A textbook of Computer Based Numerical and Statiscal Techniques part 53 ppsx

evidence against the null hypothesis which may be accepted, and we conclude that there is no significant difference between the two groups in their attitude towards the new plan.. Null H

Sol Given that

n1 = 800, n2 = 1200

p1 = X

n

1 1

= 800

1000 =

4

5;

p2 = X

n

2 2

= 800

1200 =

2 3

P = p n p n

1 1 2 2

1 2

+

1 2

1 2

+ +

= 800 800

1000 1200

+

8

11; Q =

3 11

Also, Null hypothesis H0: p1 = p2

i.e.,there is no significant difference in the consumption of tea before and after increase of

excise duty

H1: p1 > p2 (right tailed test)

PQ

1 2

1 2

– +

F

= 0 8 0 6666 8

11

3 11

1 1000

1 1200

–

× FH + IK = 6.842

Conclusion: Since the calculated value of |Z| > 1.645 also |Z| > 2.33, both the significant values of z at 5% and 1% level of significance Hence H0 is rejected i.e., there is a significant

decrease in the consumption of tea due to increase in excise duty

Example 12 In two large populations there are 30% and 25% respectively of fair haired people Is

this difference likely to be hidden in samples of 1200 and 900 respectively from the two populations.

Sol P1 = proportion of fair haired people in the first population = 30% = 0.3; P2 = 25%

= 0.25; Q1 = 0.7; Q2 = 0.75

Here H0: Sample proportions are equal i.e., the difference in population proportions is likely

to be hidden in sampling

H1 : P1 ≠ P2

P Q n

P Q n

1 2

1 1 1

2 2 2

–

+

= 0 3 0 25

0 3 0 7 1200

0 25 0 75 900

−

Conclusion: |Z| > 1.96 the significant value of Z at 5% level of significance H0 is rejected

However |Z| < 2.58, the significant value of Z at 1% level of signficance, H0 is accepted At 5% level, these samples will reveal the difference in the population proportions

Example 13 500 articles from a factory are examined and found to be 2% defective, 800 similar

articles from a second factory are found to have only 1.5% defective Can it reasonably be concluded that the product of the first factory are inferior to those of second?

Sol n1 = 500

n2 = 800

p1 = proportion of defective from first factory = 2% = 0.02

p2 = proportion of defective from second factory = 1.5% = 0.015

H0: There is no significant difference between the two products i.e., the products do not

differ in quality

H1 : P1 < p2 (one tailed test)

Under H0 : z = p p

PQ

1 2

1 2

– +

F

P = n p1 1n n p n2 2

1 2

+ +

= 0 02 500 0 015 800

500 800

a f a fa f+

+

= 0.01692

Q = 1 – P = 1 – 0.01692

= 0.9830

0 01692 0 983 1

500

1 800

–

= 0.68

Conclusion: As |Z| < 1.645, the significant value of Z at 5% level of significance, H0 is

accepted i.e., the products do not differ in quality.

Example 14 Random samples of 400 men and 600 women were asked whether they would like to

have a flyover near their residence 200 men and 325 women were in favour of the proposal Test the hypothesis that proportions of men and women in favour of the proposal, are same against that they are

Sol Null Hypothesis H0: P1 = P2 = P, (say), i.e., there is no significant difference between the

opinion of men and women as far as proposal of flyover is concerned

Alternative Hypothesis, H1:P1 ≠ P2 (two-tailed)

We are given:

n1 = 400, X1 = Number of men favouring the proposal = 200

n2 = 600, X2 = Number of women favouring the proposal = 325

∴ p1 = Proportion of men favouring the proposal in the sample

= X

n

1 1 = 200

400 = 0.5

p2 = Proportion of women favouring the proposal in the sample

= X

n

2 2

= 325

600 = 0.541

Test Statistic Since samples are large, the test statistic under the Null Hypothesis, H0 is:

PQ

1 2

1 2

–

 F +

∼ N (0, 1)

1 1 2 2

1 2

+

1 2

1 2

+ +

= 200 325

400 600

+

525

1000 = 0.525

0 525 0 475 1

400

1 600

–

× 0 041×FH IK

0 525 0 475 5

1200

0 041

0 001039 =

–

0 041

0 0323 = –1.269 Conclusion: Since Z = 1.269, which is less than 1.96, it is not significant at 5% level of

significance Hence H0 may be accepted at 5% level of significance and we may conclude that men and women do not differ significantly as regards proposal of flyover is concerned

Example 15 A machine produced 16 defective bolts in a batch of 500 After overhauling it produced

3 defectives in a batch of 100 Has the machine improved ?

Sol We have,

p1 = 16

500 = 0.032; n1 = 500

p2 = 3

100 = 0.03; n2 = 100 Null Hypothesis H0: The machine has not improved due to overhauling

H0 : p1 = p2

i.e., H1 : p1 > p2 (right tailed)

∴ P = p n p n

1 1 2 2

1 2

+ + =

19

600 ≅ 0.032

Under H0, the test statistic

PQ

1 2

1 2

–

+

F

= 0 032 0 03

0 032 0 968 1

500

1 100

–

Conclusion: The calculated value of |Z| < 1.645, the significant value of Z at 5% level of significance, H0 is accepted i.e., the machine has not improved due to overhauling.

Example 16 A company has the head office at Calcutta and a branch at Bombay 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 Calcutta, 62% favoured the new plan At Bombay out of a sample of 400 workers, 41% were against the new plan Is there any significant difference between the two groups in their attitude towards the new plan at 5% level ?

Sol In the usual notations, we are given:

n1 = 500, p1 = 0.62 and n2 = 400, p2 = 1 – 0.41 = 0.59

Null hypothesis, H0 : P1 = P2, i.e., there is no significant difference between the two

groups in their attitude towards’ the new plan

Alternative hypothesis, H1 : P1 ≠ P2 (Two-tailed)

Test Statistic Under H0, the test statistic for large samples is:

1 2

1 2

– b – g = p p

PQ

1 2

1 2

–

 F +

∼ N (0, 1)

1 1 2 2

1 2

+

500 0 62 400 0 59

500 400

+

= 0.607

0 607 0 393 1

500

1 400

– × ×FH + IK

= 0 03

0 00107

0 03

0 0327

= 0.917.

Critical Region: At 5% level of significance, the critical value of Z for a two-tailed test is 1.96.

Thus the critical region consists of all values of Z ≥ 1.96 or Z ≤ –1.96

Conclusion: Since the calculated value of Z = 0.917 is less than the critical value of Z

(1.96), it is not significant at 5% level of significance Hence the data do not provide us any

evidence against the null hypothesis which may be accepted, and we conclude that there is no significant difference between the two groups in their attitude towards the new plan

Example 17 On the basis of their total scores, 200 candidates of a civil service examination are

divided into two groups, the upper 30 per cent and the remaining 70 per cent Consider the first question

of this examination Among the first group, 40 had the correct answer, whereas among the second group,

80 had the correct answer On the basis of these results, can one conclude that the first question is no good

at discriminating ability of the type being examined here?

Sol Here, we have

n = Total number of candidates = 200

n1 = The number of candidates in the upper 30% group

= 30

100×200 = 60

n2 = The number of candidates in the remaining 70% group

= 70

100×200 = 140

X1 = The number of candidates, with correct answer in the first group = 40

X2 = The number of candidates, with correct answer in the second group = 80

n

1 1

= 40

60 = 0.6666 and p2 =

X n

2 2

= 80

140 = 0.5714.

Null Hypothesis, H0: There is no significant difference in the sample proportions, i.e., P1 = P2,

i.e., the first question is no good at discriminating the ability of the type being examined here Alternative Hypothesis,

H1 : P1 ≠ P2 Test Statistic: Under H0 the test statistic is:

PQ

1 2

1 2

–

 F +

∼ N (0, 1) (since samples are large)

1 2

1 2

+

40 80

60 140

+ + = 0.6

Q = 1 – P = 0.4

0 6 0 4 1

60

1 140

× FH + IK = 0 0953

0 0756 = 1.258

Conclusion: Since |Z| < 1.96, the data are consistent with the null hypothesis at 5% level

of significance Hence we conclude that the first question is not good enough to distinguish between the ability of the two groups of candidates

(C) Testing of Significance for Single Mean: Let x1, x2, x n be a random sample of size

n from a large population X1, X2, X N (of size N) with mean µ and variance σ2 Then sample mean d iX and variance (S2) are

1

1 n

i i

x

n∑=

n

−

=

∑ b g2 1

Also, the standard error of mean of a random sample of size n from a population with

variance σ2 is σ/ n

i.e., S.E.d ix = σ2

σ

n

For large samples, the standard normal variate corresponding to X — is

Z = x

n

– /

µ σ where σ is the standard deviation of the population

Under the null hypothesis, H0 that the sample has been drawn from a population with mean

µ and variance σ2, i.e., there is no signficiant difference between the sample mean d ix and population mean (µ), for large samples the test statistic is

Z = x

n

– /

µ σ

If the population standard deviation σ is not known

Z = X

– / µ

where S is the standard deviation of the sample.

Confidence Limits for µµµµµ

1 If the level of significance is α and Zα is the the critical value then

–Zα < x

n

– /

µ

σ < Zα The limit of the population mean µ are given by

x – Zα σ

n < µ < x + Zα σ

n

2 95% confidence interval for µ at 5% level of significance are

x

— – 1.96 σ

n ≤ µ ≤ x—

+ 1.96 σ

n

Similarly, 99% confidence limits for µ at 1% level of significance are

x

— – 2.58 σ

n < µ < x—

+ 2.58 σ

n

3 In sampling from a finite population of size N, the corresponding 95% and 99% confidence

limits for µ are

x— ± 1.96 σ

n

N n N

– – 1

n

N n N

– – 1

4 The confidence limits for any parameters (P, µ etc.) are known as its fiducial limits

Example 18 A sample of 900 members has a mean 3.5 cms, and S.D 2.61 cms Is the sample from

a large population of mean 3.25 cms, and S.D 2.61 cms?

If the population is normal and its mean is unknown Find the 95% and 98% fiducial limits of true mean.

Sol Null hypothesis, (H0): The sample has been drawn from the population with mean

µ = 3.25 cms and S.D σ = 2.61 cms

Alternative Hypothesis, H1: µ ≠ 3.25 (Two-tailed)

Test Statistic Under H0, the test statistic is:

Z = x

n

– /

µ

σ ∼ N (0, 1), (since n is large) Here, we are given x— = 3.4 cms., n = 900 cms., µ = 3.25 cms and σ = 2.61 cms

Z = 3 40 3 25

2 61 900

–

0 15 30

2 61

×

= 1.73

Since |Z| < 1.96, we conclude that the data don’t provide us any evidence against the null hypothesis (H0), which may, therefore, be accepted at 5% level of significance

95% fiducial limits for the population mean µ are:

x ± 1.96 σ/ n ⇒ 3.40 ± 1.96 × 2.61/ 900

⇒ 3.40 ± 0.1705, i.e., 3.5705 and 3.2295

98% fiducial limits for µ are given by:

x ± 2.33 σ

n , i.e., 3.40 ± 2.33 × 2 61

30

⇒ 3.40 ± 0.2027, i.e., 3.6027 and 3.1973

Remark: 2.33 is the value z1 of Z from standard normal probability intergrals, such that

Pd Z >z1i = 0.98 ⇒ Pd Z >z1i = 0.49

Example 19 The average marks in Mathematics of a sample of 100 students was 51 with a S.D.

of 6 marks Could this have been a random sample from a population with average marks 50?

Sol Given

n = 100, x = 51, s = 6, µ = 50 (σ is unknown in this problem)

H0 : The sample is drawn from a population with mean 50, µ = 50

H1 : µ ≠ 50

– /

µ = 51 50

6 100

–

10

6 = 1.6666.

Conclusion: Since |Z| = 1.666 < 1.96, Zα the significant value of Z at 5% level of significance,

H0 is accepted

Example 20 An insurance agent has claimed that the average age of policyholders who insure

through him is less than the average for all agents, which is 30.5 years.

A random sample of 100 policyholders who had insured through him gave the following age distribution: Age last birthday No of persons

Calculate the arithmetic mean and standard deviation of this distribution and use these values to test his claim at the 5% level of significance You are given that Z(1.645) = 0.95.

Sol Null Hypothesis, H0, µ = 30.5 years, i.e., the sample mean d ix and population mean (µ)

do not differ significantly

Alternative Hypothesis, H1 : µ < 30.5 years (Left-tailed alternative)

Calculations for Sample Mean and S.D

Age last No of Mid-point d= x– 28

2

birthday persons (f) x

16–20 12 18 –2 –24 48 21–25 22 23 –1 –22 22

Total N = 100 Σfd = 16 Σfd2 = 164

x = 28 + 5 16

100

×

= 28.8 years

s = 5 × 164

100

16 100

2 –FHG IKJ = 6.35 years Since the sample is large, σ  j s = 6.35 years.

Test Statistic Under H0, the test statistic is:

Z = x

– /

µ

2 ∼ N (0, 1), (since sample is large)

6 35 100

–

–

1 7

0 635 = –2.681.

Conclusion: Since computed value of Z = –2.681 < –1.645 or Z = 2.681 > 1.645, it is

significant at 5% level of significance Hence we reject the null hypothesis H0 (Accept H1) at 5% level of significance and conclude that the insurance agent’s claim that the average age of policyholders who insure through him is less than the average for all agents, is valid

Example 21 A normal population has a mean of 6.8 and standard deviation of 1.5 A sample of 400

members gave a mean of 6.75 Is the difference significant?

Sol H0 : There is no significant difference between x and µ

H1 : There is significant difference between x and µ

Given µ = 6.8 σ = 1.5 x = 6.75 and n = 400

n

– /

µ

6 75 6 8

1 5 900

– = – 0 67 = 0.67

Conclusion As the calculated value of |Z| < Zα = 1.96 at 5% level of significance H0 is accepted therefore there is no significant difference between x and µ

Example 22 The mean muscular endurance score of a random sample of 60 subjects was found to

be 145 with a S.D of 40 Construct a 95% confidence interval for the true mean Assume the sample size

to be large enough for normal approximation What size of sample is required to estimate the mean within

5 of the true mean with a 95% confidence? [Calicut University B.Sc (Main State) 1989] Sol We are given: n = 60, x = 145 and s = 40

95% confidence limits for true mean (µ) are:

x ± 1.96 s/ n (σ2 = s2, since sample is large)

= 145 ± 1 96 40

60

×

= 145 ± 78 4

7 75 = 145 ± 10.12 = 134.88, 155.12

Hence 95% confidence interval for µ is (134.88, 155.12) we know that

n = Z

E

α.σ FHG IKJ2

= 1 96 40

5

2 ×

[Œ Z0.05 = 1.96, σ = s = 40 and x –µ < 5 = E]

= (15.68)2 = 245.86 j 246

Example 23 A random sample of 900 members has a mean 3.4 cms Can it be reasonably regarded

as a sample from a large population of mean 3.2 cms and S.D 2.3 cms?

Sol We have:

n = 900, x = 3.4, µ = 3.2, σ = 2.3

H0 : Assume that the sample is drawn from a large population with mean 3.2 and

S.D = 2.3

H1 : µ ≠ 3.25 (Apply two-tailed test)

n

– /

µ

3 4 3 2

2 3 900

– = 0.261

Conclusion: As the calculated value of |Z| = 0.261 < 1.96, the significant value of Z at 5% level of significance, H0 is accepted therefore the sample is drawn from the population with mean 3.2 and S.D = 2.3

Example 24 The mean weight obtained from a random sample of size 100 is 64 gms The S.D of

the weight distribution of the population is 3 gms Test the statement that the mean weight of the population

is 67 gms at 5% level of significance Also set up 99% confidence limits of the mean weight of the population.

Sol We have:

n = 100, µ = 67, x = 64, σ = 3

H0 : There is no significant difference between sample and population mean

i.e., µ = 67, the sample is drawn from the population with µ = 67

H1 : µ ≠ 67 (Two-tailed test)

n

– /

µ

64 67

3 100

– = –10 ∴|Z| = 10

Conclusion: Since the calculated value of |Z| > 1.96, the significant value of Z at 5% level

of significance, H0 is rejected i.e., the sample is not drawn from the population with mean 67 To

find 99% confidence limits It is given by x ± 2.58 σ

n

= 64 ± 2.58 × 3

100 = 64.774, 63.226 (D) Test of Significance for Difference of means of two large samples: The test statistic

is given by, in this case

12 1

22 2 –

σ + σ