Question

A popular restaurant takes a random sample n = 35 customers and records how long each occupied a table. The found a sample mean of 1.2 hours and a sample standard deviation of 0.45 hours. Find the 95 percent confidence interval for the mean.

Please show specific steps

Answer 1

Solution :

Given that,

= 1.2

s = 0.45

n = 35

Degrees of freedom = df = n - 1 = 35 - 1 = 34

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, 34} = 2.032

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

= 2.032 * (0.45 / 35)

= 0.15

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

- E < < + E

1.2 - 0.15 < < 1.2 + 0.15

1.05 < < 1.35

(1.05, 1.35)