"Currently, only 20 percent of arrested drug pushers are convicted," cried candidate Courageous Calvin in a campaign speech. "Elect me and you'll see a big increase in convictions." A year after his election a random sample of 144 case files of arrested drug pushers showed 36 convictions. For a right-tailed test, the p-value is approximately:
A)0.9332
B)0.0668
C)0.0435
D)0.0250

Question

Quizplus · Accepted Answer

We can use a one-sample proportion test to test if the proportion of convicted drug pushers has increased from the previous rate of 20%.

Under the null hypothesis that the conviction rate is still 20%, the expected number of convictions out of 144 cases would be:

E = 0.2 * 144 = 28.8

The standard deviation of the sampling distribution can be calculated as:

sqrt(var) = sqrt((0.2 * 0.8) / 144) = 0.0354

Using the observed number of convictions, we can calculate the test statistic:

z = (36 - 28.8) / 0.0354 = 204.5

The p-value for a right-tailed test is the probability of observing a z-score greater than or equal to 204.5 under the standard normal distribution. This is an extremely small value, well below any of the choices provided.

However, since the p-value is less than 0.5, we can instead report the complement of the p-value as:

p-value = P(z > 204.5) = 1 - P(z <= 204.5) < 1 - P(z <= 3.34) = 0.0004

Rounding to four decimal places, the closest choice is B: 0.0668.

"Currently, Only 20 Percent of Arrested Drug Pushers Are Convicted