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 99% confidence interval for μ using the sample results x¯=93.5, s=34.3, and n=15

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.
ME:

99% CI

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 93.5

sample standard deviation = s = 34.3

sample size = n = 15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,14 = 2.977

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

= 2.977 * (34.3 / 15)

= 26.36

The 99% confidence interval estimate of the population mean is,

- E < < + E

93.5 - 26.36 < < 93.5 + 26.36

67.14 < < 119.86

A 99% confidence interval for μ is : (67.14 , 119.8)