Question

{Exercise 8.14 Algorithmic} A simple random sample with n = 56 provided a sample mean of 21.0 and a sample standard deviation of 4.5. a. Develop a 90% confidence interval for the population mean (to 2 decimals). ( 20 ®, 22 ) b. Develop a 95% confidence interval for the population mean (to 2 decimals). c. Develop a 99% confidence interval for the population mean (to 2 decimals). d. What happens to the margin of error and the confidence interval as the confidence level is increased? They increase

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 21.0

sample standard deviation = s = 4.5

sample size = n = 56

Degrees of freedom = df = n - 1 = 56 - 1 = 55

a) At 90% confidence level

$\alpha$ = 1 - 90%

$\alpha$ =1 - 0.90 =0.10

$\alpha$/2 = 0.05

t$\alpha$/2,df = t_0.05,55 = 1.673

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

= 1.673 * ( 4.5 / $\sqrt$ 56)

Margin of error = E = 1.01

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

$\bar x$  ± E

= 21.0  ± 1.01

= ( 19.99, 22.01)

b) At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

t$\alpha$/2,df = t_0.025,55 = 2.004

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

= 2.004 * ( 4.5 / $\sqrt$ 56)

Margin of error = E = 1.21

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

$\bar x$  ± E

= 21.0  ± 1.21

= ( 19.79, 22.21)

c) At 99% confidence level

$\alpha$ = 1 - 99%

$\alpha$ =1 - 0.99 =0.01

$\alpha$/2 = 0.005

t$\alpha$/2,df = t_0.005,55 = 2.668

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

= 2.668 * ( 4.5 / $\sqrt$ 56)

Margin of error = E = 1.60

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

$\bar x$  ± E

= 21.0  ± 1.60

= ( 19.40, 22.60)