The results of a 2010 survey showed that out of 1100 randomly selec%ted facebook

Question

The results of a 2010 survey showed that out of 1100 randomly selec%ted facebook users, 29% of them said it was ok to friend his or her boss. Uing this information, I would like for you to construct a 95% confidence interval estimate for the true percentage of Facebook users who feel it is ok to friend his or her boss.

What is the critical Z value?

What is the confidence interval?

I surveyed 1100 of my closest facebook friends and found that 27% of them felt it was ok to friend the boss. How do the results of my survey compare to your calculated confidence interval?

Explanation / Answer

for alpha = 0.05   alpha / 2 = 0.025   critical value : 1.96

confidence interval

0.29 +/- 1.96 * srqt ( 0.29*0.71 / 1100 )

0.29 +/- 0.0268

0.2632 < p < 0.3168

How do the results of my survey compare to your calculated confidence interval?

since 0.27 is in the interval we can say that we are 95% sure or confident that the true proportion is between those values

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