Question

5. A report by the NCAA states that 57.6% of football injuries occur during practices. A head trainer claims that this is too high for his conference, so he randomly selects 36 injuries and finds that 17 occurred during practices.

a. What would be the trainer’s null and alternative hypotheses?

b. Justify whether the trainer can use a Z-test.

c. Apply a Z-test with the level of significance α = .01. Compute the P-value, and interpret the results in the problem context.

Answer 1

a) Null and alternative hypotheses

Ho : p = 0.576

H1 : p < 0.576

b) we have n = 36 and p = 0.576

n * p = 20.74 and n *( 1- p) = 15.26

Both n*p and n*(1-p) are greater than 10

So we can use Z test

c) test statistic Z

Z = ( p^ - p)/sqrt [ p*(1-p)/n]

Where p^ = 17/36 = 0.472

Z = (0.472 - 0.576) / sqrt [ 0.576 * 0.424/36]

Z test = -1.26

For Z = -1.26 and left tailed test

p-value = P( Z < -1.26)

Using Z table

p-value = 0.1038

We have level of significance a = 0.01

Here p-value = 0.1038 > 0.01

Conclusion : Fail to reject the null hypothesis Ho , there is no sufficient evidence to support the trainer claim that this is too high for this conference