Posted: December 13th, 2022

A development economist is studying income growth in

Question 12 ptsA development economist is studying income growth in a rural area of a developing country. The last census of the population of this area, several years earlier,showed that mean household annual income was 425 dollars, and the variance of household income was 2500 (dollars­squared). A current random sample of 100households yields a sample mean income of $433.75. Assume that household annual incomes are approximately normally distributed, and that the populationvariance is known still to be 2500. Test the null hypothesis that population mean income has not increased against the alternative hypothesis that it has increased,at a 1% level of significance.What rejection region should be used to conduct this test at a significance level of 1%? I.e., what rejection region controls the maximum probability of Type I errorat 1%?RejectifRejectifRejectifRejectifor ifQuestion 22 ptsWhat is the critical value for question 1?425433.75436.63437.88Question 32 ptsWhat is the conclusion of the hypothesis test?Fail to reject the null hypothesis.Reject the null in favor of the alternative hypothesis.Question 40 ptsIf the population mean is $420, what is the probability of rejecting the null hypothesis?Please give your answer in the following format: 0.0450 (if it is 0.045) or 0.2345 (if it is for example 0.23452...).1/8Question 50 ptsIf the population mean is $430, what is the power of the test?Please give your answer in the following format: 0.0450 (if it is 0.045) or 0.2345 (if it is for example 0.23452...).Question 60 ptsIf the population mean is $435, what is the power of the test?Please give your answer in the following format: 0.0450 (if it is 0.045) or 0.2345 (if it is for example 0.23452...).Question 72 ptsA firm manufactures metal wheels. Diameters of metal wheels produced by this process are approximately normally distributed.The variance of wheel diameters characteristic of the firm’s old production process is .01 (inches­squared). The firm’s engineers have proposed a new process.They claim that the variance of wheel diameters characteristic of the new process is less.An evaluation team wants to test the null hypothesis that the variance of the new process is no less than the variance of the old process, against the alternative thatit is less. In a random sample of size 51, the variance is 0.0042 (inches­squared).What rejection region controls the probability of Type I error at 1%?What is the form of the rejection region?Reject H0 if s2 = cvReject H0 if s2 < cv1 or if s2 > cv2Reject H0 if s2 < cvReject H0 if s2 > cvQuestion 82 ptsWhen calculating the rejection region that controls the probability of Type I error at 1%, the table value you need is the _____ percentile of the _____ randomvariable.99th percentile;1st percentile;1st percentile;99th percentile;2/8Question 92 ptsWithout looking it up, what is the table value? (You can look it up, but you should not need to.)76.155029.71Question 102 ptsWhat is the conclusion of the test?Fail to reject the null.Reject H0 in favor of the alternative hypothesis.Question 112 ptsIf the population variance of the new process is 0.0025, what is the power of the test?Question 122 ptsAs part of a study of informal financial markets in developing countries, you are investigating whether moneylenders charge “usurious” interest rates. You take arandom sample of 61 loans made by moneylenders to farmers, for seeds and fertilizer. The sample mean interest rate is 37.4 (percentage points), and the samplevariance is 16.8 (squared percentage points). Farmers who are able to obtain bank loans for seeds and fertilizer pay 35.9 percentage points.Preliminary analysis shows it is reasonable to assume that interest rates charged by moneylenders on loans for seeds and fertilizer are approximately normallydistributed. Test the null hypothesis that the mean interest rate charged by moneylenders is no greater than 35.9 percent, against the alternative that it is greater.Give the test statistic and rejection region, as well as the conclusion of the test. Use a 5% level of significance.Assume there is no difference between the creditworthiness of farmers who get bank loans for seeds and fertilizer and those who borrow from moneylenders.Otherwise, this hypothesis test would not make sense, because any observed difference in interest rates then could be due to a difference in the riskiness of theloans.What are the test statistic and its sampling distribution?by the CLT since n=61 is large.since the population is normal and is unknown.by the CLT since n=61 is large, and is unknown.3/8 since the population is normal.Question 130 pts(Contd.)Before conducting this hypothesis test using the rejection region approach, let’s conduct it using the p­value approach.What is the p­value?Please give your answer in the following format: 0.0450 (if it is 0.045) or 0.2345 (if it is for example 0.23452...).Question 140 ptsNow taking the rejection region approach, what is the critical value for the test in Question 9?Please give your answer in the following format: 0.04 (if it is 0.042) or 234.23 (if it is for example 234.228...).Question 152 ptsWhat is the conclusion of the test?Reject the null hypothesis in favor of the alternative hypothesis.The test fails to reject the null.Question 162 ptsHistorically, a real estate agency has found houses for 0.7 of its clients. The agency wants to test whether its new web site has increased the proportion of clientswho find houses (this ratio is denoted by p). The agency plans to test the null hypothesisagainst the alternativeon the basis of its salesrecord in a random sample of 15 clients.Suppose the rejection region is X>12, where X is the number of clients in the sample who find a house. What is the maximum probability of Type I error?0.12210.170040.1270.03534/8Question 172 ptsIf p = 0.8, what is the probability of Type II error (using the rejection region X > 12)?0.63720.83290.8730.602Question 182 ptsWhat rejection region controls the maximum probability of Type I error at 5%?x>14X>13NoneX>11Question 192 ptsA medical study showed that girls under sixteen who are treated for Hodgkin’s disease face an exceptionally high risk of developing a second type of cancer later inlife (New York Times, 3/21/96). The study followed 1,380 girls after their original treatment, and 88 of them, or 0.0638, developed second cancers. Only four ofthem, or 0.0029, would have been expected based on data for the population at large.Test the null hypothesis that girls treated for Hodgkin’s disease are not at increased risk for a second cancer (H0: p ≤ 0.0029) against the alternative hypothesis thatthey are at increased risk (H0: p > 0.0029). Use a 10% level of significance.What is the test statistic and its sampling distribution?Question 200 ptsWhat is your critical value?Question 210 ptsWhat is the result of your test?5/8The test fails to reject the null.Reject the null.Question 222 ptsA random sample of 1562 undergraduates enrolled in a marketing course was asked to respond on a scale from one (strongly disagree) to seven (strongly agree)with the proposition: “Advertising helps raise our standard of living.” The sample mean response was 4.27, and the sample standard deviation was 1.32. Test thenull hypothesis that the population mean is 4, against the alternative hypothesis that it is not 4. Use a 1% level of significance.The test statistic and its sampling distribution arebecause the population is normal and is unknown.by the CLT since n=1562 is large and is unknown.by the CLT since n=1562 is largeby the CLT, since n=1562 is large, and is unknown.Question 23The 99% CI interval for is2 pts. Thereforethe test fails to reject the null hypothesis.the test rejects the null in favor of the alternative.Question 242 ptsWhat is the shape of the rejection probability curve?6/8Question 252 ptsAn engineer is designing a new production process to reduce energy consumption per unit produced. The old process had a mean level of energy consumption of50 per unit, and a standard deviation of 6. The relevant hypothesis isthe same for the new process as it was for the old (. Assume the variance of energy consumption per unit is). Also, assume energy consumption per unit is approximately normally distributed.The engineer decides to use a significance level of 10%. The engineer also decides the probability should be 90% of detecting that H0 is false in the event that μequals 48.What is the form of the rejection region?RejectifRejectifRejectifRejectiforQuestion 262 ptsWhat sample size should be utilized? (Collecting a sample is costly.)n=55n=50n=60n=65Question 272 ptsWhat is the shape of the rejection probability curve?7/88/8

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