# Chapter 10: One- and Two-Sample Tests of Hypotheses by gdDK26

VIEWS: 71 PAGES: 86

• pg 1
```									                                                              10.1
Chapter 10: One- and Two-Sample Tests of Hypotheses
Take Sample

Inference                        Population
Sample

Many of the same types of things we did in Chapter 9 will be
done here as well, but through more formal methods
(hypothesis tests). Below is a summary of the whole
process:
1) Define a population and parameter(s) of interest.
2) State a hypothesis about the parameter’s value.
3) Take a representative sample from the population.
4) Calculate the statistic(s) using the sample.
5) Make inferences from the sample to the population by
using hypothesis tests. The hypothesis tests will allow
us to make a decision about hypotheses of interest with
a certain level of confidence.

From this chapter, it is important to learn the following:
 Null and alternative hypotheses
 Type I and II errors
 Hypothesis tests procedures for a variety of problems
 2005 Christopher R. Bilder
10.2
10.1-10.7: The Basics of Hypothesis Testing and Testing
Hypotheses Regarding 

Hypothesis: A statement that something is true.

Below is an example to help introduce hypothesis testing:

Example: Light Bulbs (light_bulbs.xls from Chapter 9)

Suppose that General Electric is interested in estimating
the mean lifetimes of its light bulbs. It hypothesizes that
=250 (this could be what is stated on the package). How
can this be checked?

General Electric takes a random sample of 16 light
bulbs and finds they last on average for 299.2 hours
with a standard deviation of 80 hours. The 95% C.I. for
 is 264.14 <  < 334.26.

Is =250? Since 264.14 <  < 334.26 with a 95% level
of confidence,  appears to be greater than 250.
Therefore, reject the hypothesis of =250.

Suppose before the sample was conducted, General
Electric hypothesized that =270. Is this correct?

Again, the sample was taken and the C.I. above was
obtained. Since  could be 264.15, 268, 270, 272,
 2005 Christopher R. Bilder
10.3
300,…, =270 may be correct. Therefore, do not reject
the hypothesis of =270.

There is not sufficient evidence from the sample to
prove the hypothesized value of =270 to be incorrect.

Finally, suppose before the sample was conducted,
General Electric hypothesized that =350. Is this correct?

Again, the sample was taken and the confidence
interval above was obtained. Since 264.14 <  <
334.26 with a 95% level of confidence,  appears to be
less than 350. Therefore, reject the hypothesis of
=350.

The above is an informal example of a hypothesis test. In
many real life situations, there is a hypothesis about the
population mean or other population parameters. A sample
from the population is taken to investigate the hypothesis.

For the first hypothesis of =250 in the light bulb
example, two hypotheses were considered:
 Null Hypothesis, Ho:=250
 Alternative Hypothesis, Ha:250

One of two possible decisions were made:
 Reject Ho - This indicates  is not 250
 2005 Christopher R. Bilder
10.4
 Don't Reject Ho - This indicates there is not sufficient
evidence from the sample to say  is different from
250. You can not say "Accept Ho"; i.e., can not say Ho
Comment [CRB1]: C.I. gave a range of
is true. See the reason in the following (and previous)       possible value for . One of those is the
hypothesized value of the population mean (if
example.                                                      we don’t reject Ho). Thus, the null hypothesis
may or may not be true.

Note: Some people will use the terminology “Fail to
reject Ho” instead of “Don’t reject Ho”. Both are fine
to use.

Example: Jury Trials

Juries are asked to consider two hypotheses:

Ho:Defendant is innocent
Ha:Defendant is guilty

The defendant is assumed innocent until proven guilty. In
hypothesis testing, we assume Ho is true until there is
enough evidence to prove otherwise.

The jury listens to the prosecution and the defense to
make a judgment. This is like taking a SAMPLE.
 If there is ENOUGH evidence (beyond a reasonable
doubt) to convict - Reject Ho, the defendant is “guilty”.
 If there is NOT ENOUGH (reasonable doubt) evidence
to convict - Don't Reject Ho, the defendant is “not guilty”.
Notice, this does not mean the defendant is innocent.
 2005 Christopher R. Bilder
10.5

Types of errors in hypothesis test decisions
 Type I - Reject Ho, but in reality Ho is true
 Type II - Don't reject Ho, but in reality Ha is true
(reality=population)

These errors indicate that the sample led us to believe
something about the population that is incorrect.

Example: Jury Trials

 Type I: Reject Ho = jury says the defendant is guilty,
but Ho is really true = defendant is innocent.

Send an innocent person to jail

 Type II: Don't reject Ho = jury says the defendant is not
guilty, but Ha is really true = defendant is guilty.

Let a criminal go free

Probability of making errors

A type I error is the more serious error in the jury trial
example and in statistics. Thus, the P(Type I error) is

 2005 Christopher R. Bilder
10.6
controlled in a hypothesis test at a specified level
denoted by . Therefore,

P(Type I error) = P(Reject Ho | Ho is TRUE) = .

This is also called the “level of significance”.

A type II error is generally not as serious, so it is usually
not controlled at a fixed level. We can still define the
probability of committing this error:

P(Type II error) = P(Don’t reject Ho | Ha is TRUE) = 

Table describing the two errors:
Based on Sample
Reject Ho        Don’t Reject Ho
Ho is TRUE       Type I Error     Correct Conclusion
Population
Ha is TRUE    Correct Conclusion     Type II Error

Same table, but with the conditional probabilities:
Based on Sample
Reject Ho         Ha is TRUE
P(Reject Ho |       P(Reject Ho |
Ho is TRUE
Ho is TRUE) =     Ha is TRUE) = 1-
Population
P(Don’t reject Ho | P(Don’t teject Ho |
Ha is TRUE
Ho is TRUE) = 1-    Ha is TRUE) = 

Use the basic definitions of conditional probabilities from
Chapter 2 to help interpret the table! Remember that
P(A|B) + P(A|B) = 1

 2005 Christopher R. Bilder
10.7
Power

In hypothesis testing, we will make the assumption that
Ho is true and then try to prove it to be incorrect using
the evidence gathered in the sample. Thus, it is
important to define

P(Reject Ho | Ha is TRUE).

This is called the power of the test. Notice where this
result falls in the above table and it has a probability of
1-.

Question: Do you want this probability to be small or
large?

 2005 Christopher R. Bilder
10.8
Three Methods for performing a hypothesis test
1) Confidence interval – Section 10.6
2) Test statistic – Sections 10.5, 10.7
3) P-value – Sections 10.4, 10.5, 10.7

All three provide the same answer when testing the
population mean! Note that there may be slightly different
conclusions when testing a population proportion, p.

 2005 Christopher R. Bilder
10.9
1) The confidence interval method - 4 Steps

1. State Ho:=0
Ha:0 where 0 is some number
2. Find the C.I. for 
3. Reject or do not reject Ho – Check if the hypothesized
value of  is inside the interval.
4. Conclusion – Describe what 3. means in terms of the
original problem

Example: GPA Example.

Test the hypothesis that the mean GPA of UNL students
is 3.0. Suppose P(Type I error)==0.05, x =2.9, n=16,
and s=0.1.

1. Ho:=3.0
Ha:3.0
s                              s
2. x  t / 2,n1          x  t / 2,n1
n                                n
0.1                   0.1
 2.9  2.131       2.9  2.131
4                     4
 2.847 <  < 2.953

3. Reject Ho since =3.0 is not in the interval.
4. The average GPA of UNL students is not 3.0.

 2005 Christopher R. Bilder
10.10
Remember: The probability of incorrectly rejecting =3.0
is 5% (probability of making a type I error). Thus, if the
whole process of taking a sample and doing the
hypothesis is repeated 1,000 times WITH  = 3.0, we
would expect 0.051,000 = 50 times to incorrectly reject
Ho:=3.0.

Example: Volleyball quality control
(hyp_volleyball_data.xls)

Suppose Mikasa, a volleyball manufacturer, is
concerned about whether their volleyballs are being
produced with the correct radius of 11.6cm. A sample of
36 volleyballs is taken with x =11.5 and s=1. Part of the
data set is below.

11.38
12.78
10.61

10.10

Is there evidence to show the volleyballs are being made
incorrectly? Conduct a hypothesis test with =0.05.

1. Ho:=11.6
Ha:11.6

 2005 Christopher R. Bilder
10.11
s                               s
2. x  t / 2,n1          x  t / 2,n1
n                                n
1                   1
 11.5  2.03       11.5  2.03
6                   6
 11.16 <  < 11.84
3. Do not reject Ho since =11.6 is in the interval.
4. There is not sufficient evidence to prove the

OR

There is not sufficient evidence to conclude the
population mean radius is different from 11.6.

Notes:
 What should Mikasa do? Continue with production of
the volleyballs.
 I did not say, "The volleyballs are being produced
correctly." THIS IS WRONG because of the probability
of committing a Type II Error is NOT controlled ( was
not stated). Compare this to the GPA example!

 2005 Christopher R. Bilder
10.12
2) The test statistic method - 5 Steps

1. State Ho and Ha
x  0
2. Find the test statistic: t 
s/ n

Does this look familiar??? See Chapters 8 and 9.

The test statistic examines how far the sample mean is
from the hypothesized mean. The numerator of t, x -0,
is divided by s / n to account for the variation of x .

Provided Ho is true and X1, X2, …, Xn is a random
sample from population with a normal PDF with E(X) =
0 and Var(X) = 2, the random variable version of the
test statistic:

X  0
T
S/ n

has a t-distribution with  = n-1 degrees of freedom. In
part 1 of the Chapter 9 (p. 12) notes, we showed that

P(  t / 2,n1  T  t / 2,n1)  1  
X  0
 P(  t / 2,n1                t / 2,n1)  1  
S/ n

 2005 Christopher R. Bilder
10.13

Thus, we have a range of probable values for T for a
given . If we observe t to be outside of this range, this
gives us evidence that our initial assumption of “Ho is
true” is incorrect!

3. Find the critical values: ±t/2, n-1

There are two critical values which define the range of
probable values for T. Again, remember that

P(  t / 2,n1  T  t / 2,n1)  1  
X  0
 P(  t / 2,n1                t / 2,n1)  1  
S/ n

If we observe t to be outside of this range, this may give
us evidence that our initial assumption of “Ho is true” is
incorrect!

4. Reject or do not reject Ho
i) Draw the t-distribution
 2005 Christopher R. Bilder
10.14
ii) Plot the critical value
iii) Label the graph with reject and don't reject regions
iv) Plot the test statistic
v) Write reject or don’t reject Ho and provide a reason

t-distribution
h(t)

Reject Ho
Reject Ho
Don't Reject Ho

Critical Value        0        Critical Value     t

5. Conclusion – Describe what 4. means in terms of the
original problem.

Example: Volleyball quality control (hyp_volleyball_data.xls
and hyp_1sample_pic.xls); remember n=36 and =0.05

1. Ho:=11.6
Ha:11.6
x  0
2. t          = (11.5 - 11.6)/(1/6) = -0.6
s/ n
3. ± t/2, n-1=±2.03
4.
 2005 Christopher R. Bilder
10.15

t-distribution
h(t)

Reject Ho
Reject Ho           Don't Reject Ho

-2.03       -0.6        0              2.03               t

Since –2.03<-0.6<2.03, do not reject Ho

5. There is not sufficient evidence to prove the volleyballs

OR

There is not sufficient evidence to conclude the
population mean radius is different from 11.6.

To help better understand the test statistic method,
suppose we had a different x and everything else
remained the same. Below is a table showing what would
happen with the hypothesis test.

Case           x               t              Decision
 2005 Christopher R. Bilder
10.16

Case         x               t           Decision
1          11.2           -2.4          Reject Ho
2         11.4           -1.2     Don't Reject Ho
3         11.6            0       Don't Reject Ho
4         11.8           1.2      Don't Reject Ho
5         12.0           2.4           Reject Ho

t-distribution
h(t)

Reject Ho
Reject Ho
Don't Reject Ho

-2.4 -2.03    -1.2        0      1.2 2.03   2.4     t

See the file, hyp_1sample_pic.xls, for how some of the
calculations can be done in Excel.

 2005 Christopher R. Bilder
10.17

 2005 Christopher R. Bilder
10.18

 2005 Christopher R. Bilder
10.19
All values in red can be changed by the user to see the
effect on the test statistic, critical values, and the
hypothesis test decision. Make changes on your own so
that you familiarize yourself with what happens if the
sample size increases, standard deviation changes,…
Note that this file can be used to help perform ANY
hypothesis test for a population mean.

P-values will be discussed later in this chapter.

Notes:
 The way hypothesis testing is set up is to try to find
evidence (through a sample) against the null
hypothesis (Ho). If enough evidence is found, we can
conclude that the alternative hypothesis (Ha) to be true
(with the probability of a type I error of ). Since 
(probability of type II error) is not controlled, we can
not set up hypothesis testing to go the other way.
 Notice that in the formula for t we put in the
hypothesized value of , 0. We assume the null
hypothesis to be true by doing this (remember the jury
trial example). We put values from the sample ( x , s,
n) into the test statistic to see if the sample mean is far
enough from the hypothesized mean to conclude that
the null hypothesis is incorrect.
 Why was Ho:=11.6 vs. Ha:11.6?
o In order for the theory behind all of this to work, we

 2005 Christopher R. Bilder
10.20
o If some kind of new "action" is to be taken when a
hypothesis is proved to be true, this hypothesis
typically should be in Ha. This is because we can
control the probability of making an error in our
decision (i.e.  is specified).
 What would happen if Mikasa's volleyballs did
not have an average radius of 11.6?
Production of volleyballs would be stopped and
the manufacturing process would be
investigated to find the problem. This implies we
should use Ha:11.6.
 What would happen if Mikasa's volleyballs have
an average radius of 11.6? The production of
volleyballs would continue.
 The book uses the normal PDF instead of the t-
distribution when n30 and  is known (Section 10.5)
o The same problems with using the normal PDF
version of the C.I. occur here;  is unknown in real-
life applications.
o Remember that for large samples (n30) the t-
distribution is approximately a standard normal PDF.
o IN THIS CLASS, WE WILL ONLY USE THE t-
DISTRIBUTION!

 2005 Christopher R. Bilder
10.21
Example: Volleyball quality control (hyp_volleyball_data.xls
and hyp_1sample_pic.xls)

There are a few different ways Excel can be used to help
do the hypothesis test. One way has already been
shown in hyp_1sample_pic.xls. Another way is to select
TOOLS > DATA ANALYSIS from the main Excel menu
bar and select the appropriate analysis tool.

Before this can be done, a new column of data must be
entered into the spreadsheet for the hypothesized value
of . Below is part of the spreadsheet with this
information entered.

After this is entered, select TOOLS > DATA ANALYSIS
from the main Excel menu bar. Select t-test: Paired Two
Sample for Means. The hypothesis test performed for
one population mean is actually a special case of
 2005 Christopher R. Bilder
10.22
another hypothesis test called “Paired Two sample for
Section 10.9. After the correct option is chosen, select
OK to bring up the t-test Paired Two Sample for Means
window. Below is the completed window.

Notice what is put in the Variable 2 Range – the column
of hypothesized values. The Hypothesized Mean
Difference value comes from 0 in Ho:-11.6=0 vs. Ho:-
11.60.

Below is the output produced after selecting OK.

 2005 Christopher R. Bilder
10.23

Mean                                      11.5000                11.6
Variance                                    1.0000                  0
Observations                                    36                 36
Pearson Correlation                  #DIV/0!
Hypothesized Mean Difference                     0
df                                              35
t Stat                                     -0.6001
P(T<=t) one-tail                            0.2762
t Critical one-tail                         1.6896
P(T<=t) two-tail                            0.5523
t Critical two-tail                         2.0301

For example, the x , s2, and n are given in their
appropriate rows and columns. The test statistic is
given in the t Stat row. The positive part of the
critical values is given in the t Critical two-tail row.

There are a few other things given in the table which
we will discuss later.

Chris Malone’s Excel Instructions - From the website
(http://www.statsteacher.com/excel), select Analyses
> Mean: One Sample to find help on how to use the
Data Analysis tool in Excel to perform the hypothesis
tests discussed in this chapter.

 2005 Christopher R. Bilder
10.24

 2005 Christopher R. Bilder
10.25
3) The p-value method – 5 steps

The test statistic method compared t and the critical
values from the t-distribution. The p-value method
compares probabilities. Thus, we have “p”-value
method.

1. State Ho and Ha
2. Find the p-value
x  0
Find the test statistic t      and compute the p-
s/ n
value as 2P(T > |t|) where T is a random variable with
 = n-1 degrees of freedom. The Excel function is
2*TDIST(ABS(t), df , 1)
t-distribution
h(t)

t
0               |t|

The p-value gives the probability of finding a value of
|t| at least this great assuming the null hypothesis is
true.
 2005 Christopher R. Bilder
10.26

The probability, P(T > |t|), is multiplied by two since the
disagreement between the data and Ho can be in two
directions; i.e., on two tails of the PDF.

Remember that the p-value is just a probability found
through integration! Here, the p-value is

     1 / 2  u2 
 ( 1) / 2

2                   1
  
du
|t|    / 2          

3. State 
4. Reject or do not reject Ho

 Reject Ho if p-value < 
 Don’t reject Ho if p-value  

Remember  is also called "the level of significance"

5. Conclusion – Describe what 4. means in terms of the
original problem.

Example: Volleyball quality control (hyp_volleyball_data.xls
and hyp_1sample_pic.xls)

1. Ho:=11.6
Ha:11.6
 2005 Christopher R. Bilder
10.27
x  0
2. t             = (11.5 - 11.6)/(1/6) = -0.6
s/ n
2P(T > |-0.6|) = 2P(T > 0.6) = 20.2762 = 0.5524

where  = n – 1 = 35. The probability of observing a test
statistic value this great in magnitude, |-0.6|, is 0.5524 if
Ho:=11.6 was true. Therefore, this is a likely event to
happen if =11.6.

Another way to think about the p-value is the following: If
 really was 11.6, then a test statistic value, t, at least
this large in absolute value (0.6) would occur about 55%
of the time if the hypothesis test process (take a new
sample and perform a new hypothesis test) is repeated a
very large number of times. In other words, this is likely
to occur if =11.6. Thus,  could be 11.6 since this is a
likely event.

Finding the p-value using integration:
  35  1 / 2 
 (35 1) / 2
 1 t 
                          2
2         
 35                 dt = 20.2762 =
|0.6|   35 / 2  35       
0.5524

> f(t):=GAMMA((nu+1)/2)/(GAMMA(nu/2) *
sqrt(Pi*nu)) * (1+t^2/nu)^(-(nu+1)/2);

 2005 Christopher R. Bilder
10.28
(  1/2  1/2 )
 1  1   1 t 
2

2                   
        2 
        
f( t ) :=
    
1

2   
     
> 2*int(eval(f(t),nu=35),t=abs(-
0.6)..infinity);
.5523713960
3. =0.05
4. Since 0.5524 > 0.05 do not reject Ho
t-distribution

0.2762
h(t)

/2=
=0.025
0.025

t

0     |-0.6|

5. The sample does not provide enough evidence to
suggest that the volleyballs are being made with the

OR
 2005 Christopher R. Bilder
10.29

There is not sufficient evidence to conclude the
population mean radius is different from 11.6.

Fill in the p-values for the table below:

Case      x       t             Decision        p-value
1     11.2    -2.4            Reject Ho
2     11.4    -1.2      Don't Reject Ho
3     11.6     0        Don't Reject Ho
4     11.8    1.2       Don't Reject Ho
5     12.0    2.4             Reject Ho

See the file, hyp_1sample_pic.xls, for how the p-value
calculations can be done in Excel. Also, make changes on
your own to the values in red so that you familiarize
yourself with what happens if the sample size increases,
standard deviation changes,…

Question: How can the power be increased for a
hypothesis test?

Make sure you can do the hypothesis test problems with
EACH method. Remember all three hypothesis test

 2005 Christopher R. Bilder
10.30
methods give the same answers for tests involving  and the
t-distribution.

Understanding the type I error rate

Suppose Ho is true and the type I error rate is denoted by
. If the hypothesis testing procedure is repeated R times
(take a new sample and perform a new hypothesis test),
we would expect R of the hypothesis tests to incorrectly
reject Ho.

Example: CI_ex.xls from Chapter 9

Using the CLT_GPA_ex.xls file, confidence intervals for
the population mean are calculated for each of the 1,000
samples. If =0.05, we would expect approximately 5%
of the confidence intervals to NOT contain the population
mean. The proportion that contain the population mean
is 94.7%. Thus, the “type I error rate” is 5.3%. If this
procedure was repeated a lot more than 1,000 times, the
type I error rate would be 5%.

 2005 Christopher R. Bilder
10.31
Notes:
 The hypothesis tests done so far are often called t-tests
since the t-distribution is used in the test.
 Use hyp_1sample_pic.xls as a template for some of the
calculations needed for the hypothesis tests!
 One-tail tests - The hypothesis tests performed so far have
been of the form:

Ho:=o vs. Ha:o

where o is just a number like 11.6cm.

In order to reject Ho, the test statistic is too big OR too
small (i.e. there are two rejection regions). These kinds of
hypothesis tests are called two-tail tests because the
rejection region falls in two tails of the PDF.

There are also ONE-TAIL tests of the form:
Test        Name
Ho:o
Left-tail
Ha:<o
Ho:o
Right-tail
Ha:>o

The discussion of these types of tests will be postponed
until p. 10.67.

 2005 Christopher R. Bilder
10.32
10.8: Two Samples: Tests on Two Means

In Sections 9.8 and 9.9, we examined two different
cases of estimating the difference between two means:
i) Independent samples from two populations
ii) Dependent (paired) samples from two populations

Both of these situations will be discussed with respect to
hypothesis testing here. Similar to Chapter 9, we will
only work with the realistic situations of unknown 1 and
2

2 and 1 possibly unequal to 2 for independent
2
2
2
samples. For the dependent samples, we will only work
with the case where the population variance is unknown
as well.

Example: Dividend Yield (div_yield.xls from chapter 9)

Is there a difference in average dividend yield of
Comment [CRB2]: Why would there possibly
companies traded on the NYSE vs. NASDAQ? Perform              be a difference – type of companies on the
stock exchanges
a hypothesis test to determine if there is a difference.

Plot from Chapter 9.

 2005 Christopher R. Bilder
10.33

Do you think there is a difference between the mean
dividend yields for all companies traded on the stock
exchanges? Try to make an initial judgement based on
the plots.

Suppose 1 = NYSE and 2 = NASDAQ.

C.I. Method using =0.05:
1) Ho:1 - 2 = 0
Ha:1 - 2  0

 2005 Christopher R. Bilder
10.34
2) The 95% C.I. is -0.0070 < 1 - 2 < 0.0245 (from
Chapter 9)
3) Do not reject Ho since 0 is in the interval.
4) There is not sufficient evidence to indicate a
difference between the mean dividend yields for
companies traded on the two stock exchanges.

Of course, you could also do the hypothesis test using the
test statistic and p-value methods.

Test statistic: From Chapter 9, we saw that

                                              


P   t / 2, 

X1  X2  1  2             
 t / 2,   1  
                     2
S1 S2 2                   
                                             
                    n1 n2                     
2
 s1 s2 
2  2

n n 
where              1   2 

                          
2        2        2        2
s1 n1             s2 n2
n1  1           n2  1

Thus, the test statistic to be used here replaces the
random variables with their observed values for what
is in the middle of the inequality above:
 2005 Christopher R. Bilder
10.35

x1  x2  (1  2 )            x1  x2  0
t                               
2   2                          2   2
s1 s2                          s1 s2
                              
n1 n2                          n1 n2

Notice that 1 - 2 is replaced with 0. This is done
since most often we will hypothesize
Ho:1 - 2 = 0 or Ho:1 - 2  0 or Ho:1 - 2  0.

P-value: 2P(T>|t|) where T has  degrees of freedom
for a two-tail test.

Example: Dividend Yield (div_yield.xls from chapter 9)

Below are the screen captures from Chapter 9:

 2005 Christopher R. Bilder
10.36

Test statistic method (using =0.05):
1) Ho:1 - 2 = 0
Ha:1 - 2  0
x1  x2  0    0.0206  0.0118
2) t                                  1.1181
2
s1 s2      0.03192 0.02892
 2              
n1 n2         30        30
3) t0.025,57 = 2.0025
4)
T Distribution
Probability

Reject Ho
Reject Ho
Don't Reject Ho

t
-2.0025                        2.0025
0     1.1181

 2005 Christopher R. Bilder
10.37

Do not reject Ho since -2.0025 < 1.1181 < 2.0025
5) There is not sufficient evidence to indicate a
difference between the mean dividend yields for
companies traded on the two stock exchanges.

P-value method:
1) Ho:1 - 2 = 0
Ha:1 - 2  0
2) 2P(T > |1.1181|) = 0.2682
3)  = 0.05
4) Do not reject Ho since 0.2682 > 0.05
5) There is not sufficient evidence to indicate a
difference between the dividend yields for companies
traded on the two stock exchanges.

Using the Data Analysis tool

Select TOOLS > DATA ANALYSIS > T-TEST: TWO-
SAMPLE ASSUMING UNEQUAL VARIANCES to bring
up the window below.

 2005 Christopher R. Bilder
10.38

Notice the window has already been filled in with the
appropriate entries for the hypothesis test. The labels
box was not checked because no labels were in the
variable 1 or 2 ranges. The reason no labels were
included was due to how the data was entered into
Excel (see the file). If the data was entered differently,
the labels option could be used.

Click OK to produce the output below. Note that
variable 1 is NYSE and variable 2 is NASDAQ. It is
usually good to label these in the output if the Labels
option was not selected.
 2005 Christopher R. Bilder
10.39
Variable 1 Variable 2
Mean                             0.0206     0.0118
Variance                         0.0010     0.0008
Observations                         30         30
Hypothesized Mean Difference           0
df                                   57
t Stat                           1.1181
P(T<=t) one-tail                 0.1341
t Critical one-tail              1.6720
P(T<=t) two-tail                 0.2682
t Critical two-tail              2.0025

Notes:
 The same problems discussed earlier in Chapter 10
regarding the critical value and p-value calculation
occur here.
 When you are doing a one-tail test, you need to be
very careful with what you are testing (right or left tail
test) and what Excel thinks you are testing. For
example, notice the Variable 1 Range contains the
NYSE data. This means Excel thinks I am testing
NYSE-NASDAQ. One can easily highlight the NASDAQ
range for the Variable 1 Range, but still think they are
testing NYSE-NASDAQ!

Chris Malone’s Excel Instructions
Help on how to perform this test in Excel is available on
the Excel Instructions website. Select Analyses > Mean:
Two Sample – Independent.
 2005 Christopher R. Bilder
10.40

 2005 Christopher R. Bilder
10.41
Hypothesis tests involving D for dependent (paired)
samples

Example: CPT (cpt.xls from Chapter 9)

A pharmaceutical company is conducting clinical trials on a
new drug used to treat schizophrenia patients. Ten
healthy male volunteers were given 3mg of the drug.
Before the drug was administered (time=0) and 4 hours
after (time=4), a psychometric test called the Continuous
Performance Test (CPT) was administered and the
number of “hits” was recorded for each patient.

From cpt.xls:

 2005 Christopher R. Bilder
10.42

Notice that (time 0 hits) – (time 4 hits) was found (Di =
Xi1- Xi2 where 1 = time 0 and 2 = time 4). Perform a
hypothesis test to determine if there was any type of an
effect on the average hits at the different times.

C.I. Method using =0.01:
1) Ho:D = 0
Ha:D  0
2) The 99% C.I. is 1.4470 < D < 4.1530
3) Reject Ho since 0 is not in the interval.
4) There is sufficient evidence to indicate a difference
between the average hits at the two time periods.

Of course, you could also do the hypothesis test using the
test statistic and p-value methods.

Test statistic: Note that

 2005 Christopher R. Bilder
10.43

              D  D            
P   t / 2,          t / 2,   1  
              SD n              

where D is the random variable for the difference in
sample means and  = n-1.

Thus, the test statistic to be used here replaces the
random variables with their observed values for what
is in the middle of the inequality above:

d  D            d0
t               
sD      n       sD          n

Notice that D is replaced with 0. This is done since
most often we will hypothesize Ho:D = 0 or Ho:D  0
or Ho:D  0.

P-value: 2P(T>|t|) where T has  = n – 1 degrees of
freedom for a two-tail test.

Example: CPT (cpt.xls from Chapter 9)

Test statistic method:
1) Ho:D = 0
Ha:D  0
 2005 Christopher R. Bilder
10.44
d       2.8
2) t                                 6.7254
sD   n 1.3166 / 10
3) t0.005,9 = 3.2498
4)

Reject Ho since 6.7254 > 3.2498.
5) There is sufficient evidence to indicate a difference
between the average hits for the two time periods.
Thus, the drug is having an effect on the patients.

P-value method:
1) Ho:D = 0
Ha:D  0
2) 2P(T > |6.7254|) = 8.6  10-5
3)  = 0.01
4) Reject Ho since 8.6  10-5 < 0.01
5) There is sufficient evidence to indicate a difference
between the average hits for the two time periods.
Thus, the drug is having an effect on the patients.

 2005 Christopher R. Bilder
10.45
Using the Data Analysis tool

Select TOOLS > DATA ANALYSIS > T-TEST: PAIRED
TWO SAMPLE FOR MEANS to bring up the window
below.

Notice the window has already been filled in with the
appropriate entries for the hypothesis test. The labels
box was checked because labels were in the variable 1
and 2 ranges. Click OK to produce the output below.

 2005 Christopher R. Bilder
10.46

t-Test: Paired Two Sample for Means
Time 0 Time 4
Mean                                49.6  46.8
Variance                           84.04 91.29
Observations                          10    10
Pearson Correlation               0.9910
Hypothesized Mean Difference           0
df                                     9
t Stat                            6.7254
P(T<=t) one-tail              4.301E-05
t Critical one-tail               2.8214
P(T<=t) two-tail              8.602E-05
t Critical two-tail               3.2498

Notes:
 The same problems discussed earlier in Chapter 10
regarding the critical value and p-value calculation
occur here.
 When you are doing a one-tail test, you need to be
very careful with what you are testing (right or left tail
test) and what Excel thinks you are testing. For
example, notice the Variable 1 Range contains the
Time 0 data. This means Excel thinks difference =
Time 0 – Time 4. One can easily highlight the Time 4
range for the Variable 1 Range, but still think they are
testing difference = Time 0 – Time 4!

 2005 Christopher R. Bilder
10.47
10.9: Choice of Sample Size for Testing Means

This section discusses how to choose a sample size in
order to have a particular level of power (1-). You are
not responsible for the material in this section.

10.10: Graphical Methods for Comparing Means

We have already been doing this starting in Section 8.3.
Box plots and dot plots are the two useful tools which
can be used here!

 2005 Christopher R. Bilder
10.48
10.11: One Sample: Test on a Single Proportion

We examined C.I.s for a proportion, p, in Section 9.10.
Now, we are going to formalize the discussion of making
decisions about p using hypothesis tests.

The hypothesis test described in this section and the one
in Section 10.12 are two places where performing a
hypothesis test using the C.I.s discussed in Chapter 9
may not result in the same answer as in Chapter 10.

Example: HCV (HCV.xls from Chapter 9)

Excerpt from Tebbs, Bilder, and Moser (Communications
in Statistics: Theory and Methods, 2003):

Hepatitis C (HCV) is a viral infection that causes
cirrhosis and cancer of the liver. Since HCV is
transmitted through contact with infectious blood,
screening donors is important to prevent further
transmission. With over 4.5 million people infected
in the United States, the World Health Organization
has projected that HCV will be a major burden on
the US health care system before the year 2020.

In our paper, we used data from Liu et al. (Transfusion,
1997) to demonstrate a new statistical procedure. The
authors reported results on 1,875 blood donors screened
 2005 Christopher R. Bilder
10.49
Comment [unl3]: "Shu-zoh"
for HCV at the Blood Transfusion Service in Xuzhou
City, China. There were 42 positive blood donors found.

Suppose Xuzhou City officials say that HCV prevalence
is 0.01. What do you think about the correctness of their
statement? Use =0.05 to perform a hypothesis test in
order to examine their statement.

Hypothesis test using the C.I. method:
1) Ho:p = 0.01
Ha:p  0.01
2) Using the Agresti-Coull C.I.,

x  (Z2 / 2 ) 2 42  (1.962 ) / 2

p                                   0.0234
n  (Z2 / 2 )
        1875  1.962

p(1  p)                    p(1  p)
p  z / 2               p  p  z / 2
n  z2 / 2
                      n  z2 / 2


0.0234(1  0.0234)                     0.0234(1  0.0234)
 0.0234  1.96                       p  0.0234  1.96
1875  1.962
1875  1.962

 0.0165 < p < 0.0302

3) Since 0.01 is not in the C.I., reject Ho.
4) There is sufficient evidence to conclude that the
proportion of people in Xuzhou City with HCV is not
0.01.
 2005 Christopher R. Bilder
10.50

Of course, you could also do the hypothesis test using the
test statistic and p-value methods.

Test statistic: Suppose our null hypothesis is:

Ho:p = p0 or Ho:p  p0 or Ho:p  p0.

From Chapter 9, we saw that

              ˆ
Pp                 
P   z / 2                z / 2   1  
            p(1  p) / n          
                                  

Thus, the test statistic to be used here replaces the
random variables with their observed values for what
is in the middle of the inequality above:

p  p0
ˆ
z
p0 (1  p0 ) / n

where p0 is the hypothesized value of p.

Question: What are the critical values?

P-value: 2P(Z>|z|) for a two-tail test.

 2005 Christopher R. Bilder
10.51
Example: HCV (HCV.xls from Chapter 9)

Suppose Xuzhou City officials say that HCV prevalence
is 0.01. What do you think about the correctness of their
statement? Use =0.05 to perform a hypothesis test in
order to examine their statement.

Hypothesis test using the test statistic method:
1) Ho:p = 0.01
Ha:p  0.01
x     42
2) Note that p  
ˆ                0.0224
n 1875
p  p0
ˆ                  0.0224  0.01
z                                            5.3964
p0 (1  p0 ) / n    0.01(1  0.01) /1875
3) z0.025 = 1.96
4)

Since 5.3964 > 1.96, reject Ho.
5) There is sufficient evidence to conclude that the
proportion of people in Xuzhou City with HCV is not
0.01.

Hypothesis test using the p-value method:

 2005 Christopher R. Bilder
10.52
1) Ho:p = 0.01
Ha:p  0.01
2) 2P(Z > |5.3964|) = 6.8010-8

Remember that a p-value is found through integration.
The p-value is

                       z2
1          
2  P(Z | 5.3964 |)  2                      e       2
dz = 6.8010-8
5.3964      2

In Maple,
> 2*int(f(z),z=5.3964..infinity);
.6799127778 10 -7
3)  = 0.05
4) Since 6.8110-8 < 0.05, reject Ho.
5) There is sufficient evidence that the proportion of
people in Xuzhou City with HCV is not 0.01.

Below are the results from HCV.xls.

 2005 Christopher R. Bilder
10.53

 2005 Christopher R. Bilder
10.54

Why can the C.I. method result in a different test outcome?

When deriving a C.I. for p, we used the following result

              ˆ
Pp                 
P   z / 2                z / 2   1  
            p(1  p) / n          
                                  

     
ˆ                             ˆ                    
P P  z / 2 p(1  p) / n  p  P  z / 2 p(1  p) / n  1  

This directly resulted in the C.I. of

p(1  p)
ˆ     ˆ                   p(1  p)
ˆ     ˆ
p  z / 2
ˆ                      p  p  z / 2
ˆ
n                         n

p(1  p)
Notice that we needed to replace p in                     ˆ
with p .
n
This was because we could not use the parameter value
ˆ
to calculate the C.I. for the parameter itself! Using p in
p(1  p)
can cause problems so we ended up using the
n
Agresti-Coull C.I. of

p(1  p)                    p(1  p)
p  z / 2               p  p  z / 2
n  z2 / 2
                      n  z2 / 2


 2005 Christopher R. Bilder
10.55
x  (z2 / 2 ) 2

where p                  .
n  (z2 / 2 )


For hypothesis testing, we will specify a hypothesized
value of p. Thus, we can now use the interval of

p  z / 2 p0 (1  p0 ) / n  p  p  z / 2 p0 (1  p0 ) / n
ˆ                                 ˆ

to do the hypothesis test where p0 is the hypothesized
value of p. Using this interval will result in the same
hypothesis test outcomes as the test statistic and p-
value methods.

Example: HCV (HCV.xls)

In the previous output under “Hypothesis test” the
p  z / 2 p0 (1  p0 ) / n  p  p  z / 2 p0 (1  p0 ) / n interval is
ˆ                                ˆ
calculated to be 0.0179 < p < 0.0269. Since 0.01 is not
in the interval, reject Ho.

 2005 Christopher R. Bilder
10.56
10.12: Two Samples: Tests on Two Proportions

We examined C.I.s for the difference of two proportions,
p1-p2, in Section 9.11. Now, we are going to formalize
the discussion of making decisions about p1-p2 using
hypothesis tests.

The hypothesis test described in this section and the one
in Section 10.11 are two places where performing a
hypothesis test using the C.I.s discussed in Chapter 9
may not result in the same answer as in Chapter 10.

Example: Larry Bird (bird_ch9.xls from Chapter 9)

Second
First
Missed 48      5            53
Total 299    39           338

Consider the first free throw made as one population and
first free throw missed as the second population. We
are interested in estimating the second free throw
probability of success (made) for each first free throw
outcome population. Then

x1 251                   x2 48
p1 
ˆ             0.8807 and p2 
ˆ          0.9057
n1 285                   n2 53
 2005 Christopher R. Bilder
10.57

Perform a hypothesis test to determine if the probability
of success for the second free throw attempt is
dependent on the outcome of the first free throw. Use
=0.05 to perform a hypothesis test in order to examine
the statement.

Hypothesis test using the C.I. method:
1) Ho:p1-p2 = 0
Ha:p1-p2  0
2) Using the Agresti-Caffo C.I.:

x  1 251  1
p1  1             0.8780 and
n1  2 285  2
x  1 48  1
p2  2             0.8909 .
n2  2 53  2

0.8780(1  0.8780) 0.8909(1  0.8909)
0.8780  0.8909  1.96                                         p1  p2
285  2            285  2
0.8780(1  0.8780) 0.8909(1  0.8909)
 0.8780  0.8909  1.96                         
285  2            285  2

-0.1035 < p1-p2 < 0.0778

3) Since 0 is in the C.I., do not reject Ho.
4) There is not sufficient evidence to conclude that the
outcome on the first throw has an effect on the
outcome of the second free throw.
 2005 Christopher R. Bilder
10.58

Of course, you could also do the hypothesis test using the
test statistic and p-value methods.

Test statistic: Suppose our null hypothesis is:

Ho:p1-p2 = 0 or Ho:p1-p2  0 or Ho:p1-p2  0

In Section 9.11, we used the following result for the
C.I. called “Large sample C.I. for p1-p2” (first C.I. in
that section):

                                                

P   z / 2 
            
P1  P2  p1  p2 
ˆ ˆ                          
 z / 2   1  
               p1(1  p1) p2 (1  p2 )          
                                               
                  n1          n2                

Thus, the test statistic to be used here replaces the
random variables with their observed values for what
is in the middle of the inequality above:

z
p1  p2   p1  p2 
ˆ ˆ
p1(1  p1) p2 (1  p2 )

n1          n2

Notice that we still need to substitute the
hypothesized parameter values for p1-p2. Since the
null hypothesis always include a p1-p2=0, this means
 2005 Christopher R. Bilder
10.59
p1=p2! Call this common value of proportions, pc
(the book just calls it p, but this could be confused
with p in Section 10.11). The test statistic then
becomes:

z
p1  p2   0
ˆ ˆ

p1  p2   0
ˆ ˆ
pc (1  pc ) pc (1  pc )                                 1 1
                                pc (1  pc )   
n1           n2                                        n1 n2 

where 0 was substituted for p1-p2 in the numerator
and pc was substituted for p1 and p2 in the
denominator.

Now, we need to estimate pc. Since p1=p2 under Ho,
we could say that this means both populations are
the same (since this is the only parameter for both).
Thus, we could then “pool” the results from both
samples to form an estimator for pc:

x1  x2
pc 
ˆ
n1  n2

The final test statistic becomes,

z
p1  p2   0
ˆ ˆ

ˆ 
1 1
pc (1  pc )   
ˆ
 n1 n2 
 2005 Christopher R. Bilder
10.60

Question: What are the critical values?

P-value: 2P(Z>|z|) for a two-tail test.

Example: Larry Bird (bird_ch9.xls)

Hypothesis test using the test statistic method:
1) Ho:p1-p2 = 0
Ha:p1-p2  0
x1  x2 251  48
2) Note that pc 
ˆ                       0.8846 . Then
n1  n2 285  53
z
p1  p2   0
ˆ ˆ

0.8807  0.9057  0

ˆ 
1 1                          1     1 
pc (1  pc )   
ˆ                       0.8846(1  0.8846)         
 n1 n2                        285 53 
 0.5222

3) z0.025 = 1.96
4)

Since -1.96 < -0.5222 < 1.96, do not reject Ho.

 2005 Christopher R. Bilder
10.61
5) There is not sufficient evidence to conclude that the
outcome on the first throw has an effect on the
outcome of the second free throw.

Hypothesis test using the p-value method:
1) Ho:p1-p2 = 0
Ha:p1-p2  0
2) 2P(Z > |-0.5222|) = 0.6015
3)  = 0.05
4) Since 0.6015 > 0.05, do not reject Ho.
5) There is not sufficient evidence to conclude that the
outcome on the first throw has an effect on the
outcome of the second free throw.

Below are the results from bird_ch9.xls.

 2005 Christopher R. Bilder
10.62

 2005 Christopher R. Bilder
10.63

 2005 Christopher R. Bilder
10.64
Questions:
 Are first and second free throw attempts independent?
 Suppose the purpose of the problem was changed to:

Perform a hypothesis test to determine if the
probability of success for the second free throw
attempt decreases if the first throw is missed rather

Note that this is what most basketball fans probably
think. How would you perform the test?

1) Ho:p1-p2  0
Ha:p1-p2 > 0

Notice that in Ha, p1>p2. Thus, probability of
success on the second free throws given first one
is made is greater than the probability of success
on the second free throws given first one is missed.
If we reject Ho to conclude Ha is true, the probability
of this being incorrect (type I error) is .

2) z = -0.5222
3) +z0.05 = +1.645
4)

Since -0.5222 < 1.645, do not reject Ho
 2005 Christopher R. Bilder
10.65
5)There is not sufficient evidence to conclude that the
probability of success on the second throw
decreases when the first free throw is missed

The p-value for the above test would be
P(Z > -0.5222) = 0.6992 which is found with the Excel
function of 1-NORMDIST(-0.5222,0,1,TRUE).

 2005 Christopher R. Bilder
10.66
10.13: One- and Two-Sample Tests Concerning
Variances

These lecture notes are available on the schedule web
page of the course website.

10.14: Goodness-of-Fit Test

These lecture notes are available on the schedule web
page of the course website. Note that the confidence
interval method is not available for this hypothesis test!

 2005 Christopher R. Bilder
10.67
Back to Sections 10.1-10.7

One-tail tests

The hypothesis tests done so far have been of the form:

Ho:=o vs. Ha:o

where o is just a number like 11.6cm.

In order to reject Ho, the test statistic is too big OR too
small (i.e. there are two rejection regions). These kinds of
hypothesis tests are called two-tail tests because the
rejection region falls in two tails of the PDF.

Now we are going to discuss ONE-TAIL Tests:
Test      Name
Ho:o
Left-tail
Ha:<o
Ho:o
Right-tail
Ha:>o

 To reject Ho for left-tail tests, the test statistic must be
<, i.e. on the left side of the PDF.
 To reject Ho for right-tail tests, the test statistic must be
>, i.e. on the right side of the PDF.

 2005 Christopher R. Bilder
10.68
1) The Confidence Interval Method - 4 Steps

1. State Ho and Ha
2. Find the “one-sided” C.I. for 

Test        Name            (1-)100% C.I.
Ho:o                                   s
Left-tail          x  t,n1
Ha:<o                                    n
Ho:o                                   s
Right-tail         x  t,n1
Ha:>o                                    n

For example, the C.I. for the left-tail test gives an
Comment [b4]: What happens if reversed the
interval such as (-,2). Therefore, we have an              interval? One could never reject! That's why
the interval is in this direction.
upper bound on the value of . If Ho:3 and
Ha:<3, then the C.I. says that  is less than 3.

3. Reject or do not reject Ho – Check if the hypothesized
value of  is inside the interval.
4. Conclusion – Describe what 3. means in terms of the
original problem

Example: Tire life

A consumer group is concerned about a manufacturer's
claim that their tires last on average a least 22,000
miles. A sample of 100 tires are taken and the number
of miles each lasted is recorded. The sample mean was
 2005 Christopher R. Bilder
10.69
21,819 miles and the sample standard deviation was
1,295 miles. Perform a hypothesis test to see if there is
evidence to disprove the manufacturer's claim using a
type I error rate of 0.01.

1. Ho:22,000
Ha:<22,000

In order to disprove the claim, <22,000 needs to be
in Ha. Then


= P(reject Ho | Ho is true)
= P(sample says <22,000 | 22,000)
= 0.01.

Thus, I am controlling the probability of making this
type of error!

2. Find the “one-sided” C.I. for 
s = 21,819 + 2.3641,295/ 100
x  t,n1
n

where TINV(0.012, 99) is used to find the critical
value. Notice that 0.01 is NOT used in the TINV()
function. See Chapter 9’s discussion on the use of
this function.

Therefore, the 99% confidence interval is
 2005 Christopher R. Bilder
10.70
- <  < 22,125.14

3. Since 22,000 is in the C.I., do not reject Ho.
4. There is not sufficient evidence to disprove the
manufacturer’s tire life claim.

2) The Test Statistic Method - 5 Steps

1.State Ho and Ha
2.Find the test statistic
3.Find the critical value
Test        Name           Critical Value
Ho:o
Left-tail                -t, n-1
Ha:<o
Ho:o Right-tail
+t, n-1
Ha:>o

4. Reject or do not reject Ho

Left-tail:

 2005 Christopher R. Bilder
10.71

Right – tail:

Write reject or don’t reject Ho and provide a reason.

5. Conclusion

Example: Tire life

1.Ho:22,000
Ha:<22,000
x  0 21,819  22,000
2. t                             1.4
s/ n         1,295 / 100
3.-t, n-1 = -t0.01, 99 = -2.364
4.

 2005 Christopher R. Bilder
10.72

Since -1.4 > -2.364 don't reject Ho.

5. There is not sufficient evidence to disprove the
manufacturer’s tire life claim.

3) The P-value Method - 5 Steps

1.State Ho and Ha
2.Find the p-value
a)Compute test statistic
b)Find the p-value

Test             Name                   p-value
Ho:o
Ha:<o         Left-tail                P(T<t)
Ho:o
Ha:>o        Right-tail                P(T>t)
 2005 Christopher R. Bilder
10.73

For right-tail (left-tail) tests, this gives the
probability of finding a value of t at least this great
(small) assuming Ho is true.

Note: These are one-tail tests so only the probability
for one-tail is needed.

3.State 
4.Reject or do not reject Ho

Reject Ho if p-value <  and do not reject if p-value  

Example of don’t reject for right-tail test:

5. Conclusion

 2005 Christopher R. Bilder
10.74
Example: Tire life

1.Ho:22,000
Ha:<22,000
2.P(T < -1.4) = 0.0823

To find this value in Excel, use the symmetry property of
the t-distribution because Excel will not find t-distribution
probabilities associated with negative values of the test
statistic. The function used is TDIST(1.4, 99, 1).

Remember that the p-value is found through integration:
1.4    99  1 / 2  
 (99 1) / 2
 1 t 
2

                        99                 dt = 0.0823
   99 / 2  99           

> f(t):=GAMMA((nu+1)/2)/(GAMMA(nu/2) *
sqrt(Pi*nu)) * (1+t^2/nu)^(-(nu+1)/2);
(  1/2  1/2 )
 1  1   1 t 
2

2                   
        2 
        
f( t ) :=
    
1 

2 
     

> int(eval(f(t),nu=99),t=-infinity..-1.4);
.08231967967
3. = 0.01
4.

 2005 Christopher R. Bilder
10.75

0
Since 0.0823 > 0.01, don't reject Ho.

5.There is not sufficient evidence to disprove the
manufacturer’s tire life claim.

Note: P-value interpretation: If  is really  22,000 in the
population, then a test statistic value, t, no smaller than
what this was observed would occur about 8.23% of time if
the hypothesis test process (take a new sample and
perform a new hypothesis test) is repeated a large number
of times. Thus, it may occur about 8 times out of 100.
This is borderline with regard to it being a likely event, and
it is why the p-value is close to the level of significance,
=0.01. Often, people will say this is “marginal evidence”
against Ho.

Example: Volleyball quality control (hyp_volleyball_data.xls)

Be VERY careful with the Excel calculation here!
 2005 Christopher R. Bilder
10.76
Suppose Ho:11.6 vs. Ha:<11.6 is being tested with
=0.05. The same Excel output as before would be
produced using the t-test Paired Two Sample for Means
window.

Mean                                       11.5000               11.6
Variance                                    1.0000                  0
Observations                                    36                 36
Pearson Correlation                  #DIV/0!
Hypothesized Mean Difference                     0
df                                              35
t Stat                                     -0.6001
P(T<=t) one-tail                            0.2762
t Critical one-tail                         1.6896
P(T<=t) two-tail                            0.5523
t Critical two-tail                         2.0301

The “one-tail test” p-value is given by Excel to be 0.2762
and the critical value given is 1.6896. How does Excel
know if we want a left-tail or right-tail test? We never
specified it!

This is what Excel does:
 Excel will always give a positive critical value. You
need to realize that it is a left-tail test and a negative
critical value is needed.
 Excel calculates the p-value as P(T<t) if t<0 and
P(T>t) if t>0. In this case, Excel calculates the p-
value correctly for this left-tail test. However, if the
hypotheses were switched to Ho:11.6 vs.
 2005 Christopher R. Bilder
10.77
Ha:>11.6, Excel would calculate the p-value still as
0.2762. The correct p-value would be 1-0.2762 =
0.7238 since we would want P(T>-0.6001) = 1-P(T<-
0.6001).
 Be careful on relying too much on the Excel output!!!

See the file, hyp_1sample_pic.xls, for how some of the
one-tail calculations can be done in Excel without the Data
Analysis tool.

 2005 Christopher R. Bilder
10.78

 2005 Christopher R. Bilder
10.79

As mentioned before, all values in red can be changed by
the user to see the effect on the test statistic, critical value,
and the hypothesis test decision. Make changes on your
own so that you familiarize yourself with what happens if
the sample size increases, standard deviation changes,…

Example: Cavaliers – this is an old test problem

Chevrolet has been advertising a 3-year, 36,000-mile
warranty for its Cavaliers. The warranty covers the
engine, transmission, and drive train for all new Cavaliers
up to 3 years or 36,000 miles, whichever comes first. One
Chevrolet dealer believes the drivers tend to reach 36,000
miles before 3-years of ownership. The dealer takes a
random sample of 32 Cavalier owners producing the
following statistics on number of miles driven after 3 years:
x =39,900 and s=1,866.

1. State the Type I and II errors for the hypotheses below.

Ho: 36,000
Ha:> 36,000

Type I: Reject Ho, but Ho is true.

Reject  36,000, but  really is  36,000.

 2005 Christopher R. Bilder
10.80
The sample leads you to believe that the average miles
driven is greater than 36,000 (i.e. 36,000 is incorrect),
but in actuality the average miles driven is less than or
equal to 36,000.

Notes:
 The probability of this happening is set at a level of
.
 The probability of correctly rejecting Ho is 1-.

Type II: Do not reject Ho, but Ha is really true.

Do not reject   36,000, but  really is > 36,000.

The sample does not give you enough evidence to
conclude that the average miles driven is greater than
36,000, but the average miles driven really is greater
than 36,000.

Notes:
 The probability of this happening is . This
probability is not controlled. Thus given that  really
is > 36,000, the probability of committing this Type II
error could be 0, 0.1, 0.2, 0.99, …, or 1.
 The probability of correctly concluding do not reject
Ho when Ho is really true is 1-.

Notice how specifying  and not specifying  controls
what goes into Ho and Ha!
 2005 Christopher R. Bilder
10.81

2. Perform a hypothesis test at the significance level of
(=) 0.01 using the test statistic or p-value method.

Test statistic method:
1. Ho: 36,000
Ha: >36,000
39,900  36,000
2. t                    11.8230
1,866 / 32
3. t0.01, 31=2.453
4. Since 11.8230>2.453 reject Ho.
5. There is sufficient evidence to show that the average
miles driven in 3-years is greater than 36,000 miles.

Make sure you can draw a picture of the t-
distribution for this example!

P-value method:
2. Ho: 36,000
Ha: >36,000
39,900  36,000
3. t                    11.8230
1,866 / 32

p-value = P(T>t) = P(T>11.8230)=2.544x10-13

P-value interpretation: If  is really  36,000 in the
population, then a test statistic value, t, at least this
 2005 Christopher R. Bilder
10.82
large (11.82) would occur 0.0000000002544% of the
time if the hypothesis test process (take a new
sample and perform a new hypothesis test) is
repeated a large number of times. This is VERY,
VERY unlikely! Therefore, most likely  really is
NOT  36,000.

4. =0.01
5. Since 2.544x10-13 < 0.01, reject Ho.
6. There is sufficient evidence to show that the average
miles driven in 3-years is greater than 36,000 miles.

C.I. method: Do on your own!

3. From the customers’ viewpoints, should they expect to
have their Cavaliers to be under warranty for 3 years?
Explain.

No, since the average number of miles driven is greater
than 36,000. Thus, their warranty will expire before 3-
years of ownership is reached on average.

Note: Students usually say deciding what goes into Ho and
Ha is the toughest part of hypothesis testing! Note that the
equality part always leads to what is in Ho. Also, let the
control of a type I error guide you to what goes in Ho or Ha.

 2005 Christopher R. Bilder
10.83
From Section 10.8

Example: CPT (cpt.xls in Chapter 9)

Suppose the problem asked you to determine if the
average hits decrease. If this happened, it could mean:
o drug causes drowsiness
o drug causes blurred vision
o some other effect

In this case, we would have a one-tail test.

Test statistic method:
1) Ho:D  0
Ha:D > 0; average hits go down over time

Notice that the purpose of the problem was to
determine if the hits decrease. The only way we can
determine this is to put it in Ha. Think about what the
possible conclusions could be:

 Reject Ho: There is sufficient evidence to indicate
the average hits decrease over the two time
periods. The probability of making a type I error
here (reject Ho, but Ho is true) is  = 0.01.
 Do not reject Ho: There is not sufficient evidence
to indicate the average hits have decreased.
Notice that does NOT say 1 - 2  0.

 2005 Christopher R. Bilder
10.84
d       2.8
2) t                                 6.7254
sD  n 1.3166 / 10
3) +t0.01,9 = +2.8214
4)

Reject Ho since 6.7254 > 2.8214.
5) There is sufficient evidence to indicate the average
hits decrease over the two time periods.

P-value method:
1) Ho:D  0
Ha:D > 0
2) P(T > 6.7254) = 4.3  10-5
3)  = 0.01
4) Reject Ho since 4.3  10-5 < 0.01
6) There is sufficient evidence to indicate the average
hits decrease over the two time periods.

 2005 Christopher R. Bilder
10.85
From Section 10.11-10.12

testing procedures in these sections for one-tail tests?

 2005 Christopher R. Bilder
10.86
Summary of hypothesis testing steps

For the test statistic method:

1) State Ho and Ha
2) Calculate the test statistic
3) State the critical value
4) Decide whether or not to reject Ho
5) State a conclusion in terms of the problem

For the p-value method:

1) State Ho and Ha
2) Calculate the p-value
3) State 
4) Decide whether or not to reject Ho
5) State a conclusion in terms of the problem

For the C.I. method:

1) State Ho and Ha
2) Calculate the C.I.
3) Decide whether or not to reject Ho.
4) State a conclusion in terms of the problem

 2005 Christopher R. Bilder

```
To top