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 [tex]\hat p = 0.70[/tex]
The sample size n = 1800
At 95% confidence interval level:
The level of significance = 1 - 0.95 = 0.05
Critical value: [tex]Z_{0.05/2} = Z_{0.025} = 1.96[/tex]
Thus, the Margin of Error E = [tex]Z_{\alpha/2} \times \sqrt{\dfrac{\hat p ( 1- \hat p)}{n}}[/tex]
[tex]= 1.96 \times \sqrt{\dfrac{ 0.70 ( 1-0.70)}{1800}}[/tex]
[tex]= 1.96 \times \sqrt{\dfrac{ 0.70 ( 0.30)}{1800}}[/tex]
[tex]= 1.96 \times \sqrt{\dfrac{0.21}{1800}}[/tex]
[tex]= 1.96 \times \sqrt{1.16666667\times 10^{-4}}[/tex]
the Margin of Error E = 0.021
At 95% C.I for the population proportion will be:
[tex]= \hat p \pm Z_{\alpha/2} \sqrt{\dfrac{\hat p(1- \hat p)}{n}}[/tex]
= 0.70 ± 0.021
= (0.70 - 0.021, 0.70 + 0.021)
= 0.679, 0.721)
