Need answer for PartC. I got answer as (859.8 - 1098.2), but it is wrong. When a

Question

Need answer for PartC. I got answer as (859.8 - 1098.2), but it is wrong.

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 1,700 clicks coming into a company's site during a week identified 230 of these clicks as fraudulent. Complete parts (a) through (c) below Identify the necessary conditions. Select all that apply. A. The week must be randomly selected. B. The week must be representative of a typical week. C. The clicks must be selected at random D. The company must be representative of all companies that advertise on the Internet. Identify the population below A. All clicks on ads placed in search results B. All clicks into this company's site O C. All fraudulent clicks O D. All companies that advertise on the Internet (b) Show the 95% confidence interval for the population proportion of fraudulent clicks in a form suitable for sharing with a nontechnical audience. 11.9 % to 15.2 % The 95% confidence interval for the population proportion is (Round to one decimal place as needed.) (c)If a company pays the operators of a search engine S4 25 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.)

Explanation / Answer

c) p = 230/1700 = 0.135

SD, ? = sqrt(p*(1-p)/n) = sqrt(0.135*0.865/1700) = 0.0083

Pay per click by fradulent means, x = 4.25

95% CI for mean cost per click due to fraud

x +- Z*?/sqrt(n) where Z = 1.96 and n =230 as we deal with fradulent clicks

4.25 +- 1.96*0.0083/sqrt(230) = 4.25 +- 0.00107

= ( 4.24893, 4.25107)

Let me know if you need anything else, if not please don't forget to like the answer :)

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