Question

A random sample of 43 observations is used to estimate the population variance. The sample mean and sample standard deviation are calculated as 68.5 and 3.1, respectively. Assume that the population is normally distributed. (You may find it useful to reference the appropriate table: chi-square table or F table) a. Construct the 95% interval estimate for the population variance. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) Confidence interval to b. Construct the 99% interval estimate for the population variance. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) Confidence interval to c. Use your answers to discuss the impact of the confidence level on the width of the interval. The width of the interval increases with the confidence level. The width of the interval decreases with the confidence level.

Answer 1

ANSWER:

Given that,

Sample size = n = 43

Sample mean = $\bar{x}$ = 68.5

Sample standard deviation = s = 3.1

(a) Construct the 95% interval estimate for the population variance

Degree of freedom = df = n-1 = 43-1 = 42

c = 95% = 95/100 = 0.95

$\alpha$ = 1-c = 1-0.95= 0.05

$\alpha$/2 = 0.05/2 = 0.025

1 - $\alpha$ /2 = 1 - 0.05/2 = 0.975

Critical values,

>> X1-a/2,df
= $\chi _{0.975,42}^{2}$ = 25.99866 and

2 Xa/2,df
= $\chi _{0.025,42}^{2}$ = 61.77675

Confidence interval

(n-1)$s^{2}$ /
2 Xa/2,df
< $\sigma ^{2}$ < (n-1)$s^{2}$ /
>> X1-a/2,df

(43-1)3.1^2 / 61.77675 < $\sigma ^{2}$ < (43-1)3.1^2 / 25.99866 ​

6.5335259624373247 < $\sigma ^{2}$ < 15.5246462702308504

6.5335 < $\sigma ^{2}$ < 15.5246 (Rounded to four decimal place)

Confidence innterval 6.5335 to  15.5246

(b) Construct the 99% interval estimate for the population variance

Degree of freedom = df = n-1 = 43-1 = 42

c = 99% = 99/100 = 0.99

$\alpha$ = 1-c = 1-0.99= 0.01

$\alpha$/2 = 0.01/2 = 0.005

1 - $\alpha$ /2 = 1 - 0.01/2 = 0.995

Critical values,

>> X1-a/2,df
= $\chi _{0.995,42}^{2}$ = 22.13846 and

2 Xa/2,df
= $\chi _{0.005,42}^{2}$ = 69.33599

Confidence interval

(n-1)$s^{2}$ /
2 Xa/2,df
< $\sigma ^{2}$ < (n-1)$s^{2}$ /
>> X1-a/2,df

(43-1)3.1^2 / 69.33599 < $\sigma ^{2}$ < (43-1)3.1^2 / 22.13846 ​

5.8212192542429985 < $\sigma ^{2}$ < 18.2316204469506912

5.8212 < $\sigma ^{2}$ < 18.2316 (Rounded to four decimal place)

Confidence innterval 5.8212   to  18.2316

(c) The width of the interval increases with the confidence interval.

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