# Testing of Hypothesis

Document Sample

```					                           Testing of Hypothesis
P ROCEDURE   FOR   S OLUTION

Dr. Rohit Vishal Kumar
For Marketing / Financial Management (2nd Trimester) Students
Xavier Institute of Social Service
Ranchi, Jharkhand, India

November 11, 2008

1     Steps for Solution
1. Null Hypothesis: Set up the Null hypothesis H0 . It must be noted that the null
hypothesis H0 is always of ‘=’ type.

2. Alternative Hypothesis: Set up the Alternative hypothesis H1 from the problem
provided. The alternate hypothesis H1 can either be of Not Equal (=), Less than
(<) or Greater than (>) type.

3. Determine Level of Signiﬁcance: Choose the appropriate level of signiﬁcance
(α) depending on the reliability of the estimates and permissible risk. This is to
be decided before the sample is drawn — as because it is a key determininant of
the sample size to be selected.
Specify the level of signiﬁcance at which the testing needs to be done. The level
of signiﬁcance is usually provided in the problem. If the level of signiﬁcance is
not speciﬁed it is preferable to use 95% level rather than 99% because the null
hypothesis that is accepted at 95% level will necessarily be accepted at 99% level.
The alternative may not necessarily hold true — a null hypothesis that is accepted
at 99% level may or may not be accepted at 95% level. It is wiser to err on the
side of caution.

4. Compute the Test Statistic: Specify the test statistic (say θ) to be followed for
testing. Calculate the value of the test statistic using the problem given in the
data. This value is also termed as the observed value or the calculated value θcalc .

5. Find the Tabulated Value of the Statistic: Find out the value of test statistic
from the distribution table provided. This value is also termed as tabulated value
or θtab . In the case of examinations - either the θtab will be provided with the
problem and / or you will be allowed to consult the tables. However the values of
Z statistic at 99% and 95% level may not always be provided. You should always
learn the value(s) of the Z statistic by heart.

1
6. Compare and Conclude: Compare θcalc and θtab . If |θcalc | < |θtab | then accept H0
else reject H0 and accept H1 .
Based on the result of the comparison write down your conclusion. The conclusion
should always be written in probabilistic terms and NEVER in deterministic terms.

2     Common Test Statistics
2.1     Mean of the population has a speciﬁed value µ0
Null Hypothesis (H0 ) : (µ = µ0 )

CASE A: Population standard deviation is known and is equal to σ0
√
n(X − µ0 )
Z =
σ0
where n = sample size
X = sample mean
Sampling distribution : Standard Normal

CASE B: Population standard deviation is not known but sample is large (> 30)
√
n(X − µ0 )
Z =
s
where n = sample size
s = sample standard deviation
Sampling distribution : Standard Normal

CASE C: Population standard deviation is not known but sample is small (≤ 30)
√
n(X − µ0 )
t =
σ0
where n = sample size
X = sample mean
Sampling distribution : t distribution with (n-1) degrees of freedom. Also known as
“Students T”.

2.2     Mean of the two population are equal
Null Hypothesis (H0 ) : (µ1 = µ2 )

CASE A: Population are independent and their standard deviation are known and are
equal to σ1 and σ2 respectively
X1 − X2
Z =
2
σ1        2
σ2
n1
+   n2

where n1 n2 = sample sizes
X 1 X 1 = sample means

2
Sampling distribution : Standard Normal

CASE B: Population are independent and their standard deviation are not known and
the sample is large (n1 , n2 > 30) respectively
X1 − X2
Z =
2
S1        2
S2
n1
+   n2

where n1 n2 = sample sizes
X 1 X 1 = sample means
2 2
S1 S2 = sample standard deviations
Sampling distribution : Standard Normal

CASE C: Population are independent and their standard deviation are unknown but
equal; and the sample is small (n1 , n2 ≤ 30) respectively
X1 − X2
t =           1        1
S    n1
+   n2
where n1 n2 = sample sizes
X 1 X 1 = sample means
2            2
2   (n1 − 1)S1 + (n2 − 1)S2
S =
n1 + n2 − 2
(Pooled standard deviation)
Sampling distribution : t distribution with (n1 + n2 − 2) degrees of freedom. This
distribution is also known as “Fischer’s Non Paired T”.

CASE D: Population are correlated and the sample is small (n1 , n2 ≤ 30) respectively
√
nU
t =
SU
where n1 , n2 = sample sizes
U = (X1 − X2 )and U and SU
are calculated w.r.t U
Sampling distribution : t distribution with (n − 1) degrees of freedom. This distribution
is also known as “Fischer’s Paired T”.

2.3   Standard deviation of a population has a speciﬁed value σ0
Null Hypothesis (H0 ) : (σ = σ0 )

CASE A: Population mean is known and is equal to µ0
n
2             i=1 (xi −   µo )2
χ       =                2
σ0
where n         = sample size

3
Sampling distribution : χ2 distribution with n degrees of freedom

CASE B: Population mean is unknown but the sample is large (N > 30)
s − σ0
Z =
σ0 / (2n)
where n      = sample size
where s      = sample standard deviation

Sampling distribution : Standard Normal distribution

CASE C: Population mean is unknown but the sample is small (N ≤ 30)

(n − 1)s2
χ2 =
σ0
n          2
i=1 (Xi − X )
=
σ0
where X       = sample mean
where s      = sample standard deviation

Sampling distribution : χ2 distribution with (n − 1) degrees of freedom

2.4    Standard deviation of two populations are equal
Null Hypothesis (H0 ) : (σ0 = σ1 )

CASE A: Population are independent and the sample size is large (n1 , n2 > 30)
S1 − S2
Z =             1         1
S(    2n1
−   2n2
)
where S      = standard deviation in the
Fischer’s T distribution

Sampling distribution : Standard Normal distribution

CASE B: Population are independent and the sample sizes are small(n1 , n2 ≤ 30)
2
S1
F =       2
S2
where S1 S2      = are standard deviation of the
respective populations

Sampling distribution : F distribution with (n1 − 1), (n2 − 1) degrees of freedom

2.5    Population proportion of some attribute has a value p0
Null Hypothesis (H0 ) : (p = p0 )

4
CASE A: Population are independent and the sample size is large (n1 , n2 > 30)
p − p0
Z =
p0 (1−p0 )
n
where n      = sample size
where p      = sample proportion
Sampling distribution : Standard Normal distribution

2.6    Population proportions of two populations are equal (p0 = p1 )
Null Hypothesis (H0 ) : (p0 = p1 )

CASE A: Population are independent and the sample size is large (n1 , n2 > 30)
p1 − p2
Z =                  1      1
p(1 − p)( n1 +   n2
)
where n1 , n2  = sample sizes
where p1 , p2  = sample proportions
(n1 p1 + n2 p2 )
p =
(n1 + n2 )
Sampling distribution : Standard Normal distribution

3     Things to Remember
• All the Greek symbols always stand for parameters of population whereas all the
non-Greek symbols always stand for the parameters of the sample.
• Sample standard deviation is calculated as follows:
n
1
S =                    (xi − x)2
(n − 1) i=1

• Alternative hypothesis of = type are always two tailed tests. In such a case both
sides of the distribution needs to be taken into account.
• Alternative hypothesis of < or > are always one tailed test. In such a case only
one side of the distribution is taken into account.
• Critical region or region of acceptance will be generally given in the question
paper. However critical regions of Z test may not be provided in the question
paper. The following values of Z distribution should always be remembered:

Z Values      Two Tailed      One Tailed
5% Level        1.960           1.645
1% Level        2.576           2.362
5% Level of signiﬁcance = 95% Conﬁdence Limit
1% Level of signiﬁcance = 99% Conﬁdence Limit

5
• If Fcalc value comes out as negative (-ve), then the negative sign is to be ignored
and only the absolute value(s) needs to be compared

4    A Solved Example
BeeTEL - a renowned television manufacturing company of India - is considering pur-
chasing picture tubes from independent producers; rather than manufacturing them
in-house. They maintain high quality standards; BeeTEL will not consider buying a TV
tube unless convinced that the average life expectancy of the tube is more than 500
hours. Elektra Tubes Ltd. - a potential supplier - has supplied 9 tubes for testing. The
tests provided the following data: Mean life of the tube = 600 hours and variance
2500 hours. Consider yourself to be in charge of recommending tube suppliers to the
top management. Based on the performance results of Elektra Tubes as a supplier of
picture tubes to BeeTEL?

Null Hypothesis H0 : (µ = 500)
Alternative Hypothesis H0 : (µ > 500)
Data provided:
Sample size = n = 9
Mean = µ = 600 hours
Variance = 2500 hours √
⇒ standard deviation = 2500 = 50 hours
Test statistic to be used:
√
n(X − µ0 )
t =
σ0

Sampling distribution : t distribution with (n-1) degrees of freedom.
Calculations:
√
9(600 − 500)
t =
50
= 6

Ttab at 8 degrees of freedom and 99% Conﬁdence Level = 2.8965
Conclusions: As Tcalc > Ttab we reject H0 and may conclude that the mean life of
TV tubes supplied by Elektra Tubes Ltd. is greater than 500 hours. Therefore we may
recommend Elektra Tubes Ltd. as a potential TV tube supplier for BeeTEL Ltd.

5    Problems for Solution
Qn 1. In a sample of 1000 people in Maharashtra 540 are rice eaters and the rest are
wheat eaters. Can we assume that both rice and wheat are equally popular in the state.
Test the claim at 1% level of signiﬁcance. Hint: See Section 2.5, Zcalc = 2.532

6
Qn 2. Before an increase in excise duty on cigarettes, 800 persons out of a sample of
1000 were found to be smokers. After the increase in excise duties, 800 persons were
found to be smokers in a sample of 1200 people. Would you conclude that there has
been a signiﬁcant reduction in smoking habits of the people after the increase in excise
duty? Hint: See Section 2.6, Zcalc = 6.842
Qn 3. An insurance agent claims that the average age of policy holders who insure
through him is less than the average for all agents — which is 30.5 years. A random
sample of 100 policy holders who had insured through him gave the following age
distribution. Calculate the arithmetic mean and standard deviation of this distribution
and use these values to test his claim at 5% level of signiﬁcance. You are given that
Z(1.645) = 0.95 Hint: See Section 2.1, Zcalc = -2.681

Age Last Birthday      No of persons
16 – 20                12
21 – 25                22
26 – 30                20
31 – 35                30
36 – 40                16

Qn 4. In a survey of buying habits, 400 women shoppers are chosen at random in
a super market ‘A’ located in a certain section of the city. Their average weekly food
expenditure is Rs. 250 with a standard deviation of Rs. 40. For 400 women shoppers
chosen at random in super market ‘B’ the average weekly food expenditure is Rs. 220
with a standard deviation of Rs. 55. Test at 1% level of signiﬁcance whether the average
weekly food expenditure of the two population are equal? Would your conclusion
change if you test the above at 5% level? Hint: See Section 2.2, Zcalc = 8.82
Qn 5. The following are the values in thousands of an inch obtained by two engineers in
10 successive measurement with the same micrometer. Is one engineer more consistent
than the other? Hint: the lower the dispersion; the better the consistency. Refer Section
2.4. Fcalc = 2.4

Engineer A       Engineer B
503              502
505              497
497              492
505              498
495              499
502              495
499              497
493              496
510              498
501               –

Qn 6. A certain stimulus administered to each of the 12 patients resulted in the follow-
ing increase in blood pressure: 5, 2, 8, -1, 3, 0, -2, 1, 5, 0, 4 and 6. Can it be concluded

7
that the stimulus will, in general, be accompanied by an increase in blood pressure?
Hint: Use Paired T. tcalc = 2.89
Qn 7. A survey conducted by an NGO on the daily wages in Rs. of unskilled workers
in two cities gave the following data. Test at 5% level the equality of variance of the
wages distributed in the cities. Given: F12,15 ≥ 0.95 = 0.025.

No. of Workers   S.D. of Wages
City            Sampled       in the sample
Alipurduar         16           Rs. 25.00
Bankura            13           Rs. 32.00

Qn 8. In a year there are 956 births in town A, of which 52.5% were males. In town
A and B combined this proportion in a total of 1406 births was 0.496. Is there any
signiﬁcant difference in the proportion of male births in the two towns? Hint: Zcalc =
3.368

This document can be obtained from:
Dr. Rohit Vishal Kumar
Department of Marketing
Xavier Institute of Social Service
P.O. Box No. 7, Purulia Road
Ranchi – 834 001, Jharkhand, India
Phone: (91-651) 2200-873 Ext. 308
Email: rohitvishalkumar@gmail.com
Final Print on: November 11, 2008

8

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 41 posted: 6/14/2010 language: German pages: 8