Mathematics: On the basis of a survey of 545 television viewers, a statistician has constructed a confidence interval and estimated that the proportion of people who watched the season premiere of Glee is between .16 and .24. What level of confidence (to the nearest percent, not a proportion) did the statistician use in constructing this interval

feelsbadmanStep-by-step explanation:...

On the basis of a survey of 545 television viewers,

Ответ:

zarzuba

17.03.2022

Mathematics

Confidence level of 98%.

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 $1 - \frac{\alpha}{2}$ .

Estimate:

The estimate $\pi$ is the mean of the two endpoints of the confidence interval. So

$\pi = \frac{0.16+0.24}{2} = \frac{0.40}{2} = 0.20$

Survey of 545 television viewers

This means that n = 545

What level of confidence (to the nearest percent, not a proportion) did the statistician use in constructing this interval

First we find z. The upper end is 0.24. So

$0.20 \pm z\sqrt{\frac{0.2*0.8}{545}} = 0.24$

$z\sqrt{\frac{0.2*0.8}{545}} = 0.04$

0.017z = 0.04

$z = \frac{0.04}{0.017}$

z = 2.33

z = 2.33 has a pvalue of 0.99.

$1 - \frac{\alpha}{2} = 0.99$

$\frac{\alpha}{2} = 0.01$

$\alpha = 0.02$

$1 - \alpha = 1 - 0.02 = 0.98$

So a confidence level of 98%.

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