Aresearcher 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?

P-value of the test is 0.000434

Step-by-step explanation:

: average time in jail for robbery is same as it was several years ago

: average time in jail for robbery has increased from what it was several years ago

Test statistic can be calculated using

z= 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= ≈3.33

Corresponding one tailed p-value of the test is 0.000434

It will take him 5 days

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post