Question

A simple random sample of size n is drawn. The sample​ mean, x overbarx​, is found to be 18.5​, and the sample standard​ deviation, s, is found to be 4.3.

(a) Construct a 95​% confidence interval about muμ if the sample​ size, n, is 34.

Lower​ bound: ____

Upper​ bound: ____

​(b) Construct a 95​% confidence interval about muμ if the sample​ size, n, is 51.

Lower​ bound: ____

Upper​ bound: ____

​(c) Construct a 99​% confidence interval about muμ if the sample​ size, n, is 34.

Lower​ bound: ____

Upper​ bound: ____

Answer 1

Solution :

Given that,

a)

Point estimate = sample mean = = 18.5

sample standard deviation = s = 4.3

sample size = n = 34

Degrees of freedom = df = n - 1 = 33

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,33 = 2.035

Margin of error = E = t_/2,df * (s /n)

= 2.035 * ( 4.3 / 34)

= 1.501

The 95% confidence interval estimate of the population mean is,

- E < < + E

18.5 - 1.501 < < 18.5 + 1.501

16.999 < < 20.001

Lower bound = 16.999

Upper bound = 20.001

b)

sample size = n = 51

Degrees of freedom = df = n - 1 = 50

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,50 = 2.009

Margin of error = E = t_/2,df * (s /n)

= 2.009 * ( 4.3 / 51)

= 1.210

The 95% confidence interval estimate of the population mean is,

- E < < + E

18.5 - 1.210 < < 18.5 + 1.210

17.29 < < 19.71

Lower bound = 17.29

Upper bound = 19.71

c)

sample size = n = 34

Degrees of freedom = df = n - 1 = 33

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,33 = 2.733

Margin of error = E = t_/2,df * (s /n)

= 2.733 * ( 4.3 / 34)

= 2.015

The 99% confidence interval estimate of the population mean is,

- E < < + E

18.5 - 2.015 < < 18.5 + 2.015

16.485 < < 20.515

Lower bound = 16.485

Upper bound = 20.515