Question

a.) Use the t-distribution to find a confidence interval for a mean u given the relevant sample results. Give the best point estimate for u, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed.

A 95% confidence interval for u using the sample results x-bar=74.7, s=7.4, and n=42

Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.

point estimate =

margin of error =

The 95% confidence interval is

b.) Use the t-distribution to find a confidence interval for a mean u given the relevant sample results. Give the best point estimate for u, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed.

A 90% confidence interval for u using the sample results x-bar=140.8, s=49.9, and n=50.

Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.

point estimate =

margin of error =

The 90% confidence interval is

Answer 1

Solution :

Given that,

a.)

Point estimate = sample mean = = 74.7

sample standard deviation = s = 7.4

sample size = n = 42

Degrees of freedom = df = n - 1 = 42 - 1 =41

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,41 = 2.020

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

= 2.020 * (7.4 / 42)

= 2.31

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

- E < < + E

74.7 - 2.31 < < 74.7 + 2.31

72.39 < < 77.01

(72.39 , 77.01)

The 95% confidence interval is 72.39 , 77.01

b.)

Point estimate = sample mean = = 140.8

sample standard deviation = s = 49.9

sample size = n = 50

Degrees of freedom = df = n - 1 = 50 - 1 = 49

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,49 = 1.677

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

= 1.677 * (49.9 / 50)

= 11.83

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

- E < < + E

140.8 - 11.83 < < 140.8 + 11.83

128.97 < < 152.63

(128.97 , 152.63)

The 90% confidence interval is 128.97 , 152.63