Mathematics: Someone please help me with this

The 95% confidence interval for the true proportion of internet users who are women in Europe is (0.3159, 0.3841).Step-by-step explanation:In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.In whichz is the zscore that has a pvalue of .A recent study of 750 Internet users in Europe found that 35% of Internet users were women.This means that 95% confidence levelSo , z is the value...

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ariqueen

26.01.2022

Mathematics

The 95% confidence interval for the true proportion of internet users who are women in Europe is (0.3159, 0.3841).

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

A recent study of 750 Internet users in Europe found that 35% of Internet users were women.

This means that $n = 750, \pi = 0.35$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.35 - 1.96\sqrt{\frac{0.35*0.65}{750}} = 0.3159$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.35 + 1.96\sqrt{\frac{0.35*0.65}{750}} = 0.3841$

The 95% confidence interval for the true proportion of internet users who are women in Europe is (0.3159, 0.3841).

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