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¯=89.9, s=9.3, 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

Answer 1

Solution :

Given that,

= 89.9

s =9.3

n =42

point estimate=89.9

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

a ) 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 * (9.3/42)

= 2.90

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

- E < < + E

89.9 - 2.90 < < 89.9+ 2.90

87.00 < < 92.80

( 87.00 , 92.80)