Hinton Motors is testing an experimental battery­powered engine

QUESTION

1. Hinton Motors is testing an experimental battery­powered engine for its passenger cars. Hintonhas tested the range of the engine in five different trials, recording the distance traveled in eachtrial before the battery needed recharging. Trial results are given in the table:TrialDistance (in miles)18402820379048505700Produce the 90% confidence interval of the average distance for the population represented here.Assume that the population distribution is approximately normal. Report the upper bound for your interval.831.46768.92799.36857.612. Plainfield Telemarketing plans to estimate the average number of telephone contacts made by its245 sales reps over the past month. A sample of reps will be randomly selected for the estimate.To construct a 99% confidence interval estimate in which the margin of error will be no more than20 contacts, how many reps should be included in the sample? Assume a similar study showed astandard deviation of 120 contacts.2681451691212403. In a random sample of 5,000 recent government contracts, 19.3 percent of the contracts involvedcost overruns. Build a 99% confidence interval estimate of all recent government contracts thatinvolve cost overruns. Report the upper bound for your interval..087.235.199.1464.You want to estimate the proportion of households in the Portland area that have no land­linetelephone service. Suppose you plan to take a simple random sample of Portland householdsand target a margin of error no larger than plus or minus .02 for a proposed 90% confidenceinterval estimate. How large a sample would you recommend? (Assume you have no priorinformation about the likely population proportion.)102617023425132229635. Estimates vary widely as to the prevalence of cheating on campus. You plan to administer ananonymous survey to estimate the proportion of students who have cheated on a test in the pastyear. At a 95% confidence level, you would like your estimate to have a margin of error no greaterthan 3%. How large a sample should you take to ensure that you achieve your target margin oferror?267 students757 students1068 studentsInsufficient information is given to determine the necessary sample size.6. Greg would like to test whether his running pace differs between the morning and the afternoon.In a sample of 10 morning runs, his average pace was 7 minutes 50 seconds per mile (490seconds/mile), with a sample standard deviation of 35 seconds per mile. In a sample of 12afternoon runs, his average pace was 7 minutes, 30 seconds per mile (450 seconds/mile), with asample standard deviation of 32 seconds per mile.Use the information above to construct a 95% confidence interval for the difference between Greg’smorning and afternoon pace. Assume that the population standard deviations are equal. Complete thefollowing sentence: "I am 95% confident that Greg’s afternoon pace is between__________________________________ faster than his morning pace."10.2 seconds/mile and 69.8 seconds/mile12.0 seconds/mile and 68.0 seconds/mile15.3 seconds/mile and 64.7 seconds/mile34.8 seconds/mile and 45.2 seconds/mile7. A Seattle Times poll of 503 voters found that 30% of those surveyed favored the incumbentmayor in Seattle’s upcoming mayoral election. Using a 95% confidence level, the margin of errorfor the poll result was reported to be 4.5%. Choose the best interpretation of this poll result.The share of the sample that favors the incumbent mayor is between 25.5% and 34.5%,with 95% confidence.The share of voters that favors the incumbent mayor is between 25.5% and 34.5%, with95% confidence.There is a 95% chance that support for the mayor will be between 25.5% and 34.5% in theupcoming election.There is a 95% chance that 30% of voters currently favor the mayor in the upcomingelection.8. Bargain.com claims to have an average of 637 visitors to its site per hour. You take a randomsample of 100 hours and find the average number of visitors for the sample is 632 per hour. Canwe use this sample result to reject a µ ? 637 visitors null hypothesis at the 5% significancelevel? Assume the population standard deviation is 18 visitorsYes. Since the sample mean of 632 is less than the test’s critical value of 634, we can rejectthe null hypothesis.No. Since the sample mean of 632 is greater than the test’s critical value of 628, we can’treject the null hypothesis.Yes. Since the sample mean of 632 is less than the test’s critical value of 639, we can rejectthe null hypothesis.No. Since the sample mean of 632 is greater than the test’s critical value of 614, we can’treject the null hypothesis.9. The competing hypotheses for a hypothesis test are as follows:Ho: µ ? 1000 (The proportion mean is less than or equal to 1000)Ha: µ > 1000 (The population mean is greater than 1000)Assume the population standard deviation is known to be 80. A random sample of size 64 has asample mean of 1020. Use the p­value approach, and a significance level of 5%, to decidewhether you can reject the null hypothesis.The p-value for this sample result is .0228. Since this p-value is less than the significancelevel of .05, we can’t reject the null hypothesis.the p-value for this sample result is .0428. Since this p-value is less than the significancelevel of .05, we can’t reject the null hypothesis.The p-value for this sample result is .0428. Since this p-vlaue is less than the significancelevel of .05, we can reject the null hypothesis.The p-value for this sample result is .0228. Since this p-value is less than the significancelevel of .05, we can reject the null hypothesis.10. Suppose you are testing the following hypotheses:Ho: µ = 650Ha: µ ? 650Sample size is 100. The sample mean is 635. The population standard deviation is 140. Thesignificance level is .10. Compute the sample test statistic, z, and use it to decide whether toreject the null hypothesis.z = -1.87. Since this value is outside the interval -1.65 to +1.65, we can reject the nullhypothesis.z = -2.17. Since this value is outside the interval -1.65 and +1.65, we can reject the nullhypothesis.z = -1.37. Since this value is between -1.65 and +1.65, we can’t reject the null hypothesis.z = -1.07. Since this value is between -1.65 and +1.65, we can’t reject the null hypothesis.11. Trend Home Products wants its online service agents to respond to posted customer complaintsin an average of 28 minutes. You track a random sample of 12 complaints. The averageresponse time for the sample is 30.2 minutes. The sample standard deviation is 14.3 minutes. Isthis sample result sufficient to reject a null hypothesis that the average response time for allcustomer complaints is exactly 28 minutes? Use a 5% significance level. Assume the populationdistribution is approximately normal.t = .53. Since t is between -2.201 to +2.201, we can’t reject the null hypothesis.t = 2.73. Since t is outside the interval -2.332 to + 2.332, we can reject the null hypothesis.t = 2.73. Since t is outside the interval -2.201 to +2.201, we can reject the null hypothesis.t = 1.33. Since t is between -2.201 to +2.201, we can’t reject the null hypothesis.13. BarnabyToys.com claims that the average delivery time for items purchased on its website is 3.65days. A random sample of 45 recent deliveries has an average delivery time for the sample of4.03 days.Compute the p­value for the sample result and use it to test the null hypothesis that averagedelivery time for the population of BarnabyToys.com items purchased online is not more than3.65 days. Use a 5% significance level for the test and assume that the population standarddeviation is .92 days. is the sample result sufficient to reject the null hypothesis?Since the p-value of .0328 is less than the significance level of .05, we can reject the nullhypothesis.Since the p-value of .0028 is less than the significance level of 05, we can reject the nullhypothesis.Since the p-value of .0328 is less than the significance level of .05, we can’t reject the nullhypothesis.Since the p-value of .0028 is less than the significance level of .05, we can’t reject the nullhypothesis.13 . At a local coffee shop last year, the average amount spent per customer (APC) was $5.24. In orderto determine whether there has been a change in the APC, the coffee shop took a sample of 500 recenttransactions. For the 500 transactions, the APC was $5.39 with a sample standard deviation of $1.20.The median transaction was $5.15.The hypotheses test was constructed as follows:Ho: µ = 5.24Ha: µ ? 5.24At a 5% significance level, which of the following is the correct conclusion?Reject the null hypothesis because the test statistic of 2.80 is not between -1.65 and 1.65.There is not sufficient sample evidence of a change in the average amount of thetransaction.Reject the null hypothesis because the test statistic of 2.80 is not between -1.96 and 1.96.There is not sufficient evidence of a change in the average transaction amount.Reject the null hypothesis because the test statistic of 2.80 is not between -1.65 and 1.65.There is sufficient sample evidence to argue that the average transaction amount haschanged since last year.Reject the null hypothesis because the test statistic for 2.80 is not between -1.96 and 1.96.There is sufficient sample evidence to argue that the average transaction amount haschanged since last year.

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