A 95% confidence interval for the population proportion of professional tennis players who earn more than 2 million dollars a ye

Question

7 days ago

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A 95% confidence interval for the population proportion of professional tennis players who earn more than 2 million dollars a ye

ar is found to be between .82 and .88. Given this information, the sample size that was used was approximately: a. 382 b. 233 c. 387 d. 480 e. 545

Mathematics

1 answer:

Svet_ta [12.7K] · Answer 1 · 2026-04-09T00:05:04+03:00

Response:

e. 545

Detailed explanation:

In a survey sample containing n individuals, with a success probability of $\pi$ , and a confidence level of $1-\alpha$ , the ensuing confidence interval for proportions is established.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

Wherein

z denotes the z-score corresponding to a probability value of $1 - \frac{\alpha}{2}$ .

For this scenario, we find:

The estimate averages the two bounds. Thus $\pi = \frac{0.82+0.88}{2} = 0.85$

95% confidence level

Consequently, z represents the z value corresponding to the p-value of $1 - \frac{0.05}{2} = 0.975$ , hence $Z = 1.96$ .

The lower boundary of this interval is:

$L = \pi - z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this query, L = 0.82. Therefore

$0.82 = 0.85 - 1.96\sqrt{\frac{0.85*0.15}{n}}$

$1.96\sqrt{\frac{0.85*0.15}{n}} = 0.03$

$0.03\sqrt{n} = 1.96\sqrt{0.85*0.15}$

$\sqrt{n} = \frac{1.96\sqrt{0.85*0.15}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.85*0.15}}{0.03})^{2}$

$n = 544.23$

Thus, the accurate response is:

e. 545