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¯=15.2, s=6.9, and n=30

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 = Enter your answer; point estimate

margin of error = Enter your answer; margin of error

The 95% confidence interval is Enter your answer; The 95%confidence interval, value 1 to Enter your answer; The 95%confidence interval, value 2 .

Answer 1

ANSWER:

Given that,

= 15.2

s = 6.9

n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

Point Estimate () = 15.2

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,29 = 2.045

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

= 2.045 * (6.9 / 30)

    Margin of error (E) = 2.58

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

- E < < + E

15.2 - 2.58 < < 15.2 + 2.58

12.62 < < 17.78

Confidence Interval (12.62, 17.78)