Question

Use the t-distribution to find a confidence interval for a mean μ given the relevant sample results. Give the best point estimate for μ, 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 μ using the sample results x= 94.6, s= 6.9, 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 =

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 94.6

sample standard deviation = s = 6.9

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 = 2.064

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

= 1.683 * (6.9 / 42)

= 5.8

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

- E < < + E

94.6 - 5.8 < < 94.6 + 5.8

88.8 < < 100.4

(88.8 , 100.4)