Answer to Question #161025 in Statistics and Probability for Sunny

Question

A SURVEY OF 100 RESPONDENTS WAS CONDUCTED TO STUDY PROPORTION PEOPLE SNORING IN THEIR SLEEP. Approximately 40% people claim that they do. Using this information, construct 90% confidence interval for the proportion of people that do snore in their sleep.

Answer 1

$n=100, p=0.4, q=1-0.4=0.6$

$x=0.4\cdot100=40$

Confidence level:

$CL=0.90, \alpha=1-0.9=0.1, \alpha/2=0.05, Z_{0.05}=1.645$

The error bound for a proportion:

$EBP=Z_{\alpha/2}\sqrt{\frac{pq}{n}}=1.645\sqrt{\frac{0.4\cdot0.6}{100}}=0.081$

$p-EBP=0.4-0.081=0.319$

$p+EBP=0.4+0.081=4.081$

The confidence interval for the true binomial population proportion:

$(0.319,4.081)$

We estimate with 90% confidence that the true percent of people that do snore in their sleep is between 31.9% and 40.81%.