Question

# Your friend tells you that the proportion of active Major League Baseball players who have a...

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: 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)