Confidence Interval

[InterserveVB]

I am having a bit of difficulty with these. I think If I could see one done, I could get an idea of what I'm supposed to do. Below is the first problem.

In the 1988-89 GSS, 901 people were asked if they had to undergo a drug test before they were hired. 98 had to undergo the test out of those asked.

A. Are the assumptions met to conduct a confidence interval?

I'v decided that its a pop proportion. As such, I got p hat by doing 98/901. Is that correct? Then I said that since n(p hat) and n(1 - p hat) were greater than 15 that they met the assumptions. Im doing the second part, but not sure how to get Z. Is it infinite on the t chart under 99%? That would make it 2.576?

B. Make a 99% CI for p, the proportion of all Americans that have to undergo a drug test before they were hired.

C. Interpret.

 

[galactus]

The z-score for 99% is 2.575, n=901, p=98/901, q=1-(98/901)

\(\displaystyle \L\\E=z\sqrt{\frac{pq}{n}}\)

Interval:

\(\displaystyle \L\\p-E<p<p+E\)


Now, put 'er all together.

 

[InterserveVB]

Thanks! So, interval is (.082, .136) and We are 99% confident that the pop proportion of All Americans that had to under go a drug test is between .082 and .136. Is that right.