Question

Your friend tells you that the proportion of active Major League Baseball players who have a batting average greater than .300 is different from 0.77, a claim you would like to test. The hypotheses for this test are Null Hypothesis: p = 0.77, Alternative Hypothesis: p ≠ 0.77. If you randomly sample 24 players and determine that 19 of them have a batting average higher than .300, what is the test statistic and p-value?

Question 9 options:

1)	Test Statistic: 0.252, P-Value: 0.4

2)	Test Statistic: 0.252, P-Value: 0.199

3)	Test Statistic: 0.252, P-Value: 0.801

4)	Test Statistic: -0.252, P-Value: 0.801

5)	Test Statistic: 0.252, P-Value: 0.599

Answer 1

Answer: 3) Test Statistic: 0.252, P-Value: 0.801

Solution:

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

H0: p = 0.77 versus Ha: p ≠ 0.77

This is a two tailed test.

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 19

n = sample size = 24

p̂ = x/n = 19/24 = 0.791666667

p = 0.77

q = 1 - p = 1 - 0.77 = 0.23

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.791666667 - 0.77)/sqrt(0.77*0.23/24)

Z = (0.791666667 - 0.77)/ 0.0859

Z = 0.2522

Test statistic = 0.252

P-value = 0.801

(by using z-table)