Sem-ii Testing Of Hypothesis

Testing of Hypothesis P ROCEDURE

FOR

S OLUTION

Rohit Vishal Kumar For 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 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 (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 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 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 Fcalc . 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 Ftab . In the case of examinations - either the Ftab 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 Fcalc and Ftab . If Fcalc ≤ Ftab then accept H0 else reject H0 . 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 2.1

Common Test Statistics 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 = r σ12 σ22 + 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 = r S22 S12 + n1 n2 where n1 n2 = sample sizes X 1 X 1 = sample means S12 S22 = 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 t =

X1 − X2 S

q

1 n1

+

1 n2

where n1 n2 = sample sizes X 1 X 1 = sample means (n1 − 1)S12 + (n2 − 1)S22 2 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 χ

2

where n

Pn

i=1 (xi − σ02

=

µo )2

= 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) Z =

s − σ0 q

σ0 / (2n) where n where s

= sample size = sample standard deviation

Sampling distribution : Standard Normal distribution CASE C: Population mean is unknown but the sample is small (N ≤ 30) (n − 1)s2 σ0 Pn 2 i=1 (Xi − X ) = σ0 = sample mean = sample standard deviation

χ2 =

where X where s

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 = where S

S1 − S2 q

S(

1 2n1

−

1 ) 2n2

= 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) S12 S22 = are standard deviation of the respective populations

F = where S1 S2

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 = q

p0 (1−p0 ) n

where n where p

= sample size = 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 = q

p(1 − p)( n11 +

1 ) n2

where n1 , n2 where p1 , p2

= sample sizes = 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: S

v u u = t

n X 1 (xi − x)2 (n − 1) i=1

• Alternative hypothesis of 6= 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 5% Level 1% Level

Two Tailed 1.960 2.576

One Tailed 1.645 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 purchasing 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?

Answer: 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:

√ 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 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 16 – 20 21 – 25 26 – 30 31 – 35 36 – 40

No of persons 12 22 20 30 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 503 505 497 505 495 502 499 493 510 501

Engineer B 502 497 492 498 499 495 497 496 498 –

Qn 6. A certain stimulus administered to each of the 12 patients resulted in the following 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.

City Alipurduar Bankura

No. of Workers Sampled 16 13

S.D. of Wages in the sample Rs. 25.00 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: Rohit Vishal Kumar Reader 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: [email protected] Final Print on: February 8, 2007

8