Question

A study done by researchers at a university concluded that ​% of all student athletes in this country have been subjected to some form of hazing. The study is based on responses from athletes. What are the margin of error and​ 95% confidence interval for the​ study?

Answer 1

The missing digits is highlighted in bold form

A study done by researchers at a university concluded that ​70% of all student-athletes in this country have been subjected to some form of hazing. The study is based on responses from 1800 athletes. What are the margin of error and​ 95% confidence interval for the​ study?

Answer:

Step-by-step explanation:

Given that:

The sample proportion $\hat p = 0.70$

The sample size n = 1800

At 95% confidence interval level:

The level of significance = 1 - 0.95 = 0.05

Critical value: $Z_{0.05/2} = Z_{0.025} = 1.96$

Thus, the Margin of Error E = $Z_{\alpha/2} \times \sqrt{\dfrac{\hat p ( 1- \hat p)}{n}}$

$= 1.96 \times \sqrt{\dfrac{ 0.70 ( 1-0.70)}{1800}}$

$= 1.96 \times \sqrt{\dfrac{ 0.70 ( 0.30)}{1800}}$

$= 1.96 \times \sqrt{\dfrac{0.21}{1800}}$

$= 1.96 \times \sqrt{1.16666667\times 10^{-4}}$

the Margin of Error E = 0.021

At 95% C.I for the population proportion will be:

$= \hat p \pm Z_{\alpha/2} \sqrt{\dfrac{\hat p(1- \hat p)}{n}}$

= 0.70 ± 0.021

= (0.70 - 0.021, 0.70 + 0.021)

= 0.679, 0.721)