Question

A random sample of 9 automobiles of a particular model has a mean of 32 mpg and a standard deviation of 10 mpg. Assume gas mileages are normally distributed. Use this information to compute a 95% confidence interval for the mean miles per gallon obtained by all automobiles of this type.

Round your confidence interval values to the nearest whole number (Do NOT use any decimals in your answer).

The left endpoint of the confidence interval is  , and the right endpoint is  .

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 32

sample standard deviation = s = 10

sample size = n = 9

Degrees of freedom = df = n - 1 = 9 - 1 = 8

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,8 = 2.306

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

= 2.306 * (10 / 9)

Margin of error = E = 8

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

- E < < + E

32 - 8 < < 32 + 8

24 < < 40

The left endpoint of the confidence interval is 24 , and the right endpoint is 40