大学课件-商务与经济统计习题答案（第8版中文版）SBE8-SM09.doc

Hypothesis TestingChapter 9Hypothesis TestingLearning Objectives1.Learn how to formulate and test hypotheses about a population mean and/or a population proportion.2.Understand the types of errors possible when conducting a hypothesis test.3.Be able to determine the probability of making various errors in hypothesis tests.4.Know how to compute and interpret p-values.5.Be able to determine the size of a simple random sample necessary to keep the probability of hypothesis testing errors within acceptable limits.6.Know the definition of the following terms:null hypothesisone-tailed testalternative hypothesistwo-tailed testtype I errorp-valuetype II erroroperating characteristic curvecritical valuepower curvelevel of significanceSolutions:1.a.H0:m 600Managers claim.Ha:m > 600b.We are not able to conclude that the managers claim is wrong.c.The managers claim can be rejected. We can conclude that m > 400.2.a.H0:m 14Ha:m > 14Research hypothesisb.There is no statistical evidence that the new bonus plan increases sales volume.c.The research hypothesis that m > 14 is supported. We can conclude that the new bonus plan increases the mean sales volume.3.a.H0:m = 32Specified filling weightHa:m 32Overfilling or underfilling existsb.There is no evidence that the production line is not operating properly. Allow the production process to continue.c.Conclude m 32 and that overfilling or underfilling exists. Shut down and adjust the production line.4.a.H0:m 220Ha:m < 220Research hypothesis to see if mean cost is less than $220.b.We are unable to conclude that the new method reduces costs.c.Conclude m < 220. Consider implementing the new method based on the conclusion that it lowers the mean cost per hour.5.a.The Type I error is rejecting H0 when it is true. In this case, this error occurs if the researcher concludes that the mean newspaper-reading time for individuals in management positions is greater than the national average of 8.6 minutes when in fact it is not.b.The Type II error is accepting H0 when it is false. In this case, this error occurs if the researcher concludes that the mean newspaper-reading time for individuals in management positions is less than or equal to the national average of 8.6 minutes when in fact it is greater than 8.6 minutes.6.a.H0:m 1The label claim or assumption.Ha:m > 1b.Claiming m > 1 when it is not. This is the error of rejecting the products claim when the claim is true.c.Concluding m 1 when it is not. In this case, we miss the fact that the product is not meeting its label specification.7.a.H0:m 8000Ha:m > 8000Research hypothesis to see if the plan increases average sales.b.Claiming m > 8000 when the plan does not increase sales. A mistake could be implementing the plan when it does not help.c.Concluding m 8000 when the plan really would increase sales. This could lead to not implementing a plan that would increase sales.8.a.H0:m 220Ha:m < 220b.Claiming m < 220 when the new method does not lower costs. A mistake could be implementing the method when it does not help.c.Concluding m 220 when the method really would lower costs. This could lead to not implementing a method that would lower costs.9.a.z = -1.645Reject H0 if z < -1.645b.Reject H0; conclude Ha is true.10.a.z = 2.05Reject H0 if z > 2.05b.c.Area from z = 0 to z = 1.36 is .4131p-value = .5000 - .4131 = .0869d.Do not reject H011.Reject H0 if z < -1.645a. Reject H0b. Do not reject H0 c. Do not reject H0d. Reject H012.a.p-value = .5000 - .4656 = .0344Reject H0b.p-value = .5000 - .1736 = .3264Do not reject H0c.p-value = .5000 - .4332 = .0668Do not reject H0d.z = 3.09 is the largest table value with .5000 - .4990 = .001 area in tail. For z = 3.30, the p-value is less than .001 or approximately 0. Reject H0.e.Since z is to the left of the mean and the rejection region is in the upper tail, p-value = .5000 + .3413 = .8413. Do not reject H0.13.a.H0:m 1056Ha:m < 1056 b.Reject H0 if z < -1.645c.d.Reject H0 and conclude that the mean refund of “last minute” filers is less than $1056.e.p-value = .5000 - .4664 = .033614.a.z.01 = 2.33Reject H0 if z > 2.33b.c.Reject H0; conclude the mean television viewing time per day is greater than 6.70.15.a.z.05 = 1.645Reject H0 if z < -1.645b.c.Reject H0; conclude that the mean sales price of used cars is less than the national average.16.a.H0:m 13Ha:m < 13b.z.01 = 2.33Reject H0 if z < -2.33c.d.Reject H0; conclude Canadian mean internet usage is less than 13 hours per month. Note: p-value = .00217.a.H0:m 15Ha:m > 15b.c.p-value = .5000 - .4985 = .0015d.Reject H0; the premium rate should be charged.18.a.H0:m 5.72Ha:m > 5.72b.c. p-value = .5000 - .4830 = .0170d.p-value < a; reject H0. Conclude teens in Chicago have a mean expenditure greater than 5.72. 19.a.H0:m 181,900Ha:m < 181,900b.c.p-value = .5000 - .4983 = .0017d.p-value < a; reject H0. Conclude mean selling price in South is less than the national mean selling price. 20.a.H0:m 37,000Ha:m > 37,000b.c. p-value = .5000 - .4292 = .0708d.p-value > a; do not reject H0. Cannot conclude population mean salary has increased in June 2001.21.a.Reject H0 if z < -1.96 or z > 1.96b.Reject H0; conclude Ha is true.22.a.Reject H0 if z < -2.33 or z > 2.33b.c.p-value = (2) (.5000 - .3708) = .2584d.Do not reject H023.Reject H0 if z < -1.96 or z > 1.96a. Reject H0b. Do not reject H0c. Do not reject H0d. Reject H024.a.p-value = 2 (.5000 - .4641) = .0718Do not reject H0b.p-value = 2(.5000 - .1736) = .6528Do not reject H0c.p-value = 2(.5000 - .4798) = .0404Reject H0d.approximately 0Reject H0e.p-value = 2(.5000 - .3413) = .3174Do not reject H025. a.z.025 = 1.96Reject H0 if z < -1.96 or z > 1.96b.c.Do not reject H0. Cannot conclude a change in the population mean has occurred.d.p-value = 2(.5000 - .4382) = .123626.a.H0:m = 8Ha:m 8Reject H0 if z < -1.96 or if z > 1.96b.c.Do not reject H0; cannot conclude the mean waiting time differs from eight minutes.27.a.H0:m = 16Continue productionHa:m 16Shut downReject H0 if z < -1.96 or if z > 1.96b.Reject H0 and shut down for adjustment.c.Do not reject H0; continue to run.d.For = 16.32, p-value = 2 (.5000 - .4857) = .0286For = 15.82, p-value = 2 (.5000 - .3907) = .218628.a.H0:m = 1075Ha:m 1075b.c.p-value = 2(.5000 - .4236) = .1528d.Do not reject H0. Cannot conclude a change in mean amount of charitable giving.29.a.H0:m = 15.20Ha:m 15.20b.p-value = 2(.5000 - .3554) = .2892c.Do not reject H0; the sample does not provide evidence to conclude that there has been a change.30.a.H0:m = 26,133Ha:m 26,133b.c.p-value = 2(.5000 - .4817) = .0366d.p-value < a; reject H0. Conclude population mean wage in Collier County differs from the state mean wage.31.a.935 25 or 910 to 960Since 900 is not in the interval, reject H0 and conclude m 900.b.Reject H0 if z < -1.96 or if z > 1.96Reject H0c.p-value = 2(.5000 - .4970) = .006032.a.The upper 95% confidence limit is computed as follows:Thus, we are 95% confident that m is $14.66 per hour or less.b.Since $15.00 is not in the interval $14.66 per hour or less, we reject H0.Conclude that the mean wage rate is less than $15.00.33.a.With 15 degrees of freedom, t.05 = 1.753Reject H0 if t > 1.753b.Do not reject H034.a. = 108 / 6 = 18b.c.Reject H0 if t < -2.571 or t > 2.571d.e.Reject H0; conclude Ha is true.35. Reject H0 if t < -1.721a.Do not reject H0 b.Reject H0c.Do not reject H0d.Do not reject H036.Use the t distribution with 15 degrees of freedoma.p-value = .01Reject H0b.p-value = .10Do not reject H0c.p-value is between .025 and .05Reject H0d.p-value is greater than .10Do not reject H0e.p-value is approximately 0Reject H037.a.H0:m = 3.00Ha:m 3.00b.t.025 = 2.262Reject H0 if t < -2.262 or if t > 2.262c.d.e.f.Do not reject H0; cannot conclude the population means earning per share has changed.g.t.10 = 1.383p-value is greater than .10 x 2 = .20Actual p-value = .391638.a.t.025 = 2.064 24 degrees of freedomReject H0 if t < -2.064 or if t > 2.064b.c.Do not reject H0; cannot conclude the mean expenditure in Corning differs from the U.S. mean expenditure.39.a.t.05 = 1.895 7 degrees of freedomb.c.d.c.Reject H0; conclude that the mean number of hours worked exceeds 55.40.a.H0:m = 4000Ha:m 4000b.t.05 = 2.160 13 degrees of freedomReject H0 if t < -2.160 or if t > 2.160c.d.Do not reject H0; Cannot conclude that the mean cost in New City differs from $4000.e.With 13 degrees of freedomt.05 = 1.771t.10 = 1.3501.63 is between 1.350 and 1.771. Therefore the p-value is between .10 and .20.41. a.H0:m 280Ha:m > 280b.286.9 - 280 = 6.9 yardsc.t.05 = 1.860 with 8 degrees of freedomd.e.Reject H0; The population mean distance of the new driver is greater than the USGA approved driver.f.t.05 = 1.860t.025 = 2.306p-value is between .025 and .05Actual p-value = .036142.a.H0:m 2Ha:m > 2b.With 9 degrees of freedom, reject H0 if t > 1.833c. = 24 / 10 = 2.4d.e.f.Reject H0 and claim m is greater than 2 hours. For cost estimating purposes, consider using more than 2 hours of labor time.g.t.025 = 2.262, t.01 = 2.821p-value is between .025 and .01.43.a.Reject H0 if z > 1.645b.Reject H044.a.Reject H0 if z < -1.96 or z > 1.96b.c.p-value = 2 (.5000 - .3944) = .2122d.Do not reject H0.45.Reject H0 if z < -1.645a.p-value = .5000 - .4974 = .0026Reject H0.b.p-value = .5000 - .3849 = .1151Do not reject H0.c.p-value = .5000 - .4772 = .0228Reject H0.d.p-value = .5000 + .2881 = .7881Do not reject H0.46.a.H0:p .40Ha:p > .40b. Reject H0 if z > 1.645c.= 188/420 = .4476d.Reject H0. Conclude that there has been an increase in the proportion of users receiving more than ten e-mails per day. 47.a.z.05 = 1.645Reject H0 if z < -1.645b.= 52/100 = .52c.Reject H0. Conclude less than 64% of shoppers believe supermarket ketchup is as good as the national brand.d.p-value = .5000 - .4938 = .006248.a.= 285/500 = .57b.c.z = 3.13 is not in the table. Closest value is z = 3.09. Thus, p-value is approximately.5000 - .4990 = .001d.p-value < .01, Reject H0. Over 50% prefer Burger King fries.e.Yes; the statistical evidence shows Burger King fries are preferred. The give-away was a good way to get potential customers to try the new fries. 49.a.H0:p = .48Ha:p .48b. z.025 = 1.96Reject H0 if z < -1.96 or if z > 1.96c.= 368/800 = .45d.d.Do not reject H0. Cannot conclude the proportion of drivers who do not stop has changed. 50.a.= 67/105 = .6381 (about 64%)b.c.p-value = 2(.5000 - .4977) = .0046d.p-value < .01, reject H0. Conclude preference is for the four ten-hour day schedule. 51.a.H0:p = .44Ha:p .44b. = 205/500 = .41p-value = 2(.5 - .4115) = .1770Do not reject H0. Cannot conclude that there has been a change in the proportion of repeat customers.c.= 245/500 = .49p-value = 2(.5 - .4878) = .0244Reject H0. conclude that the proportion of repeat customers has changed. The point estimate of the percentage of repeat customers is now 49%.52.a.b.p-value = .5000 - .3849 = .1151c.Do not reject H0. Cannot conclude the managers claim is wrong based on this sample evidence.53.H0:p .15Ha:p > .15Reject H0 if z > 2.33= 88/500 = .176Do not reject H0; p .15 cannot be rejected. Thus the special offer should not be initiated.p-value = .5000 - .4484 = .051654.a.H0:p .047Ha:p < .047b.= 35/1182 = .0296c.d.p-value = .5000 - .4976 = .0024e.p-value < a, reject H0. The error rate for Brooks Robinson is less than the overall error rate. 55.H0:p .20Ha:p < .20Reject H0 if z < -1.645= 83/596 = .1393p-value 0Reject H0; conclude that less than 20% of workers would work for less pay in order to have more personal and leisure time.56.c = 10 - 1.645 (5 /) = 9.25Reject H0 if < 9.25a.When m = 9,Prob (H0) = (.5000 - .2088) = .2912b.Type II errorc.When m = 8,b = (.5000 - .4969) = .003157.Reject H0 if z < -1.96 or if z > 1.96c1 = 20 - 1.96 (10 /) = 18.61c2 = 20 + 1.96 (10 /) = 21.39a.m = 18b = .5000 - .3051 = .1949b.m = 22.5b = .5000 - .4418 = .0582c.m = 21b = .5000 +.2088 = .708858.a.H0:m 15Ha:m > 15Concluding m 15 when this is not true. Fowle would not charge the premium rate even though the rate should be charged.b.Reject H0 if z > 2.33Solve for = 16.58Decision Rule:Accept H0 if 16.58Reject H0 if > 16.58For m = 17,b = .5000 -.2324 = .2676c.For m = 18, b = .5000 -.4821 = .017959.a.H0:m 25Ha:m < 25Reject H0 if z < -2.05Solve for = 23.88