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 pvalue?
Question 9 options:









Answer: 3) Test Statistic: 0.252, PValue: 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
Pvalue = 0.801
(by using ztable)
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