Mathematics: In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television.

Their is not another shape so you cannot calculate the minimum but you can calculate the maximum, 6*11, which is 66 km....

In a random sample of 200 people, 154 said that

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29.01.2022

Mathematics

The 90% confidence interval of the true proportion of people who watched educational television is (0.7117, 0.8283)

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:

$n = 200, \pi = \frac{154}{200} = 0.77$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so Z = 1.645 .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.96\sqrt{\frac{0.77*0.23}{200}} = 0.7117$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.96\sqrt{\frac{0.77*0.23}{200}} = 0.8283$

The 90% confidence interval of the true proportion of people who watched educational television is (0.7117, 0.8283)

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In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television.

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