Question

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

Answer 1

Answer:

e. 545

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

The estimate is the mean of the two bounds. So $\pi = \frac{0.82+0.88}{2} = 0.85$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

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

In this question, L = 0.82. So

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

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

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

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

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

$n = 544.23$

So the correct answer is:

e. 545

Answer 2

The sample size should be option e. 545.

Calculation of the sample size:

Since A 95% confidence interval for the population proportion of professional tennis players who earn more than 2 million dollars a year is found to be between .82 and .88.

Here the z score should be 1.96

Now the margin of error should be

$= (0.88-0.82)\div 2$

= 0.03

Now the estimation of the point is

= 0.88-0.03

= 0.85

Now the sample size should be

$0.03 = 1.96 \times \sqrt{\frac{0.85(1-0.85)}{n} }$

n = 545

hence, The sample size should be option e. 545.

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