Question

Complete a-d

9.2.17 O of 1 Point Question Help A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 106. and the sample standard deviation, s, is found to be 10. (a) Construct a 96% confidence interval about if the sample size, n, is 24. (b) Construct a 96% confidence interval about u if the sample size, n, is 28. (c) Construct a 98% confidence interval about if the sample size, n, is 24. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution. (a) Construct a 96% confidence interval about u if the sample size, n, is 24. Lower bound: : Upper bound: (Use ascending order. Round to one decimal place as needed.)

Answer 1

(a)

n = 24

$\bar{x}$ = 106

s = 10

SE = s/

п

= 10/$\sqrt{24}$

= 2.0412

$\alpha$ = 0.04

ndf = 24 - 1 = 23

From Table, critical values of t = $\pm$ 2.1770

Confidence Interval:

106 $\pm$ (2.1770 X 2.0412)

= 106 $\pm$ 4.4438

= ( 101.5562 ,110.4438)

Confidence Interval:

Lower bound = 101.6; upper bound = 110.4

(b)

n = 28

$\bar{x}$ = 106

s = 10

SE = s/

п

= 10/$\sqrt{28}$

= 1.8898

$\alpha$ = 0.04

ndf = 24 - 1 = 23

From Table, critical values of t = $\pm$ 2.1770

Confidence Interval:

106 $\pm$ (2.1770 X 1.8898)

= 106 $\pm$ 4.1141

= ( 101.8859 ,110.1141)

Confidence Interval:

Lower bound = 101.9; upper bound = 110.1

(c)

n = 24

$\bar{x}$ = 106

s = 10

SE = s/

п

= 10/$\sqrt{24}$

= 2.0412

$\alpha$ = 0.02

ndf = 24 - 1 = 23

From Table, critical values of t = $\pm$ 2.4999

Confidence Interval:

106 $\pm$ (2.4999 X 2.0412)

= 106 $\pm$ 5.1028

= ( 100.8972 ,111.1028)

Confidence Interval:

Lower bound = 100.9; upper bound = 111.1

(d)

We could not have computed the confidence intervals in part (a) - (c) if the population is not normally distributed because the population standard deviation $\sigma$ is not provided and it has to be estimated from sample standard deviation s and the sample size = n < 30 Small Sample and so Central Limit Theorem cannot applied.