Question

When a company advertises on the Internet, the company pays the operators of search engines each time an ad for the company appears with search results and someone clicks on the link. Click fraud is when a computer program pretending to be a customer clicks on the link. An analysis of 1,100 clicks coming into a company's site during a week identified 165 of these clicks as fraudulent. Complete parts (a) through (c) below. YA. Tne week must pe representative or a typicar week B. The company must be representative of all companies that advertise on the Internet. C. The week must be randomly selected. D. The clicks must be selected at random. ldentify the population below. A. All clicks on ads placed in search results O B. All companies that advertise on the Internet OC. All fraudulent clicks D.All clicks into this company's site (b) Show the 95% confidence interval for the population proportion of fraudulent clicks in a form suitable for sharing with a nontechnical audience The 95% confidence interval for the population proportion is [ 12.9% to 17.1 %] Round to one decimal place as needed.) (c) If a company pays the operators of a search engine $3.75 for each click, give a 95% confidence interval for the mean cost per click due to fraud. The 95% confidence interval for the mean cost is (Round to the nearest cent as needed.)

(Round to the nearest cent as​ needed.)(Round to the nearest cent as​ needed.)(Round to the nearest cent as​ needed.)

Answer 1

(a)

The population is: "All clicks into this company's site".

(b)

Sample proportion is, $\hat{p}$ =x/n =165/1100 =0.15

At 95% confidence level, critical value of Z is: Z =1.96 (for two-tailed case)

95% confidence interval for the population proportion of fraudulent clicks, P= $\hat{p}\ \pm [Z. \sqrt{\hat{p} (1-\hat{p}/n}]$ =$0.15\ \pm [1.96. \sqrt{0.15 (1-0.15/1100}]$ =[0.13, 0.17]​​​​​​ =[13%, 17%]

(c)

Thus, 95% confidence interval for the mean cost per click due to fraud is: (3.75*13%/15%, 3.75*17%/15%) =[$3.25, $4.25]