Fifty-six percent of respondents to an online poll said that they were Perry Como fans. If 982 randomly selected people responded to this poll, what is the true proportion of all local residents who are Perry Como fans? Estimate at the 95% confidence level

Expert Answers

An illustration of the letter 'A' in a speech bubbles

For 95% confidence interval we can calculate the true proportion as;

`P = barp+-1.96sqrt((barp(1-barp))/n)`

Where;

P = true proportion of perry como fans

`barp` = Observed proportion of perry como fans

n = sample size

In the above equation 1.96 stands for 95% confidence interval. 

If you need 90% confidence...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

For 95% confidence interval we can calculate the true proportion as;

`P = barp+-1.96sqrt((barp(1-barp))/n)`

Where;

P = true proportion of perry como fans

`barp` = Observed proportion of perry como fans

n = sample size

In the above equation 1.96 stands for 95% confidence interval. 

If you need 90% confidence interval you should use 1.65 and for 99% confidence interval 2.58 respectively.

 

According to our data;

`barp = 56% = 0.56`

`n = 982`

 

`P = 0.56+-1.96sqrt((0.56(1-0.56))/982)`

For upper range P = 0.576 = 57.6%

For lower range P = 0.544 = 54.4%

 

So the true proportion of perry como fans is between 54.4%-57.6%.

Approved by eNotes Editorial Team