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1. The Fox TV network is considering replacing one of its prime-time crime investigation shows with a new family-oriented comedy show. Before a final decision is made, network executives commission a sample of 450 viewers. After viewing the shows, 200 indicated they would watch the new show and suggested it replace the crime investigation show.?Estimate the value of the population proportion.?(Round your answers to 3 decimal places.)Sample size = n = 450Number of success = x = 200Population proportion = p-hat = x/n = 200/450 = 0.444Answer: 0.444??Develop a 99% confidence interval for the population proportion. (Use ?z Distribution Table.)?(Round your answers to 3 decimal places.)Confidence interval for proportion is:p-hat z?/2(p-hat(1-p-hat)/n) where p-hat = x/n = 200/450 = 4/9z?/2= z0.01/2=z0.005 = 2.576So CI is: 4/9 2.576(4/9(5/9)/450) Answer: (0.384,0.505)?2. The estimate of the population proportion should be within plus or minus 0.06, with a 95% level of confidence. The best estimate of the population proportion is 0.17. How large a sample is required? (Use?z Distribution Table.)?(Round up your answer to the next whole number.)Error = E <= 0.06, ? = 0.05 , z?/2 = z0.025 = 1.96 , p-hat = 0.17E = z?/2(p-hat(1-p-hat)/n) = 1.96(0.17(0.83)/n) <= 0.06 0.17(0.83)/n <=(0.06/1.96)2 n >= 0.17(0.83)/ (0.06/1.96)2 = 150.57So sample size must be 151 or greaterAnswer: 151 3. In a poll to estimate presidential popularity, each person in a random sample of 1,510 voters was asked to agree with one of the following statements:?The president is doing a good job.The president is doing a poor job.I have no opinion.A total of 575 respondents selected the first statement, indicating they thought the president was doing a good job.?Construct a 98% confidence interval for the proportion of respondents who feel the president is doing a good job.?(Round your answers to 3 decimal places.)Confidence interval for proportion is:p-hat z?/2(p-hat(1-p-hat)/n) where p-hat = x/n = 575/1510 = 115/302z?/2= z0.02/2=z0.01 = 2.326So CI is: 115/302 2.326(115/302(187/302)/1510) Answer: (0.352,0.410)?Based on your interval in part (a), is it reasonable to conclude that a majority of the population believes the president is doing a good job? Yes or NoNo because the interval is below 0.54. Suppose the U.S. president wants to estimate the proportion of the population that supports his current policy toward revisions in the health care system. The president wants the estimate to be within 0.05 of the true proportion. Assume a 99% level of confidence. The president’s political advisors found a similar survey from two years ago that reported that 45% of people supported health care revisions. (Use ?z Distribution Table.)?How large of a sample is required??(Round up your answer to the next whole number.)?Error = E <= 0.05, ? = 0.01 , z?/2 = z0.005 = 2.576 , p-hat = 0.45E = z?/2(p-hat(1-p-hat)/n) = 2.576(0.45(0.55)/n) <= 0.05 0.45(0.55)/n <=(0.05/2.576)2 n >= 0.45(0.55)/ (0.05/2.576)2 = 656.94So sample size must be 657 or greaterAnswer: 657?How large of a sample would be necessary if no estimate were available for the proportion supporting current policy??(Round up your answer to the next whole number.)?Error = E <= 0.05, ? = 0.01 , z?/2 = z0.005 = 2.576 , p-hat = 0.5E = z?/2(p-hat(1-p-hat)/n) = 2.576(0.5(0.5)/n) <= 0.05 0.5(0.5)/n <=(0.05/2.576)2 n >= 0.5(0.5)/ (0.05/2.576)2 = 663.58So sample size must be 664 or greaterAnswer: 6645. HighTech, Inc. randomly tests its employees about company policies. Last year in the 470 random tests conducted, 28 employees failed the test.?Develop a 95% confidence interval for the proportion of applicants that fail the test.?(Round your answers to 3 decimal places.)??Confidence interval for proportion is:p-hat z?/2(p-hat(1-p-hat)/n) where p-hat = x/n = 28/470 = 14/235z?/2= z0.05/2=z0.025 = 1.96So CI is: 14/235 1.96(14/235(221/235)/235) Answer: (0.038,0.081)Would it be reasonable to conclude that 8% of the employees cannot pass the company policy test? Yes or NoYes, because 8% = 0.08 is in the interval6. Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 112 randomly sampled television viewers, 14 indicated that they asked their physician about using a prescription drug they saw advertised on TV.?Develop a 95% confidence interval for the proportion of viewers who discussed a drug seen on TV with their physician.?(Round your answers to 3 decimal places.)???Confidence interval for proportion is:p-hat z?/2(p-hat(1-p-hat)/n) where p-hat = x/n = 14/112 = 1/8z?/2= z0.05/2=z0.025 = 1.96So CI is: 1/8 1.96(1/8(7/8)/112) Answer: (0.064,0.186)Is it reasonable to conclude that 28% of the viewers discuss an advertised drug with their physician? Yes or NoNo, because 0.28 is not in the interval?7. A sample of 42 observations is selected from one population with a population standard deviation of 4.5. The sample mean is 100.5. A sample of 56 observations is selected from a second population with a population standard deviation of 3.8. The sample mean is 98.5. Conduct the following test of hypothesis using the 0.10 significance level.??H0?:?μ1?=?μ2H1?:?μ1?≠?μ2?Is this a one-tailed or a two-tailed test?One-tailed testTwo-tailed test?State the decision rule.?(Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)? Reject H0 if z < -1.65 or z >1.65Compute the value of the test statistic.?(Round your answer to 2 decimal places.)Z = 2.32?What is your decision regarding?H0?Reject?H0Do not reject?H08. A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.29 cups per day and 1.44 cups per day for those drinking decaffeinated coffee. A random sample of 55 regular-coffee drinkers showed a mean of 4.41 cups per day. A sample of 47 decaffeinated-coffee drinkers showed a mean of 5.16 cups per day.?Use the 0.03 significance level.?Is this a one-tailed or a two-tailed test??One-tailed test.Two-tailed test.?State the decision rule.?(Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)Z < -1.88??Compute the value of the test statistic.?(Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)Z = -2.75??What is the?p-value?p-value 0.003??What is your decision regarding?H0??Reject?H0.Do not reject?H0.9. The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginning of the growing season to protect against various insects and diseases. Two new insecticides have just been marketed: Pernod 5 and Action. To test their effectiveness, three long rows were selected and sprayed with Pernod 5, and three others were sprayed with Action. When the grapes ripened, 435 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 330 vines sprayed with Action were checked. The results are:?InsecticideNumber of Vines Checked (sample size)Number of Infested VinesPernod 543522Action33042?At the 0.10 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action??Hint: For the calculations, assume the Pernod 5 as the first sample.?State the decision rule.?(Negative amounts should be indicated by a minus sign. Do not round the intermediate values. Round your answers to 2 decimal places.)Reject Ho if z < -1.65 or z>1.65?Compute the pooled proportion.?(Do not round the intermediate values. Round your answer to 2 decimal places.)p-hat = 0.08??Compute the value of the test statistic.?(Negative amount should be indicated by a minus sign. Do not round the intermediate values. Round your answer to 2 decimal places.)Z =-3.79?What is your decision regarding the null hypothesis??RejectFail to reject10. A study was conducted to determine if there was a difference in the humor content in British and American trade magazine advertisements. In an independent random sample of 270 American trade magazine advertisements, 56 were humorous. An independent random sample of 203 British trade magazines contained 52 humorous ads. Do these data provide evidence at the 0.05 significance level that there is a difference in the proportion of humorous ads in British versus American trade magazines??State the null and alternate hypotheses.?H0?:??1?=??2H1?:??1?≠??2Make the decision rule.?(Negative amounts should be indicated by a minus sign.?Round your answers to 2 decimal places.)Reject Ho if z < -1.96 or z > 1.96?Evaluate the test statistic.?(Negative amount should be indicated by a minus sign.?Round your answer to 2 decimal places.)Z = -1.25?What is the decision regarding the null hypothesis??H0 is Rejected/not rejected. The proportions could/could not be the same..11. A sample of 44 observations is selected from one population with a population standard deviation of 3.9. The sample mean is 102.0. A sample of 46 observations is selected from a second population with a population standard deviation of 5.6. The sample mean is 100.3. Conduct the following test of hypothesis using the 0.10 significance level.??H0?:?μ1?=?μ2H1?:?μ1?≠?μ2?Is this a one-tailed or a two-tailed test?One-tailed testTwo-tailed test?State the decision rule.?(Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)Reject Ho if z < -1.65 or z>1.65?Compute the value of the test statistic.?(Round your answer to 2 decimal places.)Z = 1.68?What is your decision regarding?H0?Reject?H0Do not reject?H012. A coffee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinking regular coffee is 1.33 cups per day and 1.45 cups per day for those drinking decaffeinated coffee. A random sample of 53 regular-coffee drinkers showed a mean of 4.45 cups per day. A sample of 44 decaffeinated-coffee drinkers showed a mean of 4.95 cups per day.?Use the 0.03 significance level.?Is this a one-tailed or a two-tailed test??One-tailed test.Two-tailed test.?State the decision rule.?(Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)Z < -1.88??Compute the value of the test statistic.?(Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)?? Z = -1.76What is the?p-value?p-value = 0.04??What is your decision regarding?H0??Do not reject?H0.Reject?H0.13. It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the null hypothesis?H0?:?π?≥ 0.114. The following hypotheses are given.?H0?:?π?≤ 0.82H1?:?π?> 0.82?A sample of 140 observations revealed that?p?= 0.88. At the 0.01 significance level, can the null hypothesis be rejected??State the decision rule.?(Round your answer to 2 decimal places.)Reject H0 if z > 2.33?Compute the value of the test statistic.?(Round your answer to 2 decimal places.)Z = 1.85?What is your decision regarding the null hypothesis???Do not reject?H0.Reject?H0.15. Chicken Delight claims that 88% of its orders are delivered within 10 minutes of the time the order is placed. A sample of 90 orders revealed that 74 were delivered within the promised time. At the 0.01 significance level, can we conclude that less than 88% of the orders are delivered in less than 10 minutes??What is the decision rule??(Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)Reject H0 if z < -2.33?Compute the value of the test statistic.?(Negative amount should be indicated by a minus sign. Round the intermediate values and final answer to 2 decimal places.)Z = -1.69?What is your decision regarding the null hypothesis? Fail to reject Ho?16. After a losing season, there is a great uproar to fire the head football coach. In a random sample of 280 college alumni, 112 favor keeping the coach. Test at the .05 level of significance whether the proportion of alumni who support the coach is less than 50 percent.(a)State the null hypothesis and the alternate hypothesis.?(Round your answers to 2 decimal places.)????H0: ?π?≥0.5????H1: ?π?<0.5??(b)State the decision rule for .05 significance level.?(Negative amount should be indicated by a minus sign.?Round your answer to 2 decimal places.)??Reject?H0?if?z?is <?-1.65(c)Compute the value of the test statistic.?(Negative amount should be indicated by a minus sign.?Round your answer to 3 decimal places.) ??Value of the test statistic-3.347??(d)Test at the .05 level of significance whether the proportion of alumni who support the coach is less than 50 percent.??? ?? ?reject?H0. There is??? ?? ?enough?evidence to conclude the population proportionof alumni supporting the coach is less than .50.17. Past experience at the Crowder Travel Agency indicated that 44% of those persons who wanted the agency to plan a vacation for them wanted to go to Europe. During the most recent season, a sampling of 1,000 persons was selected at random from the files. It was found that 480 persons wanted to go to Europe on vacation. Has there been a significant shift upward in the percentage of persons who want to go to Europe? Test at the 0.05 significance level.?a. State the null and alternate hypotheses.H0?:?π?≤ 0.44H1?:?π?> 0.44?Make the decision rule.?(Round your answer to 2 decimal places.)?Reject Ho if z > 1.65 Evaluate the test statistic.?(Round your answer to 2 decimal places.)?Z = 2.55What is the decision regarding the null hypothesis?Reject H0? ................
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