Question

A researcher wants to know if the average time in jail for robbery has increased from what it was several years ago when the average sentence was 7 years. He obtains data on 400 recent robberies and finds an average time served of 7.5 years. If we assume the standard deviation is 3 years, perform a significance test (calculating the P-value) at the 0.05 level.

Answer 1

Answer:

So, the interval is : (7.206,7.794)

Step-by-step explanation:

The mean μ = 7.5

Standard deviation σ=3

n = 400

At 95% confidence interval, the z score is 1.96

$7.5+1.96(\frac{3}{\sqrt{400} } )$

And $7.5-1.96(\frac{3}{\sqrt{400} } )$

7.5+0.294 and 7.5-0.294

So, the interval is : (7.206,7.794)