Test_of_Hyp

Testing of HypothesisPROCEDURE FOR SOLUTION

Rohit Vishal KumarFor Circulation to Information Management (Second Semester) Students

Xavier Institute of Social Service

Ranchi, Jharkhand, India

February 8, 2007

1 Steps for Solution

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

2. Alternative Hypothesis: Set up the Alternative hypothesis H1 from the problemprovided. The alternate hypothesis H1 can either be of Not Equal (6=), 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 tobe decided before the sample is drawn — as because it is a key determininant ofthe sample size to be selected.

Specify the level of significance at which the testing needs to be done. The levelof significance is usually provided in the problem. If the level of significance isnot specified it is preferable to use 95% level rather than 99% because the nullhypothesis 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 acceptedat 99% level may or may not be accepted at 95% level. It is wiser to err on theside of caution.

4. Compute the Test Statistic: Specify the test statistic to be followed for testing.Calculate the value of the test statistic using the problem given in the data. Thisvalue is also termed as the observed value or the calculated value Fcalc.

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

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6. Compare and Conclude: Compare Fcalc and Ftab. If Fcalc ≤ Ftab then accept H0

else reject H0.

Based on the result of the comparison write down your conclusion. The conclusionshould 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

Z =

√n(X − µ0)

σ0

where n = sample sizeX = sample mean

Sampling distribution : Standard Normal

CASE B: Population standard deviation is not known but sample is large (> 30)

Z =

√n(X − µ0)

swhere n = sample size

s = sample standard deviation


CASE C: Population standard deviation is not known but sample is small (≤ 30)

t =

√n(X − µ0)

σ0

where n = sample sizeX = 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 areequal to σ1 and σ2 respectively

Z =X1 −X2√

σ21

n1+

σ22

n2

where n1 n2 = sample sizesX1 X1 = sample means

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CASE B: Population are independent and their standard deviation are not known andthe sample is large (n1, n2 > 30) respectively

Z =X1 −X2√

S21

n1+

S22

n2

where n1 n2 = sample sizesX1 X1 = sample meansS2

1 S22 = sample standard deviations


CASE C: Population are independent and their standard deviation are unknown butequal; and the sample is small (n1, n2 ≤ 30) respectively

t =X1 −X2

S√

1n1

+ 1n2

where n1 n2 = sample sizesX1 X1 = sample means

S2 =(n1 − 1)S2

1 + (n2 − 1)S22

n1 + n2 − 2(Pooled standard deviation)

Sampling distribution : t distribution with (n1 + n2 − 2) degrees of freedom. Thisdistribution is also known as “Fischer’s Non Paired T”.

CASE D: Population are correlated and the sample is small (n1, n2 ≤ 30) respectively

t =

√n U

SU

where n1 , n2 = sample sizesU = (X1 −X2)and U and SU

are calculated w.r.t U

Sampling distribution : t distribution with (n−1) degrees of freedom. This distributionis 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

χ2 =

∑ni=1(xi − µo)

2

σ20

where n = sample size

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Sampling distribution : χ2 distribution with n degrees of freedom

CASE B: Population mean is unknown but the sample is large (N > 30)

Z =s− σ0

σ0/√

(2n)

where n = sample sizewhere s = sample standard deviation

Sampling distribution : Standard Normal distribution

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

χ2 =(n− 1)s2

σ0

=

∑ni=1(Xi −X

2)

σ0

where X = sample meanwhere 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)

Z =S1 − S2

S(√

12n1− 1

2n2)

where S = standard deviation in theFischer’s T distribution


CASE B: Population are independent and the sample sizes are small(n1, n2 ≤ 30)

F =S2

1

S22

where S1 S2 = are standard deviation of therespective 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)

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Z =p− p0√p0(1−p0)

n

where n = sample sizewhere p = sample proportion


2.6 Population proportions of two populations are equal (p0 = p1)

Null Hypothesis (H0) : (p0 = p1)


Z =p1 − p2√

p(1− p)( 1n1

+ 1n2

)

where n1, n2 = sample sizeswhere p1, p2 = sample proportions

p =(n1p1 + n2p2)

(n1 + n2)


3 Things to Remember

• All the Greek symbols always stand for parameters of population whereas all thenon-Greek symbols always stand for the parameters of the sample.

• Sample standard deviation is calculated as follows:

S =

√√√√ 1

(n− 1)

n∑i=1

(xi − x)2

• Alternative hypothesis of 6= type are always two tailed tests. In such a case bothsides of the distribution needs to be taken into account.

• Alternative hypothesis of < or > are always one tailed test. In such a case onlyone side of the distribution is taken into account.

• Critical region or region of acceptance will be generally given in the questionpaper. However critical regions of Z test may not be provided in the questionpaper. The following values of Z distribution should always be remembered:

Z Values Two Tailed One Tailed5% Level 1.960 1.6451% Level 2.576 2.362

5% Level of significance = 95% Confidence Limit1% Level of significance = 99% Confidence Limit

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• If Fcalc value comes out as negative (-ve), then the negative sign is to be ignoredand 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 themin-house. They maintain high quality standards; BeeTEL will not consider buying a TVtube unless convinced that the average life expectancy of the tube is more than 500hours. Elektra Tubes Ltd. - a potential supplier - has supplied 9 tubes for testing. Thetests provided the following data: Mean life of the tube = 600 hours and variance2500 hours. Consider yourself to be in charge of recommending tube suppliers to thetop management. Based on the performance results of Elektra Tubes as a supplier ofpicture tubes to BeeTEL?

Answer:Null Hypothesis H0: (µ = 500)Alternative Hypothesis H0: (µ > 500)

Data provided:Sample size = n = 9Mean = µ = 600 hoursVariance = 2500 hours⇒ standard deviation =

√2500 = 50 hours

Test statistic to be used:

t =

√n(X − µ0)

σ0

Sampling distribution : t distribution with (n-1) degrees of freedom.

Calculations:

t =

√9(600− 500)

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 ofTV tubes supplied by Elektra Tubes Ltd. is greater than 500 hours. Therefore we mayrecommend 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 arewheat 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

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Qn 2. Before an increase in excise duty on cigarettes, 800 persons out of a sample of1000 were found to be smokers. After the increase in excise duties, 800 persons werefound to be smokers in a sample of 1200 people. Would you conclude that there hasbeen a significant reduction in smoking habits of the people after the increase in exciseduty? Hint: See Section 2.6, Zcalc = 6.842

Qn 3. An insurance agent claims that the average age of policy holders who insurethrough him is less than the average for all agents — which is 30.5 years. A randomsample of 100 policy holders who had insured through him gave the following agedistribution. Calculate the arithmetic mean and standard deviation of this distributionand use these values to test his claim at 5% level of significance. You are given thatZ(1.645) = 0.95 Hint: See Section 2.1, Zcalc = -2.681

Age Last Birthday No of persons16 – 20 1221 – 25 2226 – 30 2031 – 35 3036 – 40 16

Qn 4. In a survey of buying habits, 400 women shoppers are chosen at random ina super market ‘A’ located in a certain section of the city. Their average weekly foodexpenditure is Rs. 250 with a standard deviation of Rs. 40. For 400 women shopperschosen at random in super market ‘B’ the average weekly food expenditure is Rs. 220with a standard deviation of Rs. 55. Test at 1% level of significance whether the averageweekly food expenditure of the two population are equal? Would your conclusionchange 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 in10 successive measurement with the same micrometer. Is one engineer more consistentthan the other? Hint: the lower the dispersion; the better the consistency. Refer Section2.4. Fcalc = 2.4

Engineer A Engineer B503 502505 497497 492505 498495 499502 495499 497493 496510 498501 –

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

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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 workersin two cities gave the following data. Test at 5% level the equality of variance of thewages distributed in the cities. Given: F12,15 ≥ 0.95 = 0.025.

No. of Workers S.D. of WagesCity Sampled in the sampleAlipurduar 16 Rs. 25.00Bankura 13 Rs. 32.00

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

This document can be obtained from:

Rohit Vishal KumarReader

Department of MarketingXavier Institute of Social Service

P.O. Box No. 7, Purulia RoadRanchi – 834 001, Jharkhand, IndiaPhone: (91-651) 2200-873 Ext. 308

Email: [email protected] Print on: February 8, 2007

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