contestada

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?

Respuesta :

dogacandu

Answer:

P-value of the test is 0.000434

Step-by-step explanation:

[tex]H_{0}[/tex]: average time in jail for robbery is same as it was several years ago

[tex]H_{a}[/tex]: average time in jail for robbery has increased from what it was several years ago

Test statistic can be calculated using

z= [tex]\frac{X-M}{\frac{s}{\sqrt{N} } }[/tex] 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= [tex]\frac{7.5-7}{\frac{3}{\sqrt{400} } }[/tex] ≈3.33

Corresponding one tailed p-value of the test is 0.000434

Otras preguntas