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Math  /  Data & Statistics

QuestionAny basketball fan knows that Shaquille O'Neal, one of the NBA's most dominant centers of the last twenty years, always had difficulty shooting free throws. Over the course of his career, his overall made free-throw percentage was 53.3%53.3 \%. During one off season, Shaq had been working with an assistant coach on his free-throw technique. During the next season, a simple random sample showed that Shaq made 26 of 39 free-throw attempts. Test the claim at the 0.05 SL that Shaq has significantly improved his free-throw shooting. H0:p=−0.533H_{0}: p=\quad-0.533 Ha\mathrm{H}_{\mathrm{a}} :p == □\square 0.533 p-hat: □\square 0.667

Studdy Solution

STEP 1

1. The null hypothesis H0 H_0 is that Shaq's free-throw percentage has not changed and is p=0.533 p = 0.533 .
2. The alternative hypothesis Ha H_a is that Shaq's free-throw percentage has improved, i.e., p>0.533 p > 0.533 .
3. The sample proportion p^ \hat{p} is calculated from the sample data.
4. A significance level (α\alpha) of 0.05 is used for the hypothesis test.
5. The sample size is sufficiently large for the normal approximation to be valid.

STEP 2

1. Calculate the sample proportion p^ \hat{p} .
2. Formulate the null and alternative hypotheses.
3. Calculate the test statistic.
4. Determine the critical value or p-value.
5. Make a decision to accept or reject the null hypothesis.

STEP 3

Calculate the sample proportion p^ \hat{p} :
p^=Number of successful free throwsTotal number of attempts=2639≈0.667 \hat{p} = \frac{\text{Number of successful free throws}}{\text{Total number of attempts}} = \frac{26}{39} \approx 0.667

STEP 4

Formulate the hypotheses:
H0:p=0.533 H_0: p = 0.533 Ha:p>0.533 H_a: p > 0.533

STEP 5

Calculate the test statistic using the formula for the z-test for proportions:
z=p^−p0p0(1−p0)n z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
where p0=0.533 p_0 = 0.533 and n=39 n = 39 .
z=0.667−0.5330.533×(1−0.533)39 z = \frac{0.667 - 0.533}{\sqrt{\frac{0.533 \times (1-0.533)}{39}}}

STEP 6

Calculate the standard error:
Standard Error=0.533×0.46739≈0.079 \text{Standard Error} = \sqrt{\frac{0.533 \times 0.467}{39}} \approx 0.079
Calculate the z-score:
z=0.667−0.5330.079≈1.696 z = \frac{0.667 - 0.533}{0.079} \approx 1.696

STEP 7

Determine the critical value for a one-tailed test at α=0.05\alpha = 0.05:
The critical value zα z_{\alpha} is approximately 1.645.

STEP 8

Compare the calculated z-score to the critical value:
Since z=1.696 z = 1.696 is greater than zα=1.645 z_{\alpha} = 1.645 , we reject the null hypothesis.
Conclusion: There is sufficient evidence at the 0.05 significance level to conclude that Shaq has significantly improved his free-throw shooting.

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