Question

simple random sample of 80 items from a population with = 8 resulted in a sample mean of 38. If required, round your answers to two decimal places. a. Provide a 90% confidence interval for the population mean. to b. Provide a 95% confidence interval for the population mean. to c. Provide a 99% confidence interval for the population mean. to

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 38

sample standard deviation = s = 8

sample size = n = 80

Degrees of freedom = df = n - 1 = 80 - 1 = 79

a.

At 95% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,79 = 1.664

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

= 1.664 * (8 / 80)

= 1.49

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

- E < < + E

38 - 1.49 < < 38 + 1.49

36.51 < < 39.49

(36.51 , 39.49)

b.

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,79 = 1.990

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

= 1.990 * (8 / 80)

= 1.78

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

- E < < + E

38 - 1.78 < < 38 + 1.78

36.22 < < 39.78

(36.22 , 39.78)

c.

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,79 = 2.640

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

= 2.640 * (8 / 80)

= 2.36

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

- E < < + E

38 - 2.36 < < 38 + 2.36

35.64 < < 40.36

(35.64 , 40.36)