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¯=94.5, s=8.5, 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,

= 94.5

s = 8.5

n = 42

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

Point estimate =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 =2.019

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

= 2.019 * (8.5 / 42 )

= 2.65

Margin of error = 2.65

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

- E < < + E

94.5 - 2.65< < 94.5 + 2.65

91.85 < < 97.15

(91.85, 97.15 )