Question

An online retailer wants to estimate the number of visitors that click on their advertisement from a particular website. Of 978 page views in a day, 8% of the users clicked on the advertisement. Create a 90% confidence interval for the population proportion of visitors that click on the advertisement.

Answer 1

Answer:

(0.0657,0.0943)  is the 90% confidence interval for the population proportion of visitors that click on the advertisement.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 978

Percentage of users that clicked on advertisement = 8%

Sample proportion:

$\hat{p} = 0.08$

90% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.10} = 1.645$

Putting the values, we get:

$0.08\pm 1.645(\sqrt{\dfrac{0.08(1-0.08)}{978}})\\\\= 0.08\pm 0.0142\\\\=(0.0658,0.0942) = (6.57\%,9.43\%)$

(0.0657,0.0943)  is the 90% confidence interval for the population proportion of visitors that click on the advertisement.