Jamaal, a player on a college basketball team, made only 50% of his free throws last season. During the off-season, he worked on developing a softer shot in the hope of improving his free-throw accuracy. This season, Jamaal made 54 of 95 free throws. We want to test the hypothesis that Jamaal's free-throw percentage p this season is significantly different from last year's percentage. The approximate P-value for an appropriate test is

1) 0.816
2) 0.091
3) 0.182
4) 0.194

fefw

.816

.194

To test the hypothesis that Jamaal's free-throw percentage this season is significantly different from last year's percentage, we can use the one-sample proportion test.

First, we need to calculate the sample proportion of successful free throws this season. From the given information, Jamaal made 54 out of 95 free throws, so the sample proportion can be calculated as:

p̂ = 54/95 ≈ 0.5684

Next, we can calculate the standard error for the proportion using the formula:

SE(p̂) = √((p̂(1-p̂))/n)

where p̂ is the sample proportion and n is the sample size. In this case, n = 95. Plugging in the values, we get:

SE(p̂) = √((0.5684(1-0.5684))/95) ≈ 0.0504

Now, to calculate the test statistic, we use the formula:

z = (p̂ - p₀)/SE(p̂)

where p₀ is the null hypothesis proportion. In this case, the null hypothesis is that Jamaal's free-throw percentage this season is equal to his percentage last year, which is 0.50. So we have:

z = (0.5684 - 0.50)/0.0504 ≈ 1.3540

Finally, we can calculate the p-value associated with this test statistic. Since this is a two-sided test, we need to calculate the probability of obtaining a test statistic at least as extreme as the observed test statistic (both in the positive and negative direction).

Using a standard normal distribution table or a calculator, we can find that the cumulative probability for z = 1.3540 is approximately 0.911, which corresponds to a p-value of 1 - 0.911 ≈ 0.089. However, since this is a two-sided test, we need to double this p-value to account for the probability in both tails.

So, the approximate p-value for this test is approximately 2 * 0.089 = 0.178.

Among the options provided, the closest approximate p-value is 0.182 (Answer 3).