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¯=82.7, s=32.9, 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.

Point estimate=

Margin of error=

The 99% confidence interval is _____ to ______.

Answer 1

Given that,

=82.7

s =32.9

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 = t0.005, 14= 2.977 ( using student t table)

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

= 2.977* (32.9 / 15) = 25.29

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

- E < < + E

82.7- 25.29< < 82.7+ 25.29

57.41 < < 107.99

( 57.41 , 107.99)