Question

Your friend tells you that the proportion of active Major League Baseball players who have a batting average greater than .300 is less than 0.8, a claim you would like to test. The hypotheses for this test are Null Hypothesis: p ≥ 0.8, Alternative Hypothesis: p < 0.8. If you randomly sample 20 players and determine that 13 of them have a batting average higher than .300, what is the test statistic and p-value?

	1) Test Statistic: 1.677, P-Value: 0.953
	2) Test Statistic: -1.677, P-Value: 0.953
	3) Test Statistic: -1.677, P-Value: 0.094
	4) Test Statistic: 1.677, P-Value: 0.047
	5) Test Statistic: -1.677, P-Value: 0.047

Answer 1

Solution:- (5) Test Statistic: -1.677, P-Value: 0.047

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P > 0.80
Alternative hypothesis: P < 0.80

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.08944
z = (p - P) / S.D

z = - 1.677

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is less than -1.677.

Thus, the P-value = 0.047

Interpret results. Since the P-value (0.047) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the proportion of active Major League Baseball players who have a batting average greater than .300 is less than 0.8.