdence Intervals (Pooled Variance)                                     Two-Sample t-Test (Pooled Variance)

A 100(1 − α)% conﬁdence interval for µ1 − µ2 is                          Hypotheses: As for unpooled two-sample t-test.
Test Statistic: Use the t-statistic
¯    ¯                ¯    ¯
X1 − X2 − tα/2(ν)SEp(X1 − X2),
¯     ¯               ¯   ¯
X1 − X2 + tα/2(ν)SEp(X1 − X2)                                                             ¯    ¯
X1 − X2 − δ0
T =
1      1
Sp   n1   + n2
¯
where SEp(X1 −                      ¯
X2) = Sp 1/n1 + 1/n2 is
standard error for                   ¯   ¯
X1 − X2 with pooling, and
ν = n1 + n2 − 2.                                                         P-Value: As for unpooled two-sample t-test (but
with ν = n1 + n2 − 2 degrees of freedom).

Example 8.12
Find a 95% conﬁdence interval for the diﬀerence in
the population mean reduction in blood pressure                          See M&M Examples 7.14, 7.15, 7.19, 7.20, 7.21;
between the treatment and control groups in                              and Exercises 7.51, 7.55, 7.57, 7.59, 7.75, 7.77 (use
Example 8.11.                                                            MINITAB where necessary).

Statistics 160 – Semester 2, 2003                               241      Statistics 160 – Semester 2, 2003                                 242

Example 8.13: SIRDS Infants                                   the population of non-surviving SIRDS infants, with
2
mean µ2 and variance σ2 . We will address the
following questions:
Severe idiopathic respiratory distress syndrome
(SIRDS) is a serious condition that can aﬀect                           (i) Is there signiﬁcant evidence that the population
newborn infants, often resulting in death. The table                        mean birth weights diﬀer for survivors and non-
below gives the birth weights (in kilograms) for two                        survivors?
samples of SIRDS infants. The ﬁrst sample contains
the weights of 12 SIRDS infants that survived, while                    (ii) What is a 95% conﬁdence interval for the
the second sample contains the weights of 14 infants                         diﬀerence between the population means?
that died of the condition. (Data from: van Vliet
& Gupta (1973), ‘Sodium bicarbonate in idiopathic                        In order to address (i) we shall test
respiratory distress syndrome’, Arch. Diseases in
Childhood, 48, 249-255.)                                                   H0 : µ 1 − µ 2 = 0                against      H 1 : µ 1 − µ2 = 0
Survived
1.130         1.680         1.930 2.090 2.700         3.160   3.640    We now need to decide whether or not to assume
1.410         1.720         2.200 2.550 3.005                          equal population variances. The sample standard
deviations are 0.758 for group 1 (survived) and 0.554
for group 2 (died). The ratio of larger to smaller is
Died
0.758/0.554 = 1.37, suggesting the an assumption
1.050         1.230         1.500     1.720   1.770   2.500   1.100
of common variance is not too unreasonable.
1.225         1.295         1.550     1.890   2.200   2.440   2.730
The analysis of the data (to answer (i) and (ii) above)
Table 20: Birth weights (in kg) of infants with SIRDS
can be conducted in MINITAB as follows. The data
It is of interest to medical researchers to know                         are available in the MINITAB worksheet SIRDS.mtw,
whether birth weight inﬂuences a child’s chance of                       in which the ﬁrst column (named weight) contains
surviving with SIRDS. Deﬁne population 1 to be                           the birth weights and the second column (named
the population of SIRDS infants that survive, with                       survived) indexes these with a 1 if the infant
2
mean µ1 and variance σ1 . Let population 2 be                            survived, and a 0 if it died. As with the unpooled

Statistics 160 – Semester 2, 2003                               243      Statistics 160 – Semester 2, 2003                                 244
two-sample t-test in Example 8.10, we obtain the                        8.3 More Than Two Samples
2-sample t dialogue box via the menu sequence
Stat → Basic Statistics → 2-sample t....
We enter the necessary information in this dialogue
in the same manner as in Example 8.10, but ﬁnish             8.3.1 Introduction
by clicking on the box Assume equal variances
2    2                                    • Sometimes have samples from more than two
(since we’re assuming σ1 = σ2 ).
populations that we wish to compare.
Two Sample T-Test and Confidence Interval
• It is unwise to start by comparing every
Twosample T for weight                                             possible pair of populations using two-sample tests
survived   N      Mean              StDev   SE Mean                because:
1         12     2.268              0.758      0.22
0         14     1.729              0.554      0.15                – it is a lot of work. E.g. there are more than
100 pairwise comparisons between 15 samples)
95% C.I. for mu 1 - mu 0: ( 0.01, 1.07)                            – problems with signiﬁcance levels in multiple
T-Test mu 1 = mu 0 (vs not =): T= 2.09 P=0.047          DF= 24       testing.
Both use Pooled StDev = 0.655
• The output conﬁrms that a pooled estimate of               ANOVA
standard deviation was used; sp = 0.655.                   Analysis of variance (‘ANOVA’) is a method for
comparing the means of several populations which
• The observed value of the t-statistic is t = 2.09,         avoids these problems.
which gives a P-value of P = 0.047. This gives
evidence in favour of H1. (We reject H0 at a 0.05
signiﬁcance level.)

• A 95% conﬁdence interval for µ1 − µ2 is
(0.01, 1.07).

Statistics 160 – Semester 2, 2003                 245        Statistics 160 – Semester 2, 2003                               246

8.3.2 One-Way Analysis of Variance                                                   More ANOVA

• The diﬀerent populations in ANOVA can be                   A ‘m-way analysis of variance’ can be used when
deﬁned by one or more factors (i.e. qualitative            there are m factors. In this unit we will restrict
explanatory variables).                                    attention to situations in which there is a single
factor.
• We can then think of ANOVA as testing for the
eﬀect of these factors on the response variable.
Data Structure for One-Way ANOVA
Example 8.14
Popn. (mean, var)               Sample (responses)       Mean
In a medical study, people are classiﬁed according to            1           2
(µ1, σ1 )                  X11, X12, . . . , X1n1    ¯
X1
sex and whether or not they smoke. Hence there are               2           2
(µ2, σ2 )                  X21, X22, . . . , X2n2    ¯
X2
.
.                                        .
.                 .
.
four populations to consider:
k           2
(µk , σk )                 Xk1, Xk2, . . . , Xknk    ¯
Xk
• female non-smokers;
Grand Mean
• female smokers;                                            The mean of all the data (irrespective of which
sample they come from) is called the grand mean,
• male non-smokers;                                                        ¯
and denoted X. The total number of observations
is denoted n.
• male smokers.

Suppose that lung function is measured for each
individual. We could consider the eﬀects of both
factors sex and smoking on the response variable
‘lung function’.

Statistics 160 – Semester 2, 2003                 247        Statistics 160 – Semester 2, 2003                               248
More ANOVA                                                          ANOVA Table

Assumptions:                                                        Xij =           X¯    +        ¯    ¯
(Xi − X)                 ¯
+ (Xij − Xi)
grand mean deviation of sample i    residual
(A6) All the observations are normally distributed;
¯    ¯              ¯     ¯
If H0 is correct then (Xi − X), and hence (Xi − X)2,
(A7) The population variances are all the same; i.e.                  should be quite close to zero for each sample. It can
2     2             2                                           be shown that
σ1 = σ 2 = · · · = σ k .

k    ni                                     k
Hypotheses:
¯
(Xij − X)2 =                              ¯    ¯
ni(Xi − X)2
In one-way ANOVA we are testing the hypotheses:
i=1 j=1                                      i=1

H0 : µ 1 = µ 2 = · · · = µ k                         Total SS                         Factor (Treatment) SS
k   ni
+                    ¯
(Xij − Xi)2
against
i=1 j=1

H1 : The µi’s are not all equal                                                                          Residual (Error) SS

where SS is an abbreviation for ‘sum of squares’.
Evidence for H1 will be provided by a substantial
degree of variability between the sample means.

Statistics 160 – Semester 2, 2003                             249   Statistics 160 – Semester 2, 2003                                      250

ANOVA Table                                                    Testing in ANOVA

The decomposition of sums of squares can be
displayed in an ANOVA table.                                        • If the populations truly have diﬀerent means (i.e.
if the factor has a real eﬀect on the responses)
Source            DF           SS           MS                  then the Factor MS should be relatively large
Factor            k−1         SSF ac   SSF ac/(k − 1)           compared to the Error MS.
Error             n−k         SSErr    SSErr /(n − k)
Total           (n − 1)       SST ot                          • Hence, an appropriate test statistic is
k      ni         ¯ 2 In this
where SST ot =      i=1    j=1(Xij − X) .                                                                      M SF ac
table, DF is ‘degrees of freedom’, and MS is ‘mean                                                       F =
M SErr
square’ (given by MS = SS/DF).
with extreme positive values of F providing
evidence against H0.

• Under the assumption that H0 is correct, F has
an F-distribution with (k − 1, n − k) degrees of
freedom. The P-value for an observed value f of
the F-statistic is

P = P (F ≥ f )

where F has the F-distribution described above.

Statistics 160 – Semester 2, 2003                             251   Statistics 160 – Semester 2, 2003                                      252
Example 8.15: Silver Content in Coins                                These data are stored in the MINITAB worksheet
silver.mtw. The sliver contents (‘responses’) are
Table 21 below shows the silver content (% Ag)                        given in the ﬁrst column (named silver), while
of coins taken from four mintings during the reign                    the second column (named minting) speciﬁes the
of King Manuel I (1143-1180), a Byzantine ruler                       minting (with 1 for minting ‘A’, 2 for minting ‘B’
who lived in what is now Cyprus. There were nine                      and so on). To perform the analysis of variance
coins from minting ‘A’, seven from ‘B’, four from                     in MINITAB, we need to go through the menu
‘C’ and seven from ‘D’. It is of historical interest to               sequence Stat → ANOVA → Oneway. The details
investigate whether or not the mean silver content                    of the analysis are then entered into the resulting
varies for minting to minting. Hence we will test                     dialogue box;. Clicking on OK should then provide
the following output in the MINITAB session window.
H0 : µ A = µ B = µ C = µ D                  against

H1 : µA, µB , µC , µD not all equal                           One-Way Analysis of Variance
where µA is the population mean silver content for
Analysis of Variance on silver
minting A (and so on).
Source    DF       SS       MS        F       p
‘A’       ‘B’   ‘C’   ‘D’                   minting    3   37.748   12.583    26.27 0.000
5.9       6.9   4.9   5.3                   Error     23   11.015    0.479
6.8       9.0   5.5   5.6                   Total     26   48.763
6.4       6.6   4.6   5.5                                              Individual 95% CIs For
7.0       8.1   4.5   5.1                                              Based on Pooled StDev
6.6       9.3         6.2                   Lvl N     Mean   StDev ---+---------+---------+-
7.7       9.2         5.8                     1 9 6.7444 0.5434                      (--*--)
7.2       8.6         5.8                     2 7 8.2429 1.0998                              (
6.9                                           3 4 4.8750 0.4500      (----*---)
6.2                                           4 7 5.6143 0.3625             (--*---)
---+---------+---------+-
Table 21: Silver content by minting.                     Pooled StDev =   0.6920     4.5        6.0       7

Statistics 160 – Semester 2, 2003                               253   Statistics 160 – Semester 2, 2003                 254

• The F test statistic is F = 26.27. (Notice that
this is equal to to MS(factor)/MS(Error); i.e.
In one-way ANOVA, conﬁdence intervals for
F = 12.583/0.479 = 26.27).
population means are based upon a pooled estimate
2
of variance, Sp (which is actually equal to
• The P-value is 0.000 (i.e. the F statistic is far
MSE(Error)).
larger than we would expect to get if H0 were
true). Since the P-value is so tiny, the data                       A 100(1 − α)% conﬁdence interval for µi is given by
provide very strong evidence against H0, and
hence in favour of a variation in mean silver                            ¯               ¯    ¯               ¯
Xi − tα/2(ν)SEp(Xi), Xi + tα/2(ν)SEp(Xi)
content from minting to minting.
¯           √
where SEp(Xi) = Sp/ ni and tα/2(ν) is a critical
Endnote:                                                              point for a t distribution with ν = n − k degrees of
The 95% conﬁdence intervals for each population                       freedom.
mean allow us to get some idea as to the way
in which the mean silver content varies between
mintings. These conﬁdence intervals suggest that                      Example 8.16
the earlier mintings, ‘A’ and ‘B’, had a rather higher                                                                   ¯
For Example 8.15, calculate the standard error for xA
mean sliver content than the later ones, ‘C’ and                      using pooling and hence compute a 99% conﬁdence
‘D’. This seems to be an early example of currency                    interval for µA.
devaluation

Statistics 160 – Semester 2, 2003                               255   Statistics 160 – Semester 2, 2003                 256
9. Comparing Two Proportions                                                             9.2 Estimation

9.1 Introduction                                                                        ˆ    ˆ
The statistic D = p1 − p2 is natural estimator of
p1 − p 2 .
Aim: to compare the proportions of individuals with
a particular characteristic in two populations.
Data: take simple random samples from both
populations, and count the number of ‘successes’                                                      p    ˆ
E[D] = E[ˆ1 − p2]
(i.e. individuals with the characteristic) in both                                                            p       p
= E[ˆ1] − E[ˆ2]
samples.
= p1 − p2

Population           Sample   Count of     Sample      Hence D is an unbiased estimator of p1 − p2.
Popn         proportion            size     success    proportion
1               p1                 n1        X1      p1 = X1/n1
ˆ                                     p    ˆ
Var(D) = Var(ˆ1 − p2)
2               p2                 n2        X2      ˆ
p2 = X2/n2
p         p
= Var(ˆ1) + Var(ˆ2)
p1(1 − p1) p2(1 − p2)
We will mostly concentrate on inference for p1 − p2.                                              =                 +
n1         n2

so the standard error of D is

p1(1 − p1) p2(1 − p2)
ˆ      ˆ    ˆ     ˆ
p p
SE(D) ≡ SE(ˆ1−ˆ2) =                                 +
n1         n2

Statistics 160 – Semester 2, 2003                            257      Statistics 160 – Semester 2, 2003                            258

Example 9.1: Who Will Go Grey First?                                  think the diﬀerence is statistically signiﬁcant or not?
example later.
Do men or women tend to get grey hair ﬁrst? In
work done by two Australian researchers in the late
1980’s (and reported in Family Circle magazine in
1990), simple random samples of 1000 men and 1000
women, all of age 25, were taken. Among the men,
290 had some grey hair, while the corresponding
ﬁgure for the women was 340.
Let p1 be the population proportion of 25 year
old men with some grey hair, and let p2 be the
corresponding population proportion for women. An
unbiased estimator of p1 − p2 is provided by the
ˆ ˆ
estimate p1 − p2 = 290/1000 − 340/1000 = −0.050.
The standard error of this estimate is

ˆ      ˆ
p1(1 − p1) p2(1 − p2)
ˆ     ˆ
p    ˆ
SE(ˆ1 − p2) =                                 +
n1         n2
0.29 × (1 − 0.29) 0.34 × (1 − 0.34)
=                     +
1000              1000
= 0.0207

The article writer for Family Circle commented that,
“this diﬀerence [between sample proportions] is so
small that it’s considered insigniﬁcant.” Do you

Statistics 160 – Semester 2, 2003                            259      Statistics 160 – Semester 2, 2003                            260
Conﬁdence Intervals                               Example 9.2: Conﬁdence Interval from
Grey Hair Data

Assume: n1p1 ≥ 10, n1(1 − p1) ≥ 10, n2p2 ≥ 10,
and n2(1 − p2) ≥ 10.                                                   Continuing from Example 9.1, ﬁnd a 95% conﬁdence
Then                                                                   interval for p1 − p2.
·
ˆ
p1 ∼ N (p1, p1(1 − p1)/n1)
and
·
ˆ
p2 ∼ N (p2, p2(1 − p2)/n2)
from section 6.2.3.
Therefore

·                       p1(1 − p1) p2(1 − p2)
D∼N              p1 − p 2 ,             +
n1         n2

See section 8.2.2.
Conﬁdence Interval:
Using exactly the same kind of arguments as in
7.1, it follows that an (approximate) 100(1 − α)%
conﬁdence interval for p1 − p2 is given by

D − zα/2SE(D), D + zα/2SE(D) .

Statistics 160 – Semester 2, 2003                           261        Statistics 160 – Semester 2, 2003                                   262

9.4 Signiﬁcance Testing                                                            Test Statistic

• Would like to use
Hypotheses:                                                                                    D−0                        D
H1 is a statement of some claim about p1 − p2, and                                     z=          =
σD               p1 (1−p1 )
+ p2(1−p2)
H0 is the negation of this statement.                                                                                n1            n2

In practice interest is almost always in detecting a                       However, the denominator depends on the
diﬀerence between p1 and p2, so we will take the null                      unknown population proportions.
hypothesis to be
• Since we assume that p1 = p2 under H0, it
H0 : p 1 − p2 = 0                         is sensible to replace both p1 and p2 in the
denominator by the pooled estimate,
The alternative will be one of                                                                                  X1 + X 2
ˆ
p=
n1 + n 2
H1 : p1−p2 > 0                 or H1 : p1−p2 < 0   or H1 : p1−p2 = 0
• We therefore use the test statistic
D
z     =
ˆ    p
p(1−ˆ)
n1
ˆ    p
+ p(1−ˆ)
n2

D
=
ˆ     ˆ
p(1 − p)(1/n1 + 1/n2)

This test statistic has a N (0, 1) sampling
distribution under the assumption that H0 is
correct.

Statistics 160 – Semester 2, 2003                           263        Statistics 160 – Semester 2, 2003                                   264
P-Value                                   Example 9.3: Do Men or Women
Really Diﬀer in How Early They Go
Grey?
Suppose that the observed value of the test statistic
is z.
In this example we take the data on grey hair and
H1               P-value
test
p1 − p 2 > 0        P (Z ≥ z)
H0 : p 1 = p2
p1 − p 2 < 0        P (Z ≤ z)
against the two-sided alternative,
p1 − p 2 = 0       2P (Z ≥ |z|)
where Z ∼ N (0, 1).                                                                                          H1 : p 1 = p2

where p1 and p2 are the population proportions as
deﬁned in Example 9.1. In order to evaluate the test
statistic here, we need to obtain a pooled estimate
of proportion:
290 + 340
ˆ
p =
1000 + 1000
= 0.315

The test statistic is
ˆ    ˆ
p1 − p 2
z     =
ˆ     ˆ
p(1 − p)(1/n1 + 1/n2)
0.05
=
0.315 × (1 − 0.315) × (1/1000 + 1/1000)
= 2.41

Statistics 160 – Semester 2, 2003                                  265   Statistics 160 – Semester 2, 2003                                   266

The P-value corresponding to this test statistic is                                               9.5 Relative Risk

P       = 2P (Z ≥ |2.41|)                since 2-sided test
= 2 × 0.0080                 from normal tables             The estimated relative risk is
= 0.016
ˆ
p1
RR =      .
We therefore conclude that the data provide evidence                                                                 ˆ
p2
that chances of developing some grey hair by the age
of 25 does diﬀer between men and women.                                  It is widely used in medical statistics because it is
easy to interpret.
End note: Notice that the numbers of men and
women with some grey were both multiples of 10. It
it quite possible that the data, as reported in Family
Example 9.4: Smoking and Heart Attacks in
Circle, were rounded to the nearest percentage
Women
point (i.e. the nearest ten individuals). Suppose
An article in the British Medical Journal (Prescott,
that the true ﬁgures were actually 294 for men,
Hippe, Schnor and Vestbo, 1998) gave the following
and 336 for women – would your conclusions from
data for random samples of Danish women:
this signiﬁcance test change? You might like to
investigate this as an exercise.                                                                                              Number of
Sample size     heart attacks
Smokers                        6461               380
Non-smokers                    5011               132
See also M&M Example 8.9, and Exercises 8.29,                            The sample proportion of heart attacks among the
8.33, 8.35.                                                                          ˆ
smokers is pS = 380/6461 = 0.0588. For non-
ˆ
smokers the proportion is pN S = 132/5011 =
0.0263. Hence the relative risk of a heart attack
for a female smoker (compared to a female non-

Statistics 160 – Semester 2, 2003                                  267   Statistics 160 – Semester 2, 2003                                   268
smoker) is                                                                            10. Inference for Regression
0.0588
RR =          = 2.24
0.0263
In other words, a smoker is more than twice as likely
to suﬀer a heart attack as a non-smoker.                                  10.1 The Simple Linear Regression Model

It is possible to obtain a conﬁdence interval                             Simple linear regression aims to model a
corresponding to this relative risk, but the details                      (quantitative) response variable, y, in terms of
are beyond the Statistics 160 unit.                                       an explanatory variable (covariate) x. The linear
regression model is

Yi ∼ N (a + bxi, σ 2)                    (i = 1, 2, . . . , n) (2)
where

• xi, Yi are the explanatory and response variables
measured on the ith individual;

• the responses, Y1, Y2, . . . , Yn are independent;

• σ 2 is the ‘error’ variance (which is not dependent
on xi).

Note that the linear regression model represents the
variability in the r.v. Yi, regarding xi as a ﬁxed
quantity.

Statistics 160 – Semester 2, 2003                                   269   Statistics 160 – Semester 2, 2003                                270

More on the Simple Linear Regression                                          Example 10.1: A Linear Regression
Model                                                                        Model

The model can be written equivalently as                                  Suppose that it is known that a response Y is
dependent upon the explanatory variable x via the
relationship
Yi = a + bxi +                 i       (i = 1, 2, . . . , n)
Y = 1 + 2x +
where has a normal distribution with zero mean
where 1, 2, . . . , n are               independent      N (0, σ 2)
and standard deviation σ = 1.5. Hence,
random ‘error’ terms.
From Eqn. 2 we have                                                                                      E[Y ] = 1 + 2x

E[Yi| xi] = a + bxi                 (3)       and the regression parameters are intercept a = 1
and slope b = 2.
• We write E[Yi|xi] to emphasise that the expected                        A computer was used to simulate 20 responses (i.e.
value of Yi depends on xi.                                              observations on Y ) when x = 2, and 20 responses
when x = 4. The regression line is plotted below,
• Eqn. 3 deﬁnes the population regression equation.                       along with a scatterplot of the simulated data.
The quantities a and b are the regression
parameters (representing intercept and slope
respectively).

Statistics 160 – Semester 2, 2003                                   271   Statistics 160 – Semester 2, 2003                                272
10.2 Inference for Regression
y
Parameters
12                                                                •
•
•
•
•
•
10.2.1 Estimation
•
•
•
•
8                                  •
•
•
Data: (x1, y1), (x2, y2), . . . , (xn, yn).
•
•
•
Aim:     use these data to estimate the true
•
•
•
•
•
(population) regression equation.
4                                  •
y = 1 + 2x              •
•
•                                           Least Squares Estimation:
The least squares intercept a and slope ˆ are
ˆ              b
estimates of the parameters a and b respectively.
0             1                2             3               4    5
x
The least squares regression line

y = a+ˆ
ˆ ˆ bx
Note:

• The regression model describes the variability in                               is often called the ﬁtted regression line (to distinguish
Y for any given x that we choose.                                               it from the true regression line).

• The regression line itself describes how the mean
of Yi depends on xi.

• The degree of scattering of points about the
regression line is controlled by the error standard
deviation, σ.

Statistics 160 – Semester 2, 2003                                       273       Statistics 160 – Semester 2, 2003                     274

Properties of Least Squares Regression                                               Example 10.2: Linear Regression for
Coeﬃcients                                                                  the Image Quality Data

The image quality data was introduced in Example
• E[ˆ] = a and E[ˆ = b (so both are unbiased).
a            b]                                                               3.2 (and revisited in Examples 3.4 and 3.7). In this
data set image quality x and reading time y are
n
• Var(ˆ = σ 2/[
b)                            i=1(xi   − x)2], so that
¯                                  available for 8 images. From Example 3.7 we see
that the ﬁtted regression equation is
σ
σˆ =
b                n                  .
i=1(xi     − x)2
¯                                                ˆ
y = 9.6378 − 0.3485x

so that the estimated slope parameter is ˆ =     b
To ﬁnd SE(ˆ we must estimate the unknown
b)                                                                  −0.3845. In this Example we shall compute the
parameter σ 2.      It turns out that an unbiased                                 standard error of this estimate by hand. This is
estimator of σ 2 is given by                                                      done for illustrative purposes only – in practice you
can use a suitable computer package to perform the
1                                    necessary calculations.
σ2 =
(n − 2)
In order to compute SE(ˆ we need to obtain an
b)
where                                                                             estimate, σ 2, of the error variance. This requires
ˆ
n
us to compute the residual sum of squares for the
RSS =                 (yi − [ˆ + ˆ i])2
a bx
regression. This calculation is facilitated by setting
i=1
out the data and other relevant quantities in a table.
is the residual sum of squares. Hence
Table 22 has columns as follows: x value, y value,
σ
ˆ                                 ﬁtted value, residual and squared residual. The ﬁtted
SE(ˆ =
b)                    n                  .                 value for the ith observation is deﬁned by
i=1(xi     − x)2
¯
yi = a + ˆ i
ˆ    ˆ bx

Statistics 160 – Semester 2, 2003                                       275       Statistics 160 – Semester 2, 2003                     276
√          √
while the residual is deﬁned by                                                                                =         0.0612/ 4.835

ei = yi − yi = yi − [ˆ + ˆ i]
ˆ         ˆ          a bx                                                                  = 0.113

x           y      yˆ      y−y ˆ      (y − y )2
ˆ            In fact this standard error is computed automatically
4.30        8.00   8.14     -0.14         0.019          by MINITAB when performing a regression analysis.
4.55        8.30   8.05      0.25         0.061          Below we reproduce some of the MINITAB output
5.55        7.80   7.70      0.10         0.009          from ﬁtting the regression in Example 3.7.
5.65        7.25   7.67     -0.42         0.175
Regression Analysis
5.95        7.70   7.56      0.14         0.018
6.30        7.50   7.44      0.06         0.003
The regression equation is
6.45        7.60   7.39      0.21         0.044
y = 9.64 - 0.349 x
6.45        7.20   7.39     -0.19         0.036
Sum                                                   0.367
Predictor    Coef                          Stdev        t-ratio        p
Table 22: x − y data, ﬁtted value, residual and                            Constant   9.6378                         0.6419          15.01    0.000
squared residual for the image data.                                       x         -0.3485                         0.1125          -3.10    0.021

From Table 22 we see that the residual sum of                              MINITAB gives SE(ˆ = 0.1125 from this output
b)
n                                                     which tallies with the value that we computed by
squares, RSS = i=1(yi − yi)2 = 0.367. Hence the
ˆ
error variance is estimated by                                             hand.

1        0.367
σ2 =
n−2       8−2
n
Now          i=1(xi     − x)2 = 4.835 for these data, so
¯

ˆ
σ
SE(ˆ =
b)                    n
i=1(xi   − x)2
¯

Statistics 160 – Semester 2, 2003                                    277   Statistics 160 – Semester 2, 2003                                       278

10.2.2 Conﬁdence Intervals                                            10.2.3 Signiﬁcance Testing for a
Regression Slope

The sampling distribution of ˆ is
b                                             Example 10.4: Predicting Tooth Age from a
Measurement of Dentine
n
A group of researchers performed a study into
ˆ ∼ N (b, σ 2/
b                      (xi − x)2)
¯                        the possibility of predicting the age in years,
i=1
y, of a tooth from the percentage, x, of the
tooth’s root with transparent dentine. The data
Using standard arguments it follows that a 100(1 −                         is displayed in the Table below. (Data source: ‘Root
α)% conﬁdence interval for b is given by                                   dentine transparency: age determination of human
teeth using computerized densitometric analysis’,
ˆ − tα/2(ν)SE(ˆ ˆ + tα/2(ν)SE(ˆ                                American Journal of Anthropology, 1991, pp. 25-
b             b), b           b)
30.)

where ν = n − 2.                                                             x      15       19      31         39       41   44    47   48   55     65
y      23       52      65         55       32   60    78   59   61     60

A scatter plot of this data is given in Figure 38
below.
Example 10.3
From Example 10.2, ﬁnd a 95% conﬁdence interval
for the regression slope b.

Statistics 160 – Semester 2, 2003                                    279   Statistics 160 – Semester 2, 2003                                       280
Signiﬁcance Testing Continued

80
If, in truth, the response is not linearly related to
the explanatory variable, then the true (population)
70

value of the slope will be b = 0.
60

Hypotheses
50
y

H0 : b = 0                i.e. no linear relationship between variables
40

against one of the alternatives
30

H1 : b = 0                     or H1 : b > 0            or H1 : b < 0
20             30       40     50    60

x
Test statistic
ˆ
b
Figure 38: Scatter plot of age, y, against percentage                                                    t=
SE(ˆb)
transparent dentine, x.
which has a t distribution with ν = n − 2 degrees of
Looking at this plot, the evidence that tooth age                    freedom, assuming that H0 is correct.
is related to percentage transparent dentine is not
entirely convincing. Is there really an increasing
linear relationship between the variables, or is the
hint of such behaviour in the scatter plot purely a
result of random chance?

Statistics 160 – Semester 2, 2003                              281   Statistics 160 – Semester 2, 2003                                 282

P-value                       Example 10.5: Attempting to Predict
Tooth Age

The P-value for an observed test statistic t is given                Following on from Example 10.4, we investigate
by                                                                   whether or not tooth age (Y ) can be predicted from
the percentage of transparent dentine (x) by ﬁtting
H1               P-value                 a linear regression
b>0              P (T ≥ t)
b<0              P (T ≤ t)                                               E[Y ] = a + bx
b=0             2P (T ≥ |t|)
where T has a t distribution with ν = n − 2 degrees                  to the data, and then testing
of freedom. These P-values can be computed by
MINITAB.                                                                          H0 : b = 0             against       H1 : b = 0.

We can ﬁt the regression model in MINITAB
via the menu sequence Stat →Regression
→Regression..., as described in Example 3.7.
Regression Analysis

The regression equation is
y = 32.1 + 0.555 x

Predictor                Coef             Stdev     t-ratio         p
Constant                32.08             13.32        2.41     0.043
x                      0.5549            0.3101        1.79     0.111

s = 14.30                    R-sq = 28.6%            R-sq(adj) = 19.7%

Statistics 160 – Semester 2, 2003                              283   Statistics 160 – Semester 2, 2003                                 284
Recall that for testing H0 against H1 we need to             11. Chi-Square Tests for Categorical
compute the observed value of the test statistic                            Data
ˆ
b
t=
SE(ˆ
b)
11.1 Testing a Probability Model
From the MINITAB output, ˆ = 0.5549 and SE(ˆ =
b                 b)         Data: Count data for a categorical variable with r
0.3101 so that t = 0.5549/0.3101 = 1.79. In fact          categories.
there is no need to do this calculation – MINITAB
also reports that the observed t statistic is 1.79        Model: A probability model assigns a probability to
(which it refers to as a ‘t-ratio’). MINITAB reports      each category.
the P-value for testing                                   Aim: Want to know whether this model is
appropriate for the data – i.e. is the model ‘a good
H0 : b = 0            ﬁt’.
Notation:
against the two-sided alternative

H1 : b = 0            • pi denotes the probability that a given individual
falls in the ith category (i = 1, 2, . . . , r).
For the data at hand this P-value is P = 0.111.
(Had we been considering the alternative H1 : p > 0       • R.v. Ni represent the number of individuals falling
then we would have had to halve this P-value to             in the ith category (i = 1, 2, . . . , r) when N1 +
give the correct value P = 0.111/2 = 0.056.) We             N2 + ... + Nr = n individuals are observed.
conclude that the data do not provide statistically
signiﬁcant evidence (at even a 0.1 level) against H0.
10.1, 10.3 (use MINITAB).

Statistics 160 – Semester 2, 2003                   285   Statistics 160 – Semester 2, 2003                          286

Example 11.1: Market Research for                                            11.1.1 Hypotheses
Toothpaste

In a ‘goodness of ﬁt test’ for a probability model,
A market research company asks a random sample of         the null hypothesis states that the model is correct.
n = 100 people which of 4 brands of toothpaste (1,        That is,
2, 3, 4) they prefer. (So ‘toothpaste brand preferred’
is a categorical variable with r = 4 categories.)                     H 0 : p 1 = π1 , p 2 = π2 , . . . , p r = πr
Anecdotal evidence suggests that each of 4 brands
of toothpaste are equally popular, but the company        where π1, π2, . . . , πr are the category probabilities
believes this may not be true. The market research        according to the model.
company therefore wants to test the adequacy of the
probability model                                         The alternative hypothesis is

p1 = p2 = p3 = p4 = 1/4                                      H1 : H0 is not correct

(where pi is the category i probability; i.e. the
probability that a randomly selected individual prefers
brand i).
Let the r.v.s N1, N2, N3, N4 represent the counts for
each category (i.e. the number of people preferring
each brand). If the probability model is correct
then E[Ni] = 25 for each category i = 1, 2, 3, 4. If
we observe category counts n1, n2, n3, n4, we will
have reason to doubt the probability model if these
observed counts diﬀer markedly from 25.

Statistics 160 – Semester 2, 2003                   287   Statistics 160 – Semester 2, 2003                          288
11.1.2 Test Statistic                                       Example 11.2: Testing the Mendelian
Theory of Inheritance

• If we observe n individuals, then the expected
number falling in category i is
The data in Table 23 below give observed counts of
diﬀerent kinds of pea seeds in crosses from plants
E[Ni] = npi                        (i = 1, 2, . . . , r)
with round yellow seeds and plants with wrinkled
green seeds. The Mendelian theory of inheritance
• Hence under the assumption that H0 is correct,                                              1
suggests that 16 of the seeds will be wrinkled green
the expected category counts are nπ1, nπ2, . . . ,                                   3                                 3
(WG), 16 will be round green (RG), 16 will be
nπr .                                                                                                                    9
wrinkled yellow (WY), and the remaining 16 will be
round yellow (RY).
• If we observe category counts n1, n2, . . . , nr then
H0 would seem plausible if                                                      Type of seed                     RY    WY     RG        WG   Total
Observed count                   93    27     32         8    160
n1 ≈ nπ1              n2 ≈ nπ2         and so on
Table 23: Pea seed counts from an experiment on
Mendelian inheritance
• On the other hand, if ni is very diﬀerent to nπi
for some categories, this will indicate evidence                             Is the Mendelian model appropriate for the data?
against H0.                                                                  We investigate by testing

Chi Square Statistic: The appropriate test statistic                                                  9                  3            3            1
is the chi square statistic                                                        H0 : p 1 =                  p2 =           p3 =         p4 =
16                 16           16           16
r
(ni − nπi)2                      where p1 = P (RY), p2 = P (WY), p3 = P (RG),
χ2         =
i=1
nπi                          p4 = P (WG) .
Greek letter chi
(There is no need to state an alternative hypothesis

Statistics 160 – Semester 2, 2003                                      289     Statistics 160 – Semester 2, 2003                                       290

in this type of goodness of ﬁt test. It is understood                                          Chi Square Test Statistic
that the alternative is simply ‘H0 is not correct’.)
If H0 is correct, then the expected number of RYs is

9                              • The larger the value of the chi square statistic,
e1 = nπ1 = 160 ×              = 90.                          the more evidence that we have against H0.
16

Similarly,                                                                     • If H0 is correct then a chi square statistic follows
a χ2 distribution with r − 1 degrees of freedom
3                           (approximately).
for WY,                    e2 = 160 ×       = 30.
16

3
for RG,                  e3 = 160 ×
= 30.
16
1
for WG,           e4 = 160 ×      = 10.
16
This can be set out neatly in table form.

Type of seed                      RY    WY       RG        WG        Total
Observed count, o                 93    27       32         8         160
Expected count, e                 90    30       30        10         160
o−e                                3    -3        2        -2          0
(o − e)2/e                        0.1   0.3     0.133      0.4       0.933

Hence, the χ2 statistic here is 0.933 (3 d.p.).

Statistics 160 – Semester 2, 2003                                      291     Statistics 160 – Semester 2, 2003                                       292
11.1.3 χ2 Distribution                     Example 11.3: Chi Square Probabilities
in MINITAB
• If r.v. X has a chi square distribution with ν
degrees of freedom, we write X ∼ χ2(ν).
Suppose that we wish to ﬁnd P (X ≤ 5.0) where
• The degrees of freedom is the number of                        X ∼ χ2(3). This can be done via the menu
categories (r) minus the number of constraints on              sequence Calc →Probability Distributions
the counts. When testing the ﬁt of a probability               →Chisquare..., but is more easily done by typing
r
model there is a single constraint:  i=1 nr = n.               in the commands directly in the MINITAB session
window. The syntax is as follows:
• Cumulative chi square probabilities, i.e. P (X ≤
x) where X ∼ χ2(ν) (see Figure 39), can be                    (i) Type in CDF 5.0; (NB the semi-colon) at MTB
computed using MINITAB.                                           prompt, then press return. (This indicates that
we wish to compute a cumulative probability,
P (X ≤ x), for x = 5.0.)

(ii) After (i) you will then see a new SUBC prompt
(standing for ‘sub-command’). Enter Chisquare
3. (NB the full-stop) at this prompt, then press
return.

The MINITAB session window should then look as
follows.
MTB > CDF 5.0;
x
0                                              SUBC>   Chisquare 3.

Figure 39: P (X ≤ x), for a χ2 distribution.              Cumulative Distribution Function

Statistics 160 – Semester 2, 2003                         293    Statistics 160 – Semester 2, 2003                294

11.1.4 P-Values for Chi Square Test
Chisquare with 3 d.f.

x                  P( X <= x)
5.0000                     0.8282                        • If χ2 is the observed value of a chi square test
statistic, then the P-value is
Hence P (X ≤ 5.0) = 0.8282.
P = P (X ≥ χ2)
Connection with P-Values
To compute P-values in a chi square test we need to                  where X ∼ χ2(r − 1).
compute probability of getting a test statistic more
extreme (i.e. larger) than the observed one. So, if              • The P-value for a chi square test is interpreted
the observed test statistic is χ2, then we need to ﬁnd             (and compared with a signiﬁcance level α) in the
P (X ≥ χ2) where X ∼ χ2(ν). We can do this by                      usual way.
using MINITAB to compute P (X < χ2) and then
using the fact that

P (X ≥ χ2) = 1 − P (X < χ2).                    Example 11.4
In Example 11.2 we observed a test statistic χ2 =
0.933. The degrees of freedom were ν = r − 1 =
So for example, if we observed a test statistic χ2 =             4 − 1 = 3. Using MINITAB it can be shown that
5.0 when there are ν = 3 degrees of freedom, the
P-value will be                                                                         P (X ≤ 0.933) = 0.1825

P (X ≥ 5.0) = 1 − P (X < 5.0)                                where X ∼ χ2(3). Hence the P-value is
= 1 − 0.8282    from above
P = P (X ≥ 0.9333) = 1 − 0.1825 = 0.8175
= 0.1718
We conclude that there is no evidence that the
Mendelian model is inappropriate for the data.

Statistics 160 – Semester 2, 2003                         295    Statistics 160 – Semester 2, 2003                296
11.1.5 Combining Categories                                                             Age Group
18-29 30-39 40-49 50-59 60-69 70-79 80+
Count   15    25    22     18    14   5     1
Table 25: Ages of sample of 100 adult South Africans
• The χ2 statistic can be unreliable if some of the
categories have expected counts (i.e. nπi = ei)                          Let pi be the probability that a randomly selected
smaller than 5.                                                          individual sampled by the company falls in age group
i (where age group 1 is 18-29 years, age group
2 is 30-39 years and so on). We can investigate
• Categories with small expected counts should be                          whether the market research company is obtaining
combined to overcome this problem.                                       an unrepresentative sample (with respect to age) by
testing
Example 11.5: A Market Survey
H0 :           p1 = 0.29             p2 = 0.24     p3 = 0.20   p4 = 0.15
A market research company is concerned that it
is not obtaining samples with a representative age                                        p5 = 0.08             p6 = 0.03     p7 = 0.01
proﬁle. One particular survey concerned the opinions                       (against the assumed alternative, H1, that ‘H0 is
of adults in South Africa. The age proﬁle of adult                         not correct’). Now under this null hypothesis, the
South Africans is described in Table 24 below.                             expected counts from a sample of size n = 100 are
as follows:
Age Group
e1       e2         e3   e4   e5   e6   e7
18-29             30-39         40-49 50-59 60-69      70-79     80+
29       24         20   15   8    3    1
%   29               24            20      15    8         3        1
Table 24: Age proﬁle for adult South Africans                         • The expected number of individuals in the last
category (80 years or older) is e7 = 1.
The company’s survey technique obtained a sample
of 100 adults, split by age group as displayed in Table                    • Since this expect count is less than 5 we should
25.                                                                          combine this group with another.

Statistics 160 – Semester 2, 2003                             297          Statistics 160 – Semester 2, 2003                                298

• It makes sense to combine groups 6 and 7 (to give                                  11.2 Testing for Association in
a single new group 6, corresponding to 70 years                                    Two-Way Contingency Tables
and older).

• The new group 6 has expected count equal to
the sum of the expected counts from the original                         11.2.1 Introduction
groups 6 and 7; that is, e6 + e7 = 3 + 1 = 4.
Recall from section 3.5 that when individuals are
• This expected count is less than 5 so we must                            grouped by two categorical variables, then we can
continue to combine categories.                                          arrange the counts in a two-way contingency table.

• We now combine the original groups 5, 6 and 7
(to make a new group of 60 years and older).
This group has expected count 8 + 3 + 1 = 12,
so there is no need for further group combination.

• After all the combinations of groups, we obtain the
following table of observed and expected counts.
Age group
18-29      30-39    40-49 50-59     60+
Expected             29         24        20     15      12
8+3+1
Observed               15       25       22    18        20
14+5+1

• The chi square test can now continue in the usual
way using this listing of expected and observed
counts. (The remainder of the test is left as an
exercise.)

Statistics 160 – Semester 2, 2003                             299          Statistics 160 – Semester 2, 2003                                300
Example 11.6: Child Beneﬁt and                                                        11.2.2 Hypotheses
Occupation

• Typically want to know if there is a relationship
A researcher investigating public attitudes to the level                between the row and column variables.
of welfare beneﬁts in the United Kingdom carried out
a pilot survey in the town where she lived. A simple
• However, the exact statement of the hypotheses
random sample of individuals was drawn from the
depend on the manner in which the data were
electoral register, and each member of the sample
collected.
was asked to complete a questionnaire. Table 26
gives the results for questions on child beneﬁt, in
which respondents are categorized by occupation.                      • Depends on whether marginal totals are ﬁxed.

Child                Non-manual    Manual      No current   Total
beneﬁt is:           occupation   occupation   occupation
Too high                 18           13            3        34
About right              29           40           13        82
Too low                  10           26           11         47
Don’t know               15           14            8        37
Total                    72           93           35        200

Table 26: Responses to a survey on child beneﬁt

These data form an example of a 4 × 3 contingency
table. (More generally, a r × c contingency table
is a two-way table with r rows and c columns.) It
is of interest to social scientists and policy makers
to investigate whether an individual’s occupation is
related to their feelings about child beneﬁt.

Statistics 160 – Semester 2, 2003                             301     Statistics 160 – Semester 2, 2003                302

Marginal Totals Fixed                                           Totals Fixed in One Margin

Suppose the data have been collected from an
• If the data have been collected from a survey
experiment (or a survey using stratiﬁed sampling)
then neither row nor column marginal totals will
so that
be ﬁxed.
• each row is a sample from a particular population;
• In this case the appropriate hypotheses are

• the marginal row totals are ﬁxed.
H0 : No association between row and column variables

In this situation the row variable is an explanatory
and
variable (or factor) and the column variable is a
response.
H1 : Association between row and column variables exists
The hypotheses are
H0 : The distribution of the response is the same
for every level of the explanatory variable
Example 11.7
From example 11.6 we will be interested in testing                    and
H1 : The distribution of the response is not the
H0 : No assoc. between occupation and opinion on c.b.                 same for every level of the explanatory variable

against

H1 : Assoc. between occupation and opinion on c.b. exists

Statistics 160 – Semester 2, 2003                             303     Statistics 160 – Semester 2, 2003                304
Example 11.8                                         11.2.3 Test Statistics and P-values

A market research company surveys 50 employed,
and 50 unemployed people, asking how they plan to                            Denote observed counts in a r × c table as follows.
vote in the next federal election. The results are as
follows.                                                                                                         Column
1        2  ...      c     Total
Labor           Liberal   Other   Total                     Row 1          o11      o12 . . .   o1c     o1·
Employed                  18               27        5      50                           2          o21      o22 . . .   o2c     o2·
Unemployed                29               17        4      50                            .
.          .
.        .
.   ...     .
.       .
.
Total                     47               44        9                                    r         or1      cr2 . . .   orc     or·
Here employment status is an explanatory variable,                                        Total         o·1      o·2 . . .   o·c      n
and voting intention is a response. It may be of                             Hence oij is the observed count in cell i, j of the
interest to test                                                             table.
H0 : he distribution of voting intentions is the same
for employed and unemployed people
Expected Counts
against                                                                      Assuming H0 is correct, the expected count in cell
H1 : The distribution of voting intentions diﬀers                            i, j is
according to employment status
oi·o·j
eij      =
n
Row i total × Column j total
Important: The test statistic, P-value etc. are                                                =
calculated in the same way no matter whether row                                                               Grand total
marginal totals are ﬁxed or not.

Statistics 160 – Semester 2, 2003                                      305   Statistics 160 – Semester 2, 2003                                  306

Chi Square Test Statistic                                           Example 11.9: Child Beneﬁt and
Occupation

The chi square test statistic is
Following on from Example 11.6, we are interested
r   c                                   in testing
(oij − eij )2
χ2 =                                                  H0: There is no association between occupation and
i=1 j=1
eij
opinion on child beneﬁt, (i.e. they are independent)
• Under the assumption that H0 is correct, this                              against
test statistic has an (approximate) chi square
H1: There is an association between occupation and
distribution with ν = (r − 1)(c − 1) degrees of
opinion on child beneﬁt.
freedom.
Solution by Hand
• The test statistic is likely to take extreme values
Suppose H0 is correct. Then the expected number
if H0 is not correct.
of non-manual workers who think that the level of
child beneﬁt is too high will be 72 × 34/200 = 12.24.
P-Value                                                                      Carrying on in a similar fashion produces the Table
The P-value for an observed test statistic χ2 is given                       27 of expected counts:
by
P = P (X ≥ χ2)                                                Child                Non-manual        Manual       No current   Total
beneﬁt is:           occupation       occupation    occupation
where X ∼ χ2(ν) and ν = (r − 1)(c − 1).                                        Too high               12.240           15.810        5.950       34
About right            29.520           38.130        14.350      82
Too low                16.920           21.855        8.225       47
Don’t know             13.320           17.205        6.475       37
Total                    72               93            35        200

Table 27: Expected cell counts.

Statistics 160 – Semester 2, 2003                                      307   Statistics 160 – Semester 2, 2003                                  308
The contributions to the χ2 statistic from each                     Solution Using MINITAB
category, (o − e)2/e, are
The ﬁrst step is to enter the columns of the table into
columns C1-C3 (say) of a new MINITAB Worksheet.
Child                   Non-manual       Manual      No current   The χ2 test can be carried out via the menu sequence
beneﬁt is:              occupation      occupation   occupation   Stat, Tables, Chisquare Test..., which provides
Too high                  2.7106          0.4994       1.4626     a dialogue box asking for the worksheet columns
About right               0.0092          0.0917       0.1270     containing the contingency table.
Too low                   2.8302          0.7861       0.9362
Don’t know                0.2119          0.5970       0.3592     The output in the session window is as follows.

The entries in this table sum to give χ2 = 10.6211.                 Expected counts are printed below observed counts
2
If H0 is correct, this statistic comes from a χ
distribution with (4 − 1) × (3 − 1) = 6 degrees of                                        C1               C2        C3     Total
freedom. Using MINITAB,                                                     1             18               13         3        34
12.24            15.81      5.95
P (X ≤ 10.6211) = 0.8993
2             29               40        13        82
29.52            38.13     14.35
where X ∼ χ2(6). Hence the P-value is
3             10               26        11        47
P = P (X ≥ 10.6211) = 1−0.8993 = 0.101                  3 d.p.                         16.92            21.85      8.23

4             15               14         8        37
Conclusion: There is insuﬃcient evidence to
13.32            17.20      6.47
conclude that an association between type of
occupation and attitude towards level of child beneﬁt
Total                    72            93        35      200
exists.
ChiSq =            2.711 +          0.499 +   1.463 +

Statistics 160 – Semester 2, 2003                           309     Statistics 160 – Semester 2, 2003                          310

0.009 +                    0.092 +   0.127 +                    Example 11.10: Heart Attacks and
2.830 +                    0.786 +   0.936 +
Smoking in Danish Women
0.212 +                    0.597 +   0.359 = 10.621
df = 6, p=0.101
Note that this output gives us the following                        Recall from Example 9.4 that we had data on over
information.                                                        11,000 Danish women, each one of who was classiﬁed
according to whether or not she smoked, and also
• Observed and expected counts for each cell.                       according to whether or not she had suﬀered a heart
attack. These data may be displayed in a 2 × 2
• Observed chi square statistic (χ2 = 10.621).                      contingency table, as below.
Heart attack
• The degrees of freedom, ν = 6.                                                                            No     Yes
Smoker            6081 380
• The P-value of P = 0.101                                                               Non-smoker        4879 132
We may analyse this data using a chi square test in
Naturally our conclusion is exactly as above.
the usual way. We can set up the hypotheses as

H0 : No assoc. between heart attacks and smoking

and

H1 : Assoc. between heart attacks and smoking

Using MINITAB we get the following output:

Statistics 160 – Semester 2, 2003                           311     Statistics 160 – Semester 2, 2003                          312
Chi-Square Test                                                against
H1 : p S = pN S
Expected counts are printed below observed counts              (where pS is the population proportion of female
smokers having heart attacks, and pN S the
C1                   C2       Total         population proportion for non-smokers), we can use
1        6081                  380        6461         the techniques from chapter 9. The test statistic is
6172.64               288.36
0.0588 − 0.0263
2        4879                  132        5011         z     =
0.0446 × (1 − 0.0446) × (1/6461 + 1/5011)
4787.36               223.64
= 8.36
Total              10960               512       11472
The P-value corresponding to this test statistic is
ChiSq =  1.361 + 29.125 +                                      p < 0.00001. Hence we again have extremely strong
1.754 + 37.553 = 69.793                               evidence against H0, i.e. in favour of a diﬀerence in
df = 1, p = 0.0000                                             risk of heart attack for smokers and non-smokers.

We can conclude that there is extremely strong                 Which Method is Better?
evidence to support the existence of an association            The chi square contingency table approach and
between having a heart attack and smoking.                     the comparison of proportions approach are both
perfectly valid. They will give the same P-values. It
We could have analyzed this data in a diﬀerent way,            is a matter of taste as to which method of analysis
by comparing the proportions of women having heart             you prefer.
attacks in the smoking and non-smoking groups.
ˆ
For the smokers, the proportion is pS = 380/6461 =
0.0588, and for the non-smokers the proportion is              See also M&M Exercises 9.1, 9.7, 9.9, 9.21, 9.23
ˆ
pN S = 132/5011 = 0.0263. For testing

H0 : p S = pN S

Statistics 160 – Semester 2, 2003                        313   Statistics 160 – Semester 2, 2003                     314

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