914 The local police chief started a crimeline program some

9.14) The local police chief started a “crimeline” program some years ago and wonders if it’s really working. The program publicizes unsolved violent crimes in the local media and offers cash rewards or information leading to arrests. Are “featured” crimes more likely to be cleared by arrest that other violent crimes? Results from random samples of both types of crimes are reported as follows:

Sample 1                                      Sample 2

(Crimeline Crimes          (Non-crimeline Crimes

Cleared by Arrest)                       Cleared by Arrest)

P_s1 = 0.35                                      P_s2 = 0.25

N₁ = 178                                        N₂ = 212

Solution

Here we have given that the local police chief started a “crimeline” program some years ago and wonders if it’s really working. The program publicizes unsolved violent crimes in the local media and offers cash rewards or information leading to arrests.

Here we want totest the hypothesis that ,

H₀ : P1 = P2 Vs H₁ : P1 > P2

This is one sided (right tailed test).

The test statistics for testing is,

Z = ( P1 - P2 ) / sqrt[ (Pbar*Qbar)/n1 + (Pbar*Qbar)/n2 ]

where Pbar = (r1 + r2) / (N1 + N2)

Qbar = 1 - Pbar

Also we have given that

proportion of crimelines crimes cleared by arrest (P1) = 0.35

proportion of non-crimelines crimes cleared by arrest (P2) = 0.25

N1 = 178

N2 = 212

We know that ,

P1 = r1 / N1

r1 = N1*P1 = 62.3

P2 = r2 / N2

r2 = N2*P2 = 53

Pbar = (62.3 + 53) / (178 + 212) = 115.3 / 390 = 0.2954

Qbar = 1 - Pbar = 1 - 0.2954 = 0.7046

First we want to check the condition that N1*Pbar,N2*Pbar,N1*Qbar and N2*Qbar all are >5

N1*Pbar = 178 * 0.2954 = 52.5812

N2*Pbar = 212 * 0.2954 = 62.6248

N1*Qbar = 178 * 0.7046 = 125.4188

N2 * Qbar = 212 * 0.7046 = 149.3752

All are greator than 5.

The test statistic is,

Z = (0.35 - 0.25) / sqrt [ ( 0.2954 * 0.7046) / 178 + (0.2954 * 0.7046) / 212 ]

= 0.1 / sqrt(0.001169 + 0.000982)

= 0.1 / sqrt(0.002151)

Z = 2.156

(level of significance) = 5% (0.05)

We can calculate critical value by using EXCEL

,Command : \"=normsinv(1-)

This will gives us critical value = 1.645

Decision : Z > critical value

Reject null hypothesis at 5% level of significance.

We have not sufficient evidence to suggest that “featured” crimes more likely to be cleared by arrest that other violent crimes.