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Example:Racial Profiling Racial profiling is the controversial practice of targe

ID: 3374794 • Letter: E

Question

Example:Racial Profiling Racial profiling is the controversial practice of targeting someone for suspicion of criminal behavior on the basis of race, national origin, or ethnicity. The table below includes data from randomly selected drivers stopped by police in a recent year (based on data from the U.S. Department of Justice, Bureau of Justice Statistics) Use a 0.05 significance level to test the claim that the proportion of blacks stopped by police is significantly greater than the proportion of whites. a) b) Construct a confidence interval that could be used to test the claim in part (a). Be sure to use the correct level of significance. What do you conclude based on the confidence interval? Race and Ethnicity Black andWhite and Non-Hispanic Non-Hispanic Drivers stopped by police Total number of observed drivers 24 200 147 1400

Explanation / Answer

p1 = 24/200 = 0.12

p2 = 147/1400 = 0.105

a) H0: P1 = P2

    H1: P1 > P2

The poopled sample proportion(P) = (p1 * n1 + p2 * n2)/(n1 + n2)

                                                        = (0.12 * 200 + 0.105 * 1400)/(200 + 1400)

                                                         = 0.11

SE = sqrt(P(1 - P) * (1/n1 + 1/n2))

       = sqrt(0.11 * (1 - 0.11) * (1/200 + 1/1400))

       = 0.024

The test statistic z = (p1 - p2)/SE

                             = (0.12 - 0.105)/0.024

                             = 0.63

P-value = P(Z > 0.63)

              = 1 - P(Z < 0.63)

               = 1 - 0.7357

               = 0.2643

As the P-value is greater than the significane level (0.2643 > 0.05), so the null hypothesis is not rejected.

So at 0.05 significance level, there is not sufficient evidence to support the claim that the proportion of blacks stopped by police is significantly greater than the proportion of whites.
b) At 95% confidence interval the critical value is z0.025 = 1.96

The confidence interval is

(p1 - p2) +/- z0.025 * SE

= (0.12 - 0.105) +/- 1.96 * 0.024

= 0.015 +/- 0.047

= -0.032, 0.062

As the confidence interval cointains 0, so the null hypothesis is not rejected.

So we can conclude that, there is not sufficient evidence to support the claim that the proportion of blacks stopped by police is significantly greater than the proportion of whites.

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