Question

A random sample of 86 observations produced a mean

x=25.8 and a standard deviation s=2.4.

a. Find a​ 95% confidence interval for μ.

b. Find a​ 90% confidence interval for μ.

c. Find a​ 99% confidence interval for μ.

a. The​ 95% confidence interval is (______,______)

Use integers or decimals for any numbers in the expression. Round to two decimal places as​needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 25.8

sample standard deviation = s = 2.4

sample size = n = 86

Degrees of freedom = df = n - 1 = 86 - 1 = 85

a. At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

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

t_{$\alpha$ /2,df} = t_0.025,85 = 1.988

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.988 * (2.4 / $\sqrt$ 86)

= 0.51

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

25.8 - 0.51 < $\mu$ < 25.8 + 0.51

25.29 < $\mu$ < 26.31

The​ 95% confidence interval is ( 25.29 , 26.31 )

b. At 90% confidence level the t is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1 / 2 = 0.05

t_{$\alpha$ /2,df} = t_0.05,85 = 1.663

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.663 * (2.4 / $\sqrt$ 86)

= 0.43

The 90% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

25.8 - 0.43 < $\mu$ < 25.8 + 0.43

25.37 < $\mu$ < 26.23

The​ 90% confidence interval is ( 25.37 , 26.23 )

c. At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

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

t_{$\alpha$ /2,df} = t_0.005,85 = 2.635

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.635 * (2.4 / $\sqrt$ 86)

= 0.68

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

25.8 - 0.68 < $\mu$ < 25.8 + 0.68

25.12 < $\mu$ < 26.48

The​ 99% confidence interval is ( 25.12 , 26.48 )