Question

A report by the NCAA states that 57.6% of football injuries occur during practices. In a recent study of 52 injuries, 32 of the injuries occurred during practice. At the 0.06 level of significance, can it be concluded that the percentage of football injuries that occur during practices is greater than 57.6%? Show your work to receive credit.

Answer 1

A report by the NCAA states that 57.6% of football injuries occur during practices, thus the population proportion is p = 0.576. In a recent study of n = 52 injuries, X = 32 of the injuries occurred during practice.

Before testing the hypothesis we need to check the normal distribution requirements which are:

1) n*p(1-p) >=10, thus 52*0.576*(1-0.576) = 12.70 which is greater than 10 hence satisfied.

So, Z distribution is applicable to hypothesis testing.

The sample proportion is calculated as:

Based on the claim the hypotheses are:

Based on the hypothesis it will be a right-tailed test.

Rejection region:

Based on the type of hypothesis and given significance level the critical value for the rejection region is computed using the excel formula for normal distribution which is =NORM.S.INV(0.94), thus the Zc is computed as 1.55.

Reject Ho if Z >1.55 and P-value <0.06

Test Statistic:

P-value:

The P-value is calculated using the excel formula for normal distribution which is =1-NORM.S.DIST(0.575, TRUE), thus the P-value is computed as 0.2828.

Conclusion:

Since P-value is greater than 0.06 and Z is less than Zc hence we failed to reject the null hypothesis and conclude that there is insufficient evidence to support the claim,

Note: Ask if any query remains