Mathematics: A study done by researchers at a university concluded that 70 70% of all student athletes in this country have been subjected to some form of hazing. The study is based on responses from 1 comma 900 1,900 athletes. What are the margin of error and 95% confidence interval for the study? The margin of error is nothing .

21.16 cups of flour just divide 127 by 6...

A study done by researchers at a university concluded

Ответ:

aperson05

28.01.2022

The margin of error is 0.0206 = 2.06 percentage points.

The 95% confidence interval for the proportion of all student athletes in this country have been subjected to some form of hazing is (0.6794, 0.7206).

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:

$n = 1900, \pi = 0.7$

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 margin of error is:

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

$M = 1.96\sqrt{\frac{0.7*0.3}{1900}} = 0.0206$

The margin of error is 0.0206 = 2.06 percentage points.

The lower limit of this interval is:

$\pi - M = 0.7 - 0.0206 = 0.6794$

The upper limit of this interval is:

$\pi + M = 0.7 + 0.0206 = 0.7206$

The 95% confidence interval for the proportion of all student athletes in this country have been subjected to some form of hazing is (0.6794, 0.7206).

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question