Question

1. Sue, a starting player for a major college basketball team, made only 38% of her free throws last year. During the summer she worked on developing a softer shot in the hope of improving her free-throw accuracy. In the first eight games of this season Sue made 22 free throws in 40 attempts. Assume that these 40 attempts represent a random sample of free throws. Let p be the probability of making each free throw she shoots this season. Does this evidence suggest that Sue has made significant improvement in her free throw shooting? Let α = 0.10. Be sure to include: the null and alternative hypothesis, the conditions, test statistic, and the conclusion in context.

1. An 80% confidence interval, for the true proportion of Sue’s made free throws, using the same point estimate, yields a lower bound of 0.449 and an upper bound of 0.650. How can you use this confidence interval to support your conclusion in problem above?

Answer 1

Population proportion, P = 0.38

Sample proportion,

Null Hypothesis, H0: P 0.38

Alternate Hypothesis, H1: P > 0.38

We use one-sample test of proportions at = 0.10

Test Statistic,

Since this a one-tailed test, Z-critical corresponding to = 0.10 is Z-crit = P(Z<=0.90) = 1.2816

Since Z-test > Z-crit, the statistic lies in the critical zone.

We reject the Null Hypothesis and conclude the difference is significant.

(b) The 80% confidence interval is [0.449, 0.65]. The confidence interval does not include assumed population proportion of 0.38. Hence we can conclude the sample proportion of free throws made is significantly different from the population proportion of 38%.