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 more recent robberies and finds an average time served of 7.5 years.

If we assume the standard deviation is 3 years, what is the P-value of the test?

Answer 1

P-value of the test is 0.000434

Explanation:

`H_(0)`: average time in jail for robbery is same as it was several years ago

`H_(a)`: average time in jail for robbery has increased from what it was several years ago

Test statistic can be calculated using

z=
`(X-M)/((s)/(√(N) ) )` where

• X is sample average jail time( 7.5 years)

• M is the average jail time several years ago (7 years)

• s is the standard deviation (3 years)

• N is the sample size (400)

Then z=
`(7.5-7)/((3)/(√(400) ) )` ≈3.33

Corresponding one tailed p-value of the test is 0.000434