Question

Nationally, 60.2% of federal prisoners are serving time for drug offenses. A warden thinks that in his prison the % is higher. he surveys 400 random prisoner records and finds that 260 are drug offenders. Use one of the worksheets I provide and evaluate whether the warden is correct at the 5% significance level. Show me your work and tell me what you think.

Consider: Is this a problem comparing averages or proportions?

Answer 1

H0: p = 0.602

Ha: p > 0.602

Sample proportion = 260 / 400 = 0.65

Test statistics

z = ( - p) / sqrt ( p ( 1 - p) / n )

= ( 0.65 - 0.602) / sqrt ( 0.602 ( 1 - 0.602) / 400)

= 1.96

p-value = P(Z > z)

= P(Z > 1.96)

= 1 - P(Z < z)

= 1 - 0.975

= 0.025

Since p-value < 0.05, Reject H0.

We have sufficient evidence to conclude that percentage of of federal prisoners are serving

time for drug offenses is higher.