Sampling and Sampling Methods 2

1

CASELETS

2. An investigation of the relative merits of two kinds of flash light batteries showed that a random sample 100 batteries of brand A lasted on the average 36.5 hrs with standard deviation of 1.8 hrs, while a random sample of 80 batteries of brand B lasted on the average 36.8 hrs with a standard deviation of 1.5 hrs. Use a Level of significance of 0.05 to test whether the observed difference between the average life time is significant.

BAH :0Null Hypothesis ,i.e., there is no significant difference between the average life of two brandsAlternative Hypothesis

(two tailed)BAH :1

Solution

2

CASELETS(Solution con’t)

nn

BA

nn

xxZ

2

2B

1

2A

2

22

1

21

21

ss

2195.1

8025.2

10024.3

8.365.36Z

Since |Z| =1.2195 is less than the tabular value, i.e., 1.96 ( 5% level of significance). Hence the null hypothesis is accepted and we conclude that there is no significant difference between the average life times of the two brands A and B of the flashlight batteries.

05.0,5.1,8.36,80

8.1,5.36,100

2

1

sBn

sAn

B

A

3

CASELETS

3. The mean height of 50 male students who showed above average participation in college athletics was 68.2 inches with a standard deviation of 2.5 inches; while 50 male students who showed no interest in such participation had a mean height of 67.5 inches with a standard deviation of 2.8 inches. Use a Level of significance of 0.05 test the hypothesis that male students who participate in college athletics are taller than other male students.

BAH :0Null Hypothesis ,i.e., there is no significant difference between the mean heights of the male students who participate and who do not participate.

Alternative Hypothesis (one tailed)

BAH :1

Solution

4


nn

BA

nn

xxZ

2

2B

1

2A

2

22

1

21

21

ss

3188.1

5084.7

5025.6

5.672.68Z

Since |Z| =1.3188 is less than the tabular value, i.e., 1.645 ( 5% level of significance). Hence the null hypothesis is accepted and we conclude that there is no significant difference between the average height of two groups of students.

05.0,8.2,5.67,50

5.2,2.68,50

2

1

sBn

sAn

B

A

5

CASELETS

4. The mean yield of wheat from a district A was 200 lbs. With S.D = 10 lbs per acre from a sample of 100 plots. In other district B, the mean yield was 220 lbs with S.D = 12 lbs from a sample of 150 plots. Assuming that the standard deviation of the yield in the entire state was 11 lbs. Test whether there is any significant difference between the mean yield of crops in the two districts. Use a Level of significance of 0.01.

BAH :0Null Hypothesis ,i.e., the mean yield of crops in two districts do not differ significantly.Alternative Hypothesis

(two tailed)BAH :1

Solution

6


nn

BA

nn

xxZ

2

2

1

2

2

2

1

221

05.7

150121

100121

220210Z

Since |Z| =7.05 is greater than the tabular value, i.e., 2.58 ( 1% level of significance). Hence the null hypothesis is rejected and we conclude that the mean yields of crop in two districts differ significantly.

11

,12,220,150

01.0,10,210,100

2

1

ss

sBn

sAn

BA

B

A

7

CASELETS

5. Given the following information relating to two places, A and B, test whether there is any significant difference between their mean wages:

BAH :0Null Hypothesis ,i.e., the mean yield of crops in two districts do not differ significantly.Alternative Hypothesis

(two tailed)BAH :1

Solution

Mean Wages (Rs.)

Standard Deviation (Rs.)

Number of workers

A

47

28

1000

A

49

40

1500

8


47.1

15001600

1000784

4947Z

Since |Z| =1.47 is less than the tabular value, i.e., 1.96 ( 5% level of significance). Hence the null hypothesis is accepted and we conclude that there is no significant difference in the mean wages at places A and B.

05.0,40,49,1500

,28,47,1000

2

1

sBn

sAn

B

A

nn

BA

nn

xxZ

2

2B

1

2A

2

22

1

21

21

ss

9

Sampling Distributions ofSample Proportions

Sampling Distributions

Sampling Distribution of

Sample Mean

Sampling Distribution of

Sample Proportion

10

Population Proportions, P

P = the proportion of the population having some characteristic

• Sample proportion ( ) provides an estimate of P:

• 0 ≤ ≤ 1

• has a binomial distribution, but can be approximated by a normal distribution when nP(1 – P) ≥ 9

size sample

interest ofstic characteri the having sample the in itemsofnumber

n

XP ˆ

P̂

P̂

P̂

11

Sampling Distribution of P

• Normal approximation:

Properties:

and

(where P = population proportion)

Sampling Distribution

.3

.2

.1 0

0 . 2 .4 .6 8 1

p)P̂E( n

P)P(1

n

XVarσ2

P

ˆ

^

)PP( ˆ

P̂

12

Z-Value for Proportions

nP)P(1

PP̂

σ

PP̂Z

P̂

Standardize to a Z value with the formula:P̂

13

CONFIDENCE INTERVAL FOR A POPULATION PROPORTION

The confidence interval for a population proportion

P̂ZσP̂

where is the standard error of the proportion:P̂σ

n

P)-P(1ZP̂ /2

Or

14

CASELETS

1. A die was thrown 9000 times and of these 3220 yielded a 3 or 4. Can the die be regarded as unbiased?

3/1:0 PHNull Hypothesis ,i.e., the die is unbiased.

Alternative Hypothesis (two tailed)

3/1:1 PH

Solution: we are given that n = 9000. The observed proportion of successes (getting 3 or 4) in 9000 throws of a die is given by

= 3220/9000 = 0.3578

P = population proportion of success

= getting 3 or 4 in throw of an unbiased die = 2/6 = 1/3 =0.333

P̂

15


Since |Z| =4.94 is greater than 3, it is significant (at all levels of significance) . Hence the null hypothesis is rejected and we conclude that the die is certainly biased.

nP)P(1

PP̂

σ

PP̂Z

P̂

94.4

9000)3333.00.3333(1

3333.03578.0Z

16

CASELETS

2. In a random sample of 400 persons from a large population, 120 are females. Can it be said that males and females are in the ratio 5 : 3 in the population? Use a 1% level of significance?

375.0:0 PHNull Hypothesis ,i.e., the males and females in the population are in the ratio 5 : 3.

Alternative Hypothesis (two tailed)

375.0:1 PH

Solution: we are given that n = 9000 and X = No. of females in the sample = 120. The observed proportion of females in the sample is given by

= 120/400 = 0.3

P = Proportion of females in the population = 3/8 = 0.375

P̂

17


Since |Z| =3.125 is greater than the tabular value, i.e., 2.58 ( 1% level of significance). Hence the null hypothesis is rejected and we conclude that the males and females in the population are not in the ration 5 : 3.

nP)P(1

PP̂

σ

PP̂Z

P̂

125.3

400)375.00.375(1

375.03.0Z

18

CASELETS3. A manufacturer claimed that at least 95% of the equipments which he supplied to a factory conformed to specifications. An examination of a sample of 200 pieces of equipment revealed that 18 were faulty. Test this claim at a significance level of 5%.

95.0:0 PHNull Hypothesis ,i.e., Proportion of pieces conforming to specifications in the sample is at least 95%.

Solution: We are given that n = 200 and X = No. of pieces conforming to specifications in the sample = 200-18 = 182.

Proportion of pieces conforming to specifications in the sample

= 182/200 = 0.91

P = Proportion of pieces conforming to specifications in the population = 0.95

P̂

19


Since |Z| =2.6 is greater than the tabular value, i.e., 1.645 ( 5% level of significance). Hence the null hypothesis is rejected and we conclude that the manufacturer claim is rejected.

nP)P(1

PP̂

σ

PP̂Z

P̂

6.2

200)95.00.95(1

95.091.0Z

95.0:1 PHAlternative Hypothesis (Left- tailed)

20

CASELETS4. In a big city 325 men out of 600 men were found to be smokers. Does this information support the conclusion that the majority of men in the city are smokers. Test this claim at a significance level of 5%.

5.0:0 PHNull Hypothesis ,i.e., no. of smokers and non-smokers are equal in the city.

Solution: We are given that n = 600 and X = No. of smokers = 325.

Sample proportion of smokers is given by

= 325/600 = 0.5417

P = Population proportion of smokers is given by

= 0.5 (No. of smokers and non smokers are equal in population)

P̂

21


Since |Z| =2.04 is greater than the tabular value, i.e., 1.645 ( 5% level of significance for single tail). Hence the null hypothesis is rejected and we conclude that the the majority of men in the city are smokers.

nP)P(1

PP̂

σ

PP̂Z

P̂

04.2

600)5.00.5(1

5.05417.0Z

5.0:1 PHAlternative Hypothesis (right- tailed)

22

CASELETS5. In sample 400 parts manufactured by a factory, the number of defective parts was found to be 30. The company, however, claimed that at most 5% of their product is defective. Is the claim tenable?

5.0:0 PHNull Hypothesis ,i.e., proportion of defectives is at most 5%.

Solution: We are given that n = 400 and X = No. of defectives in the sample = 30.

Proportion of defectives in the sample is given by

= 30/400 = 0.075P̂

23


Since |Z| =2.27 is greater than the tabular value, i.e., 1.645 ( 5% level of significance for single tailed). Hence the null hypothesis is rejected and we conclude that the the company's claim of P = 0.05 is not tenable.

nP)P(1

PP̂

σ

PP̂Z

P̂

27.2

400)05.00.05(1

05.0075.0Z

5.0:1 PHAlternative Hypothesis (right- tailed)

24

6. A random sample of 700 units from a large consignment showed that 200 were damaged. Find (i) 95% and (ii) 99% confidence limits for the proportion of damaged units in the consignment.Solution: We are given n = 700.

= proportion of damaged units in the sample

= 200/700 = 0.286

P̂

CASELETS

(i) 95% Confidence Limits for ,i.e., proportion of damaged units in the consignment are given by:

P̂

n

P)-P(1ZP̂ /2

25


)319.0,253.0(

033.0286.0700

0.286)-0.286(196.1286.0

(i) 99% Confidence Limits for ,i.e., proportion of damaged units in the consignment are given by:

P̂

)330.0,242.0(

044.0286.0700

0.286)-0.286(158.2286.0

n

P)-P(1ZP̂ /2

26

Test for Significance of difference of Proportion

Let us consider two independent samples of size n1 and

n2 from the two populations A and B and let X1and X2

be the observed no. of successes (no. of units possessing the given attribute) in these samples respectively. Then = Observed proportion of successes in the

sample from population A = X1/ n1

= Observed proportion of successes in the sample from population B = X2/ n2

P1ˆ

P2ˆ

27

Test for Significance of difference of Proportion(con’t)

Null Hypothesis: H0 : ,i.e., the population proportion are same. In other words H0

is that the sample proportions and do not differ significantly.

Hence H0 : , the test statistic (Z-value) for difference of proportions becomes:

P1ˆ = P2

ˆ = P

P1ˆ P2

ˆ

P1ˆ = P2

ˆ

)11

)(1(

P̂-P̂Z

21

21

nnPP

28


Remark: In general P, the common population proportion is not known and we used its unbiased estimate provided by both the samples taken together which is given by

nn

P̂nP̂nP

21

221 1

29


Confidence Limits for difference of Proportions

n

)P-(1P

n

)P-(1PZ)P̂P̂(

2

22

1

11/221

30

CASELETS1. A Company has the head office at Kolkata and a branch at Mumbai. The personnel director wanted to know if the workers at the two places would like the introduction of a new plan of work and a survey was conducted for this purpose. Out of a sample of 500 workers at Kolkata, 62% favored the new plan. At Mumbai, out of a sample of 400 workers, 415 were against the new plan. Is there any significant difference between the two groups in their attitude towards the new plan at 5% level?Solution: Let and denote the sample proportion of workers favoring the new plan at Kolkata and Mumbai. We are given

P1ˆ P2

ˆ

P22 59.0ˆ 400,

62.0P̂ 500, 11

==

==

n

n

31


Null Hypothesis: We set the null hypothesis that the population proportion of workers favoring the new plan in Kolkata and Mumbai is same, i.e.,

H0:

Alternative Hypothesis H1: (Two- tailed)

P1ˆ = P2

ˆ

P1ˆ P2̂

)11

)(1(

P̂-P̂Z

21

21

nnPP

32


To get the Z-value we first calculate common population proportion using both sample proportion, i.e.,

607.0400500

59.0*40062.0*005

nn

P̂nP̂nP

21

221 1

9155.0

)4001

5001

)(607.01(607.0

0.59-0.62Z

Since |Z| =0.9155 is less than the tabular value, i.e., 1.96 ( 5% level of significance for two tailed). Hence the null hypothesis is accepted and we conclude that the there is no significant difference in the two groups in their attitude.

33

CASELETS2. A machine puts out 16 imperfect articles in a sample of 500. After the machine is overhauled, it puts out 3 imperfect articles in a batch of 100. Has the machine improved? (5% level)Solution: We are given n1 = 500 and n2 = 100.

= proportion of defective in the first sample = 16/500 = 0.032

= proportion of defective in the second sample = 3/100 = 0.03

P1ˆ

P2ˆ

Null Hypothesis: H0: , i.e., there is no significant difference in the machine before overhauling and after overhauling.

Alternative Hypothesis H1: (right- tailed)

P1ˆ P2

ˆ

P1ˆ = P2

ˆ

34



032.0100500

03.0*100032.0*005

nn

P̂nP̂nP

21

221 1

04.1

)100

1500

1)(032.01(032.0

0.03-0.032Z

)11

)(1(

P̂-P̂Z

21

21

nnPP

35


Since |Z| =1.04 is less than the tabular value, i.e., 1.645 ( 5% level of significance for right tailed). Hence the null hypothesis is accepted and we conclude that the machine has not improved after overhauling.

36

CASELETS3. Before an increase in excise duty on tea, 400 people out of a sample of 500 persons were found to be tea drinkers. After an increase in duty, 400 people were tea drinkers in sample of 600 people. Using standard error of proportion, state whether there is a significant decrease in the consumption of tea. Test at 5% and 1% level of significance.Solution: We are given n1 = 500 and n2 = 600.

= Sample proportion of tea drinkers before increase in excise duty

=400/500 = 0.8

= Sample proportion of tea drinkers after increase in excise duty

=400/600 = 0.67

P1ˆ

P2ˆ

37


Null Hypothesis: H0: , i.e., there is no significant difference in the consumption of tea before and after the increase in excise duty.

P1ˆ = P2

ˆ


P1ˆ P2

ˆ


)11

)(1(

P̂-P̂Z

21

21

nnPP

38


11/8600500

67.0*6008.0*005

nn

P̂nP̂nP

21

221 1

81.4

)6001

5001

)(11/81(11/8

0.67-0.80Z

)11

)(1(

P̂-P̂Z

21

21

nnPP

39


Since |Z| =4.81 is greater than the tabular value, i.e., 1.645 ( 5% level of significance for right tailed) and 2.33 ( 1% level of significance for right tailed) . Hence the null hypothesis is rejected and we conclude that there is a significant decrease in the consumption of tea after increase in the excise duty.

40

CASELETS4. The subject under investigation is the measure of dependence of Tamil on words of Sanskrit origin. One newspaper article reporting the proceedings of the constituent assembly contained 2,025 words of which 729 words were declared by literary critic to be Sanskrit origin. A second article by the same author describing atomic research contained 1,600 words of which 640 words were declared by the same critic to be Sanskrit origin. Examine whether there is any significant difference in the dependence of this writer on words of Sanskrit origin in writing the two articles (5%).Solution: We are given n1 = 2025 and n2 = 1600.

= Sample proportion of Sanskrit origin words in first article

=729/2025 = 0.36

= Sample proportion of Sanskrit origin words in second article

=640/1600 = 0.40

P1ˆ

P2ˆ

41


Null Hypothesis: H0: , i.e., there is no significant difference in the proportion of Sanskrit words in the writers vocabulary in the two articles. In other words, Tamil does not depend on the words of Sanskrit origin.

P1ˆ = P2

ˆ

Alternative Hypothesis H1: (two- tailed)

P1ˆ P2̂


)11

)(1(

P̂-P̂Z

21

21

nnPP

42


38.016002025

4.0*160036.0*2025

nn

P̂nP̂nP

21

221 1

469.2

)1600

12025

1)(38.01(38.0

0.40-0.36Z

)11

)(1(

P̂-P̂Z

21

21

nnPP

43


Since |Z| =2.469 is greater than the tabular value, i.e., 1.96 ( 5% level of significance for right tailed). Hence the null hypothesis is rejected and we conclude that the proportion of Sanskrit words in the writer’s two articles differ significantly.

44

Small Samples

• If the sample size is small, then the distributions of standardized statistics are far from normality and consequently normal test (Z-test) cannot be applied. Hence to deal with small samples, we use student’s t-test also known as Exact Sample Test.

45

Student’s t- test

If x1, x2 , …, xn is a random sample of size n from a normal population with mean and variance then student’s t statistic is defined as:

2

n

1iiX

n

1X

n

1ii

22 )X(X1-n

1Swhere and

n

Sμ)X(

t

1-n

sμ)X(

When standard deviation of sample is given

OrWhen standard deviation of sample calculated

46

ntx

2

)2/(1S

n

Student’s t- test( con’t)

Confidence Limits for

47

CASELETS

1. Ten cartoons are taken at random from an automatic filling machine. The mean net weight of the ten cartoons is 11.8 kg. and standard deviation is 0.15 kg. Does this sample mean differ significantly from the intended weight of 12 kg? You are given that v = 9, t0.05 = 2.26

Solution: We are given that n = 10, = 11.8kg, = 0.15kg

sx

Null Hypothesis: H0: , i.e., the sample mean =11.8 kg does not differ significantly from population mean


kg12 xkg12

kg12

48

1-n

sμ)X(

n

Sμ)X(

t

The test statistic is given by

4

1-10

0.15)21(11.8

t

Since |t| =4 is greater than the tabular value of t , i.e., 2.26 (for 9 d.f at 5% level of significance). Hence the null hypothesis is rejected and we conclude that the sample mean differ significantly from the mean .

kg12

(Solution con’t)

CASELETS

49

CASELETS

2. A machine is designed to produce insulating washers for electrical devices of average thickness of 0.025 cm. A random sample of 10 washers was found to have an average thickness of 0.024 cm with a standard deviation of 0.002 cm. Test the significance of the deviation. (Value of t for 9 degree of freedom at 5% level is 2.26)Solution: We are given that n = 10, = 0.024 cm, = 0.002 cm

sx

Null Hypothesis: H0: , i.e., there is no significant deviation between sample mean = 0.024 and population mean


cm025.0x

cm025.0

cm025.0

50

1-n

sμ)X(

n

Sμ)X(

t


5.1

1-10

0.002)025.0(0.024

t

Since |t|=1.5 is less than the tabular value of t , i.e., 1.96 (for 9 d.f at 5% level of significance). Hence the null hypothesis is accepted and we conclude that there is no significant difference the sample and population mean .


51

CASELETS

3. The mean weekly sales of the chocolate bar in candy stores was 146.3 bars per store. After an advertising campaign the mean weekly sales in 22 stores for a typical week increased to 153.7 and showed a standard deviation of 17.2. Was the advertising campaign successful Test the significance of the deviation. (Value of t for 21 degree of freedom at 5% level for single tailed test is 1.721)Solution: We are given that n = 22, = 153.7, = 17.2

sx

Null Hypothesis: H0: , i.e., there is no significant deviation between sample mean = 153.7 and population mean


3.146x

3.146

3.146

52

1-n

sμ)X(

n

Sμ)X(

t


9716.1

1-22

17.2)3.146(153.7

t

Since |t|=1.97 is greater than the tabular value of t , i.e., 1.721 (for 21 d.f at 5% level of significance). Hence the null hypothesis is rejected and we conclude that the advertising campaign was successful in promoting sales. .

(Solution con’t)

CASELETS

53

CASELETS

4. A soap manufacturing company was distributing a particular brand of soap through a large number of retail shops. Before a heavy advertisement campaign, the mean sales per week per shop was 140 dozens. After the campaign, a sample of 26 shops was taken and the mean sales was found to be 147 dozens with a standard deviation 16. Can you consider the advertisement effective. (Value of t for 25 degree of freedom at 5% level for single tailed testis 1.721)Solution: We are given that n = 26, = 147 dozen, = 16 dozen

sx

Null Hypothesis: H0: dozen , i.e., there is no significant deviation between sample mean = 147 and population mean Alternative Hypothesis H1: (right- tailed)

140x

140140

54

1-n

sμ)X(

n

Sμ)X(

t


19.2

1-26

16)140(147

t

Since |t|=2.19 is greater than the tabular value of t , i.e., 1.798 (for 21 d.f at 5% level of significance for single tailed). Hence the null hypothesis is rejected and we conclude that the advertising campaign was effective in promoting sales. .


55

5. Certain pesticides is packed into bags by a machine. A random sample of 10 bags is drawn and their contents are found to weigh (in kg.) as follows:

50, 49, 52, 44, 45, 48, 46, 45, 49, 45,

Test if the average packing can be taken to be 50 kg. (Value of t for 9 degree of freedom at 5% level for two tailed testis 2.262)

CASELETS

Null Hypothesis: H0: , i.e., the average packing is 50 kg.

kg50

Alternative Hypothesis H1: (Two- tailed)

50

Solution:

56

CASELETS

3.4710

473

10

45494546484544524950X

n

1X

n

1ii

12.7

})3.4745()3.4749()3.4745()3.4746()3.4748(

)3.4745()3.4744()3.4752()3.4749()3.4750({1-10

1

)X(X1-n

1S

22222

22222

n

1ii

22

(Solution con’t)

57


n

Sμ)X(

t

2.3712.0

2.7-

10

12.7

)50(47.3t

Since |t|= 3.12 is greater than the tabular value of t , i.e., 2.262 (for 9 d.f at 5% level of significance for two tailed). Hence the null hypothesis is rejected and we conclude that the average packing cannot be taken to be 50 kg.

(Solution con’t)

CASELETS

58

6. A sample of size 9from a normal population gave and . Find a 99% interval for population mean.

CASELETS

99% confidence limits for the population mean are:

Since standard deviation is given, therefore we replace n by n-1 in the denominator in the root

Solution: We are given:

8.15X3.102 S

9,3.10,8.15 2 nSX

ntx

2

)2/(1S

n

1-ntx

2

)2/(1S

n

59


)6136.19,9864.11(

8136.38.151-9

3.1036.38.51

1-9

3.10t8.51 )2/01.0(19

Sampling Distribution for the Difference Between Two Means (small samples)

Independent Sampling

61

Small Samples

• To use the t-distribution, both sampled populations must be approximately normally distributed with equal population variances, and the random samples must be selected independently of each other.

• Since we assume that the two population variances are equal, we can construct a pooled sample estimator of for use in making inferences. 2

62

Let x1, x2, …, xn1 and y1, y2, …, yn2 be two independent random samples from the given normal populations. We set the Null Hypothesis H0 : ,i.e., the samples have been drawn from the normal populations with the same means. In other words, the sample means do not differ significantly.

21

211

)(

1

2

21

nns

xxt

p

Where is the pooled sample estimator for the two populations and it can be calculated as follows.

S P2

t-value the difference Between Two Population Means

63

2

)1()1(

)1()1(

)1()1(

21

222

211

21

222

2112

nn

snsn

nn

snsnsp

n

1ii

11

1X

n

1X

n

1ii

22

2Y

n

1X

Note: Here Degree of freedom for t test is (n1

+ n2 – 2)

t-value the difference Between Two Population Means

n

1ii

2

1

21

1)X(X

1-n

1s

n

1ii

2

2

22

2)X(X

1-n

1s

64

Student’s t- test( con’t)

Confidence Limits for Difference of Means )( 21

2

11)2/()(

1

2221 21 nn

stxx pnn

65

CASELETS

1. The nicotine content in milligrams of two samples of tobacco were found to be as follows:

Sample A: 24 27 26 21 25

Sample B: 27 30 28 31 22 36

Can it be said that two samples come from normal populations having the same mean? (5%)

Null Hypothesis: H0: , i.e., the two samples have been drawn from the normal populations with the same mean.Alternative Hypothesis H1: (two- tailed)

21

21

66

CASELETS( con’t)

6.245

123

)2521262724(5

1X

n

1X

n

1ii

11

1

296

174

)362231283027(6

1X

n

1X

n

1ii

22

2

67

CASELETS( con’t)

3.54

2.21

})6.2425()6.2421()6.2426()6.2427()6.2424({1-5

1

)X(X1-n

1

22222

n

1ii

2

1

21

1

s

6.215

108

})2936()2922()2931()2928()2930()2927({1-6

1

)X(X1-n

1

222222

n

1ii

2

2

22

2

s

68

36.149

2.129

265

6.21*53.5*4

2

)1()1(

)1()1(

)1()1(

21

222

211

21

222

2112

nn

snsn

nn

snsnsp

CASELETS( con’t)

92.1

61

51

36.14

)296.24(

211

)(

1

2

21

nns

xxt

p

Since |t|= 1.92 is less than the tabular value of t , i.e., 2.262 (for 9 d.f at 5% level of significance for two tailed). Hence the null hypothesis is accepted and we conclude that the samples come from normal populations with the same mean.

69

CASELETS

2. A group of 5 patients treated with medicine ‘A’ weigh 42, 39, 48, 60 and 41 kgs.: Second group of 7 patients from the same hospital treated with medicine ‘B’ weigh 38, 42, 56, 64, 68, 69 and 62 kgs. Do you agree with the claim that medicine ‘B’ increases the weight significantly? (The value of t at 5% level of significance for 10 degrees of freedom is 1.81 for one tail)

Null Hypothesis: H0: , i.e., there is no significant difference between the medicines A and B as regards their effect on increase in weight.

Alternative Hypothesis H1: (one- tailed)

21

21

70

CASELETS( con’t)

465

230

)4160483942(5

1X

n

1X

n

1ii

11

1

577

399

)62696864564238(7

1X

n

1X

n

1ii

22

2

71

CASELETS( con’t)

5.724

290

})4641()4660()4648()4639()4642({1-5

1

)X(X1-n

1

22222

n

1ii

2

1

21

1

s

34.1546

926

})5762()5769()5768()5764()5756()5742()5738({1-7

1

)X(X1-n

1

2222222

n

1ii

2

2

22

2

s

72

6.12110

1216

275

34.154*75.72*4

2

)1()1(

)1()1(

)1()1(

21

222

211

21

222

2112

nn

snsn

nn

snsnsp

CASELETS( con’t)

7.1

71

51

6.121

)5746(

211

)(

1

2

21

nns

xxt

p

Since |t|= 1.7 is less than the tabular value of t , i.e., 1.81 (for 10 d.f at 5% level of significance for one tailed). Hence the null hypothesis is accepted and we conclude that the medicines A and B do not differ significantly as regards their effect on increases weight

73

CASELETS

3. A random sample of 20 daily workers of State A was found to have average daily earnings of Rs. 44 with sample variance 900. Another sample of 20 daily workers from State B was found to earn on an average Rs. 30 per day with sample variance 400. Test whether the workers in State A are earning more than those in State B. (5% level of significance)

Null Hypothesis: H0: , i.e., there is no significant difference in the average daily earnings of the workers in States A and B.

Alternative Hypothesis H1: (one- tailed right)

BA

BA

Solution: We are given that

;400,30,20;900,44,20 22 sxnsxn BBBAAA

74

65038

24700

22020

400*19900*19

2

)1()1(

)1()1(

)1()1( 22222

BA

BBAA

BA

BBAAp nn

snsn

nn

snsns

CASELETS( con’t)

738.1

201

201

650

)3044(

211

)(

1

2

21

nns

xxt

p

Since |t|= 1.738 is greater than the tabular value of t , i.e., 1.645 (for 38 d.f at 5% level of significance for one tailed). Hence the null hypothesis is rejected and we conclude that the workers in the State A are earning more than those in State B.

75

CASELETS

4. The means of two random samples of size 9 and 7 are 196.42 and 198.82 respectively. The sum of the squares of the deviations from the mean are 26.94 and 18.73 respectively. Can the samples be considered to have been drawn from the same normal population? (5% level of significance)

Null Hypothesis: H0: , i.e., both the samples have been drawn from the same normal population.Alternative Hypothesis H1: (two-

tailed)

21

21


73.18)(,82.198,7

;94.26)(,42.196,9

22

22

12

11

xxxn

xxxn

76

CASELETS( con’t)

3675.38

94.26

}94.26{1-9

1

)X(X1-n

1 n

1ii

2

1

21

1

s

1217.36

73.18

}73.18{1-7

1

)X(X1-n

1 n

1ii

2

2

22

2

s

77

26.314

67.45

279

1217.3*63675.3*8

2

)1()1(

)1()1(

)1()1(

21

222

211

21

222

2112

nn

snsn

nn

snsnsp

CASELETS( con’t)

64.2

71

91

26.3

)82.19842.196(

211

)(

1

2

21

nns

xxt

p

Since |t|= 2.64 is greater than the tabular value of t , i.e., 2.15 (for 14 d.f at 5% level of significance for two tailed). Hence the null hypothesis is rejected and we conclude that the samples cannot be considered to have come from the same normal population.

78

CASELETS

5. The average number of articles produced by two machines per day are 200 and 250 with standard deviations 20 and 25 respectively on the basis of records of 25 days’ production. Can you regard both the machines equally efficient at 15 level of significance?

Null Hypothesis: H0: , i.e., both the machines are equally efficient


21

21


;25,250,25;20,200,25 222111 sxnsxn

79

5.51248

24600

22525

625*24400*24

2

)1()1(

)1()1(

)1()1(

21

222

211

21

222

2112

nn

snsn

nn

snsnsp

CASELETS( con’t)

809.7

251

251

5.512

)250200(

211

)(

1

2

21

nns

xxt

p

Since |t|= 7.809 is greater than the tabular value of t , i.e., 2.58 (for 48 d.f at 1% level of significance for two tailed). Hence the null hypothesis is rejected and we conclude that both the machines are not equally efficient at 1% level of significance.

80

If the mean of these three values is 8.0, then X3 must be 9 (i.e., X3 is not free to vary)

Degrees of Freedom (df)

Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2

(2 values can be any numbers, but the third is not free to vary for a given mean)

Idea: Number of observations that are free to vary after sample mean has been calculated

Example: Suppose the mean of 3 numbers is 8.0

Let X1 = 7

Let X2 = 8

What is X3?

Documents

Sampling and Sampling Methods 2