A Web-based retailer wishes to measure whether pay-per-click advertising using Google AdWords results in higher daily sales. To do so, it signs up for an 80-day free trial period. a) In the first 80 days of using AdWords, the retailer had average sales of $350,000 per day, with a standard deviation of $25,000. Given the sample evidence, compute the 95% and 99% confidence intervals for , the average daily sales revenue using AdWords. Solution: We have n = 80, X-bar = 350,000 and = 25,000 95% confidence interval is given by:X-bar Z (0.025)*/n 350,000 1.96*25,000/80 350,000 1.96*2795.08 350,000 5478.37 344,521.63, 355,478.37 99% confidence interval is given by:X-bar Z (0.005)*/n 350,000 2.58*25,000/80 350,000 2.58*2795.08 350,000 7211.31 342,788.69, 357,211.31 b) Before using AdWords, the retailer had an average sales revenue of $345,000 per day. Is the sample evidence sufficient to conclude that the daily sales revenue has increased? To answer this question, perform a hypothesis test at = .05. State the alternative and null hypotheses, compute the observed significance level (p-value), and state the conclusion of your test. What assumptions are necessary for the inferences of parts a and b to be valid? Solution: Null Hypothesis (Ho): 345,000 Alternative Hypothesis (Ha): > 345,000 Test statistics Z = (X-bar - )/ (/n) = (350,000 - 345,000)/ (25,000/80) = 1.789 P (Z > 1.789) = 0.0368 Since p-value is less than 0.05, we have sufficient evidence to reject the null hypothesis. We can conclude that the daily sales revenue has increased. The assumptions are:1) Sample is large (greater than 30). 2) Population standard deviation is known or sample standard deviation is unknown. c) In the first 20 days of using AdWords, the retailer had average sales of $375,000 per day, with a standard deviation of $27,000. Using only the sample evidence from these 20 days, compute the 95% and 99% confidence intervals for , the average daily sales revenue using AdWords. Solution: We have n = 20, X-bar = 375,000 and s = 27,000 95% confidence interval is given by:X-bar t (0.025, 19)*s/n 375,000 2.093*27,000/20 375,000 2.093*6037.38 375,000 12636.24 362,363.76, 387,636.24 99% confidence interval is given by:X-bar t (0.005, 19)*/n 375,000 2.861*27,000/20 375,000 2.861*6037.38 375,000 17272.94 357,727.06, 392,272.94 d) Is the sample evidence from these 20 days sufficient to conclude that the daily sales revenue has increased from the historical mean of $345,000 per day? To answer this question, perform a hypothesis test at = .05. State the alternative and null hypotheses, compute the observed significance level (p-value), and state the conclusion of your test. What assumptions are necessary for the inferences of parts c and d to be valid? Solution: Null Hypothesis (Ho): 345,000 Alternative Hypothesis (Ha): > 345,000 Test statistics t = (X-bar - )/ (/n) = (375,000 - 345,000)/ (27,000/20) = 4.969 Degrees of freedom = n - 1 = 20 - 1 = 19 P (t > 4.969) = 0.000 Since p-value is less than 0.05, we have sufficient evidence to reject the null hypothesis. We can conclude that the daily sales revenue has increased. The assumptions are:1) Sample is small (less than 30). 2) Population standard deviation is unknown or sample standard deviation known. e) The retailer has the option of extending its AdWords subscription beyond the 80-day free trial period. Based on all of the sample evidence, should the retailer continue using AdWords? Explain your answer, and resolve any apparent contradictions with your answers to parts b and d. (Hint: has the daily revenue stayed constant over time?) Solution: The retailer should continue using Ad Words because hypothesis is rejected and daily sales revenue has increased. There are no contradictions for answers to parts b and d, in both cases hypothesis is rejected and leads to the conclusion that daily sales revenue has increased. Statistical inference for the population proportion To find the effect of a recent policy change on employee morale, a large corporation decides to conduct an opinion survey, asking N randomly selected employees whether they are satisfied with the new policy. How many employees must be sampled in order to guarantee a sampling error (95% confidence interval) within +/- 2%? Here, confidence level = 95% Z /2=Z 0.025 =1.96 . Margin of error, E=2 =0.02 . Since no prior estimate for population proportion is known, assume it as The required size of the sample, ^p=0.50 Z/ 2 2 N= p^ ( 1 ^p ) E ( ) 1.96 2 ( 0.50 ) ( 10.50 ) 0.02 2,401 Therefore, 2,401 employees must be sampled in order to guarantee a sampling error (95% confidence interval) within +/- 2%. ( ) A survey is conducted, using the value of N chosen in part a. This survey reveals that 62.5% of the sampled employees are satisfied with the new policy. What is the 95% confidence interval for p, the proportion of all company employees that are satisfied with the new policy? What is the 99% confidence interval for p? Here, sample size N=2,401 . Sample proportion, ^p=62.5 =0.625 The 95% confidence interval for p, the proportion of all company employees that are satisfied with the new policy is, { ^p Z /2 }{ ^p (1 ^p ) 0.625 ( 10.625 ) = 0.625 1.96 N 2,401 { 0.625 0.019 } { 0.606, 0.644 } } The 99% confidence interval for p, the proportion of all company employees that are satisfied with the new policy is, { ^p Z /2 }{ ^p (1 ^p ) 0.625 ( 10.625 ) = 0.625 2.576 N 2,401 } { 0.625 0.025 } { 0.600, 0.650 } Is the sample evidence sufficient to conclude that more than 60% of employees are satisfied with the new policy? To answer this question, perform a hypothesis test at = .05. State the alternative and null hypotheses, compute the observed significance level (p-value), and state the conclusion of your test. The null and alternative hypotheses for this test are, H0: p 0.60 Ha: p > 0.60 = .05 Here, sample size N=2,401 . Sample proportion, ^p=62.5 =0.625 Test statistic, p-value, z= p^ p 0.6250.60 = =2.50 p ( 1 p ) / N ( 0.60 ) (10.60 ) /2,401 p=P ( Z > z )=P ( Z>2.50 )=1( 2.50 )=10.9938=0.0062 Since p = .0062 < 0.05, reject the null hypothesis. Therefore, data provides sufficient evidence to conclude that more than 60% of employees are satisfied with the new policy. What are the potential effects of Type I and Type II errors for this hypothesis test? How would the proportions of these errors be changed if you were to use a lower value of , such as = .01, instead of = .05? The Type I error for this test is concluding that more than 60% of employees are satisfied with the new policy, actually data provides no sufficient evidence to that fact. The Type II error for this test is concluding that no more than 60% of employees are satisfied with the new policy, actually data provides sufficient evidence to the fact that more than 60% of employees are satisfied with the new policy. When a lower value of , such as = .01, instead of = .05 indicates a significant decrease in the proportion of committing Type I error, it slightly increases the proportion of committing Type II error