Chapter 9.2, Problem 9.40E

To determine

To perform hypothesis test for one sample proportion $\widehat{p}$ .

Answer to Problem 9.40E

There is enough evidence to support the claim that Shaq has significantly improved his free-throw shooting at a = 0.05

Explanation of Solution

Given:

Sample size = n = 39

  $x$ = 26

Formula:

Sample proportion:

  $\widehat{p}=\frac{x}{n}$

Test statistic:

  $z=\frac{\widehat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}$

Calculation:

The sample proportion is,

  $\widehat{p}=\frac{26}{39}=0.6667$

Null and alternative hypothesis:

  $\begin{array}{l}{H}_o:p=0.533\\ H_a:p>0.533\end{array}$

This corresponding to a right-tailed test, for which a z-test for one population proportion needs to be used.

Test statistic:

  $z=\frac{0.6667-0.533}{\sqrt{0.533(1-0.533)/39}}=1.67$

Test statistic z = 1.67

Using the P-value approach:

P(Z> 2.67) = 1-P(Z<1.67)

P(Z> 2.67) = 1-0.9525 = 0.0475

Using excel formula, =NORMSDIST(1.67)

The p-value is p = 0.0475

And a = 0.05

Since p = 0.0475<0.05, it is concluded that the Null Hypothesis is rejected.

Conclusion: There is enough evidence to support the claim that Shaq has significantly improved his free-throw shooting.