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 90% confidence interval for μ using the sample results x= 134.0, s= 55.6, and n= 50

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 90% confidence interval =

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 134.0

sample standard deviation = s = 55.6

sample size = n = 50

Degrees of freedom = df = n - 1 = 50 - 1 =49

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,49 = 1.677

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

= 1.677 * (55.6/ 50)

= 13.19

The 90% confidence interval estimate of the population mean is,

- E < < + E

134.0 - 13.19 < <134.0 + 13.19

120.81 < <147.19

the 90% confidence interval = (120.81, 147.19)