Question

A sample of 41 renters of a one bedroom apartment had a mean of $1165 and a standard deviation of $150. Construct a 95% confidence interval for the true average rent. Answer the fallowing, Except for the critical value where you report the value in the table. Round to two decimal places.

The Critical value is?

The Margin of error is?

The confidence Interval is?

With 95% confidence the true average is between $_______ and $_______

Answer 1

Given:

Sample mean = 1165 , sample standard deviation s= 150 and sample size n = 41 , 95% confidence level.

As sample size is greater than 30 , Assume that data is normally distributed and population standard deviation σ is unknown we use t distribution.

First we find t0.95   , by using n – 1 d.f ; 41 – 1 = 41-1 =40

We use degrees of freedom = 40

Using t distribution table t0.95 ,40 = 2.021

The critical value is 2.021

The margin of error E = tc *  

E =2.021 *   

E = 47.34

Margin of error = 47.34

The confidence interval is

– E < µ <   + E

1165 – 47.34 < µ < 1165 + 47.34

1117.66 < µ < 1212.34

The confidence interval for population mean µ is (1117.66 ,1212.34 )

With 95% confidence the true average is between $1117.66 and $1212.34