A survey has been conducted to see how many people in a town of 40,000 people used Ebay to purchase a product last year. A simple random sample of 230 is taken and from this sample 52 people had used Ebay last year. I'm asked to estimate the total number of people who used Ebay last year and compute the 95% confidence interval for the total (we are allowed to assum the 97.5% quantile of the normal distribution is 1.96). The first part I believe is straightforward, I just find my ratio R and multiply by my population total t to get 52/230⋅40,000=9043.4783 (t^) but finding the confidence interval is a bit confusing because I'm hardly given any information here. I have formulas I would usually use to find the variance of R and then the variance of t^ and then find the confidence interval but

Zackary Duffy

Zackary Duffy

Answered question

2022-09-11

A survey has been conducted to see how many people in a town of 40,000 people used Ebay to purchase a product last year. A simple random sample of 230 is taken and from this sample 52 people had used Ebay last year.
I'm asked to estimate the total number of people who used Ebay last year and compute the 95% confidence interval for the total (we are allowed to assum the 97.5% quantile of the normal distribution is 1.96).
The first part I believe is straightforward, I just find my ratio R and multiply by my population total t to get 52/230⋅40,000=9043.4783 ( t ^ ) but finding the confidence interval is a bit confusing because I'm hardly given any information here. I have formulas I would usually use to find the variance of R and then the variance of t ^ and then find the confidence interval but all these formulas either require y ¯ (and sometimes x ¯ ) or they require y i 2 , y i x i , x i 2 and I have none of these. I only have the total population, the sample size and the ratio. Any ideas how I would obtain my 95% confidence interval for the total?
Also, there is a second part that asks what sample size would be required for the total number to be within 940 units of the true value (I'm guessing they mean the total number of people in the town who used ebay last year), with confidence 95%? On this part of the question I'm just not sure what to do.
These are both low mark questions so I'm probably just forgetting a formula or missing something but I just can't see a way to get my answers with so little information given. I've double and triple checked and this is for sure the only information that is given about this survey.

Answer & Explanation

Mateo Tate

Mateo Tate

Beginner2022-09-12Added 18 answers

EDIT: I changed "without" to "with." I mean "with" replacement so each "trial" is identical to any other "trial."
Since 230 is much smaller than 40,000, you can think of the sample as a sample with replacement. In this case, you have a probability of success p (the "true ratio of people who use Ebay") and 230 is your sample size.
Thus, you have X B i n ( 230 , p )and the observed value of X is X=53. You can then use a normal approximation to the binomial distribution using the ML estimate p ^ := X n with observed value p ^ = 53 230 0.23.
We know X N o r m ( p , n p ( 1 p ) ) so p ^ N o r m ( p , p ( 1 p ) n ) .. The 95% approximate CI for p ^ is therefore
0.23 ± 1.97 ( 0.23 ) ( 0.77 ) 230 0.23 ± 2 × 0.0277 0.23 ± 0.0555 = ( 0.1745 , 0.2855 ) .

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