Print View 1. 8/21/16, 8:19 PM Award: 10.00 points A United Nations report shows the mean family income for Mexican migrants to the United States is $25,980 per year. A FLOC (Farm Labor Organizing Committee) evaluation of 22 Mexican family units reveals a mean to be $30,540 with a sample standard deviation of $9,650. Does this information disagree with the United Nations report? Apply the 0.01 significance level. a. State the null hypothesis and the alternate hypothesis. H 0: = H 1: 25980 25980 b. State the decision rule for .01 significance level. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) Reject H0 if t is not between -2.831 and 2.831 c. Compute the value of the test statistic. (Round your answer to 2 decimal places.) Value of the test statistic 2.22 d. Does this information disagree with the United Nations report? Apply the 0.01 significance level. Do not reject Ho . This data does not contradict the report. References Worksheet 2. Difficulty: 2 Intermediate Learning Objective: 10-07 Use a t statistic to test a hypothesis. Award: 10.00 points A national grocer's magazine reports the typical shopper spends 6.5 minutes in line waiting to check out. A sample of 24 shoppers at the local Farmer Jack's showed a mean of 5.9 minutes with a standard deviation http://ezto.mheducation.com/hm.tpx Page 1 of 17 Assignment Print View 8/21/16, 8:19 PM of 2.5 minutes. Is the waiting time at the local Farmer Jack's less than that reported in the national magazine? Use the 0.025 significance level. a. What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) Reject H0: 6.5 and fail to reject H1: < 6.5 when the test statistic is greater than -2.069 . b. The value of the test statistic is -1.18 . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) c. What is your decision regarding H0? Do not reject H0 Reject H0 References Worksheet 3. Difficulty: 2 Intermediate Learning Objective: 10-07 Use a t statistic to test a hypothesis. Award: 10.00 points The amount of water consumed each day by a healthy adult follows a normal distribution with a mean of 1.44 liters. A sample of 10 adults after the campaign shows the following consumption in liters. A health campaign promotes the consumption of at least 2.0 liters per day: 1.80 1.94 1.70 1.50 1.64 1.40 1.64 1.70 1.66 1.54 At the 0.100 significance level, can we conclude that water consumption has increased? Calculate and interpret the p-value. Click here for the Excel Data File a. State the null hypothesis and the alternate hypothesis. (Round your answers to 2 decimal places.) H 0: H 1: > 1.44 1.44 b. State the decision rule for 0.100 significance level. (Round your answer to 3 decimal places.) http://ezto.mheducation.com/hm.tpx Page 2 of 17 Assignment Print View 8/21/16, 8:19 PM Reject H0 if t > 4.414 c. Compute the value of the test statistic. (Round your intermediate and final answer to 3 decimal places.) Value of the test statistic 1.383 d. At the 0.100 level, can we conclude that water consumption has increased? H0 and conclude that water consumption has increased Reject . e. Estimate the p-value. between 0.005 and 0.01 p-value is rev: 09_16_2015_QC_CS-24697 References Worksheet 4. Difficulty: 2 Intermediate Learning Objective: 10-07 Use a t statistic to test a hypothesis. Award: 10.00 points According to the Census Bureau, 3.29 people reside in the typical American household. A sample of 16 households in Arizona retirement communities showed the mean number of residents per household was 2.76 residents. The standard deviation of this sample was 1.29 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.29 persons? a. State the null hypothesis and the alternate hypothesis. (Round your answer to 2 decimal places.) H 0: H 1: < 3.29 3.29 b. State the decision rule for .05 significance level. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) Reject H0 if t < -1.753 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 http://ezto.mheducation.com/hm.tpx -1.643 Page 3 of 17 Assignment Print View 8/21/16, 8:19 PM d. Is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.29 persons? Fail to reject H0. Mean number of residents is not necessarily less than 3.29 persons. References Worksheet 5. Difficulty: 2 Intermediate Learning Objective: 10-07 Use a t statistic to test a hypothesis. Award: 10.00 points Referring to Table10-4 and with n = 100, = 400, = 9,922 and 1 = 9,900, state whether the following statement is true or false. The probability of a Type II error is 0.2912. True False References Multiple Choice 6. Diculty: 2 Intermediate Learning Objective: 10-08 Compute the probability of a Type II error. Award: 1.00 point Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shift than on the day shift. A sample of 60 day-shift workers showed that the mean number of units produced was 334, with a population standard deviation of 23. A sample of 68 nightshift workers showed that the mean number of units produced was 341, with a population standard deviation of 28 units. At the .10 significance level, is the number of units produced on the night shift larger? 1. This is a one http://ezto.mheducation.com/hm.tpx -tailed test. Page 4 of 17 Assignment Print View 8/21/16, 8:19 PM 2. The decision rule is to reject H0: d n if z < -1.28 . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) 3. The test statistic is z = -1.53 . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) 4. What is your decision regarding H0? Reject. References Worksheet 7. Difficulty: 2 Intermediate Learning Objective: 11-01 Test a hypothesis that two independent population means are equal, assuming that the population standard deviations are known and equal. Award: 1.00 point 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 decaffeinatedcoffee drinkers showed a mean of 5.16 cups per day. Use the .02 significance level. 1. This is a one -tailed test. 2. The decision rule is to reject H0: r d if z < -2.05 . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) 3. The test statistic is z = -2.75 . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) 4. What is your decision regarding H0 ? Rejected. 5. The p-value is 0.003 . References http://ezto.mheducation.com/hm.tpx Page 5 of 17 Assignment Print View 8/21/16, 8:19 PM Worksheet 8. Difficulty: 2 Intermediate Learning Objective: 11-01 Test a hypothesis that two independent population means are equal, assuming that the population standard deviations are known and equal. Award: 1.00 point Ms. Lisa Monnin is the budget director for Nexus Media Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information. Sales ($) Audit ($) 127 122 137 103 140 127 159 136 136 149 138 120 142 At the 0.01 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? Click here for the Excel Data File 1. State the decision rule. (Round your answer to 3 decimal places.) Reject H0 if t > 2. Compute the pooled estimate of the population variance. (Round your answer to 2 decimal places.) Pooled variance 3. Compute the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic 4. State your decision about the null hypothesis. (Click to select) H0 : s a 5. Estimate the p-value. (Round your answers to 3 decimal places.) p-value (Click to select) References Worksheet http://ezto.mheducation.com/hm.tpx Difficulty: 2 Intermediate Learning Objective: 11-02 Test a hypothesis that two independent population means are equal, with unknown population standard deviations. Page 6 of 17 Assignment Print View 9. 8/21/16, 8:19 PM Award: 1.00 point The null and alternate hypotheses are: H0 : d 0 H1 : d > 0 The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month. Day 1 10 9 Day shift Afternoon shift 2 11 10 3 14 13 4 18 16 At the .05 significance level, can we conclude there are more defects produced on the day shift? 1. State the decision rule. (Round your answer to 2 decimal places.) Reject H0 if t > 0.25 2. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic 5 3. What is the p-value? p-value between 0.05 and 0.1 4. What is your decision regarding H0? Reject H0 References Worksheet http://ezto.mheducation.com/hm.tpx Difficulty: 2 Intermediate Learning Objective: 11-03 Test a hypothesis about the mean population difference between paired or dependent observations. Page 7 of 17 Assignment Print View 10. 8/21/16, 8:19 PM Award: 1.00 point Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne attire with those of Calvin Klein. Assume the population standard deviations are not the same. The following is the amount ($000) earned per month by a sample of 15 Claiborne models: $4 3.9 $5 3.2 $3.4 5.8 $3.5 5.1 $5.6 6.3 $5.7 $6.8 $6.6 $3 $4.3 $4.6 $4.8 $5.2 The following is the amount ($000) earned by a sample of 12 Klein models. $3.5 5.4 $4.5 4.3 $4.1 $4.1 $3.6 $3.8 $4.5 Click here for the Excel Data File 1. Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) Degrees of freedom 2. State the decision rule for 0.1 significance level: H0: LC CK; H1: LC > CK. (Round your answer to 3 decimal places.) Reject H0 if t> 3. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic 4. Is it reasonable to conclude that Claiborne models earn more? Use the 0.1 significance level. (Click to select) H0. It is (Click to select) to conclude that Claiborne models earn more. rev: 07_28_2016_QC_CS-56420 References Worksheet http://ezto.mheducation.com/hm.tpx Difficulty: 3 Challenge Learning Objective: 11-02 Test a hypothesis that two independent population means are equal, with unknown population standard deviations. Page 8 of 17 Assignment Print View 11. 8/21/16, 8:19 PM Award: 1.00 point One of the music industry's most pressing questions is: Can paid download stores contend nose-to-nose with free peer-to-peer download services? Data gathered over the last 12 months show Apple's iTunes was used by an average of 1.73 million households with a sample standard deviation of .46 million family units. Over the same 12 months WinMX (a no-cost P2P download service) was used by an average of 2.17 million families with a sample standard deviation of .33 million. Assume the population standard deviations are not the same. 1. Find the degrees of freedom for unequal variance test. (Round down your answer to nearest whole number.) Degrees of freedom 19 2 State the decision rule for .01 significance level: H0: A = W; H1: A W . (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.) Reject H0 if t < -2.861 or t > 2.861 3. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) Value of the test statistic -2.692 4. Test the hypothesis of no difference in the mean number of households picking either variety of service to download songs. Use the .01 significance level. Do not reject H0. There is no significant difference in the mean number of households picking either variety of service to download songs. References Worksheet 12. Difficulty: 3 Challenge Learning Objective: 11-02 Test a hypothesis that two independent population means are equal, with unknown population standard deviations. Award: 1.00 point The null and alternate hypotheses are: H0: 1 2 http://ezto.mheducation.com/hm.tpx Page 9 of 17 Assignment Print View 8/21/16, 8:19 PM H1: 1 > 2 A random sample of 26 items from the first population showed a mean of 107 and a standard deviation of 9. A sample of 15 items for the second population showed a mean of 98 and a standard deviation of 7. Assume the sample populations do not have equal standard deviations. a. Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) Degrees of freedom b. State the decision rule for 0.025 significance level. (Round your answer to 3 decimal places.) Reject H0 if t> c. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic d. What is your decision regarding the null hypothesis? Use the 0.025 significance level. Null hypothesis (Click to select) References Worksheet 13. Difficulty: 3 Challenge Learning Objective: 11-02 Test a hypothesis that two independent population means are equal, with unknown population standard deviations. Award: 1.00 point A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distributions follow the normal probability distribution and the population standard deviations are not equal. The information is summarized below. Statistic Sample mean Sample standard deviation Sample size Men 24.38 5.91 33 Women 22.36 4.80 37 At the .01 significance level, is there a difference in the mean number of times men and women order takeout dinners in a month? http://ezto.mheducation.com/hm.tpx Page 10 of 17 Assignment Print View 8/21/16, 8:19 PM 1. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic 2. What is your decision regarding on null hypothesis? The decision is (Click to select) the null hypothesis that the means are the same. 3. What is the p-value? (Round your answer to 4 decimal places.) p-value rev: 01_22_2015_QC_CS-492, 10_06_2015_QC_CS-28538, 04_06_2016_QC_CS-47932 References Worksheet 14. Difficulty: 2 Intermediate Learning Objective: 11-01 Test a hypothesis that two independent population means are equal, assuming that the population standard deviations are known and equal. Award: 1.00 point Businesses, particularly those in the food preparation industry such as General Mills, Kellogg, and Betty Crocker regularly use coupons as a brand allegiance builder to stimulate their retailing. There is uneasiness that the users of paper coupons are different from the users of e-coupons (coupons disseminated by means of the Internet). One survey recorded the age of each person who redeemed a coupon along with the type (either electronic or paper). The sample of 32 e-coupon users had a mean age of 38.5 years with a standard deviation of 12.3, while a similar sample of 26 traditional paper-coupon clippers had a mean age of 38.8 with a standard deviation of 5.6. 1. Find the degrees of freedom for unequal variance test. (Round down answer to nearest whole number.) Degrees of freedom 2. State the decision rule for 0.10 significance level: H0: e = p; H1: e p. (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.) Reject H0 if t < or t > 3. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) Value of the test statistic http://ezto.mheducation.com/hm.tpx Page 11 of 17 Assignment Print View 8/21/16, 8:19 PM 4. Test the hypothesis of no difference in the mean ages of the two groups of coupon clients. Use the .10 significance level. (Click to select) H0. There is (Click to select) difference in the mean ages of the two groups of coupon clients. References Worksheet 15. Difficulty: 3 Challenge Learning Objective: 11-02 Test a hypothesis that two independent population means are equal, with unknown population standard deviations. Award: 1.00 point Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 18 people making inquiries at the first development is $165,000, with a standard deviation of $36,000. A corresponding sample of 28 people at the second development had a mean of $179,000, with a standard deviation of $25,000. Assume the population standard deviations are the same. 1. State the decision rule for .02 significance level: H0: 1 = 2; H1:1 2. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) Reject H0 if t is not between and . 2. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) Value of the test statistic 3. At the .02 significance level, can Fairfield conclude that the population means are different? (Click to select) H0. Fairfield (Click to select) conclude that the population means are different. rev: 02_17_2015_QC_CS-4586 References Worksheet http://ezto.mheducation.com/hm.tpx Difficulty: 2 Intermediate Learning Objective: 11-02 Test a hypothesis that two independent population means are equal, with unknown population standard deviations. Page 12 of 17 Assignment Print View 16. 8/21/16, 8:19 PM Award: 10.00 points Four brands of lightbulbs are being considered for use in the final assembly area of the Ford F-150 truck plant in Dearborn, Michigan. The director of purchasing asked for samples of 175 from each manufacturer. The numbers of acceptable and unacceptable bulbs from each manufacturer are shown below. At the 0.05 significance level, is there a difference in the quality of the bulbs? (Round your answers to 3 decimal places.) Manufacturer Unacceptable Acceptable Total A 49 126 B 22 153 C 21 154 D 14 161 175 175 175 175 H0: There is no relationship between quality and manufacturer. Reject H0 if X2 > H1: There is a relationship. . X2 = (Click to select) H0. There (Click to select) relationship between quality and manufacturer. References Worksheet 17. Difficulty: 2 Intermediate Learning Objective: 15-06 Perform a chisquare test for independence on a contingency table. Award: 10.00 points For many years, TV executives used the guideline that 30% of the audience were watching each of the traditional big three prime-time networks and 10% were watching cable stations on a weekday night. A random sample of 550 viewers in the Tampa-St. Petersburg, Florida, area last Monday night showed that 110 homes were tuned in to the ABC affiliate, 110 to the CBS affiliate, 165 to the NBC affiliate, and the http://ezto.mheducation.com/hm.tpx Page 13 of 17 Assignment Print View 8/21/16, 8:19 PM remainder were viewing a cable station. At the 0.05 significance level, can we conclude that the guideline is still reasonable? (Round your answers to 3 decimal places.) H0: The proportions are as stated. H1: The proportions are not as stated. Reject H0 if X2 > . X2 = (Click to select) H0. Proportion of viewers (Click to select) as stated. References Worksheet 18. Difficulty: 2 Intermediate Learning Objective: 15-03 Test a hypothesis comparing an observed set of frequencies to an expected frequency distribution. Award: 10.00 points A study regarding the relationship between age and the amount of pressure sales personnel feel in relation to their jobs revealed the following sample information. At the 0.01 significance level, is there a relationship between job pressure and age? (Round your answers to 3 decimal places.) Degree of Job Pressure Age (years) Less than 25 25 up to 40 40 up to 60 60 and older Low 17 53 56 34 Medium 28 57 68 39 H0: Age and pressure are not related. Reject H0 if X2 > High 15 50 55 38 H1: Age and pressure are related. . X2= (Click to select) H0. Age and pressure (Click to select) related. References http://ezto.mheducation.com/hm.tpx Page 14 of 17 Assignment Print View 8/21/16, 8:19 PM Worksheet 19. Difficulty: 2 Intermediate Learning Objective: 15-06 Perform a chisquare test for independence on a contingency table. Award: 10.00 points The following hypotheses are given. H0 : 0.81 H1 : > 0.81 A sample of 80 observations revealed that p = 0.86. At the 0.10 significance level, can the null hypothesis be rejected? a. State the decision rule. (Round your answer to 2 decimal places.) Reject H0 if z > b. Compute the value of the test statistic. (Round your answer to 2 decimal places.) Value of the test statistic c. What is your decision regarding the null hypothesis? (Click to select) H0. References Worksheet 20. Difficulty: 2 Intermediate Learning Objective: 15-01 Test a hypothesis about a population proportion. Award: 10.00 points Banner Mattress and Furniture Company wishes to study the number of credit applications received per day for the last 293 days. Number of Credit http://ezto.mheducation.com/hm.tpx Frequency Page 15 of 17 Assignment Print View 8/21/16, 8:19 PM Applications 0 1 2 3 4 5 or more (Number of Days) 54 69 83 43 35 9 To interpret, there were 54 days on which no credit applications were received, 69 days on which only one application was received, and so on. Would it be reasonable to conclude that the population distribution is Poisson with a mean of 3.0? Use the 0.02 significance level. Hint: To find the expected frequencies use the Poisson distribution with a mean of 3.0. Find the probability of exactly one success given a Poisson distribution with a mean of 3.0. Multiply this probability by 293 to find the expected frequency for the number of days in which there was exactly one application. Determine the expected frequency for the other days in a similar manner. (Round x2 to 3 decimal places. Round fe to 4 decimal places. Round (f0 - fe)2/fe) to 4 decimal places.) H0: Distribution with Poisson with = 3. Decision rule: If 2 > Applications H1: Distribution is not Poisson with = 3 reject H0. f0 fe (f0 fe)2/fe 0 1 2 3 4 5 or more Total H0. There (Click to select) distribution is Poisson with =3. (Click to select) sufficient evidence to reject the null that the rev: 11_30_2013_QC_41873 References Worksheet http://ezto.mheducation.com/hm.tpx Difficulty: 3 Challenge Learning Objective: 15-03 Test a hypothesis comparing an observed set of frequencies to an expected frequency distribution. Page 16 of 17 Assignment Print View http://ezto.mheducation.com/hm.tpx 8/21/16, 8:19 PM Page 17 of 17 Question 8 Sales ($)Audit ($) 127 122 137 103 140 127 159 136 136 149 138 120 142 Mean standard deviation n df a 139.5 128.4286 10.55936 15.4149 6 7 5 6 0.01 0.01 2.7180791838 1. Reject H_0 when t> 2.718 2. Pooled Variance 180.2922 180.29 3. Test statistic t= 1.482077 1.482 4. Do not reject the null hypothesis 5. P-value = Question 10 0.083205 0.083 Claibome Klein $4 $5 $3.40 $3.50 $5.60 $3.50 $4.50 $4.10 $4.10 $3.60 11 $5.70 $6.80 $6.60 $3 $4.30 3.9 3.2 5.8 5.1 6.3 Mean standard deviation n df a $3.80 $4.50 $4.60 $4.80 $5.20 5.4 4.3 $4.8133 $4.3667 1.287781 0.591352 15 12 14 11 1. Degrees of freedom t_crit 25 1.316345 2. Reject H_0 when t> 1.316 3. Test statistic t= x1-x2 Sp^2 t= $0.4467 1.08256 21.21429 0.093 4. Do not reject H_0. It is not possible to conclude that Claibone models earn more Question 12 H_0: U_1U_2 H_1: U_1>U_2 Population 1 N Mean Stdev df Population 2 26 15 107 98 9 7 25 14 39 a. Degrees of freedom b. Decision rule Reject H_0 if t> c. Test statistic t_crit 39 2.022691 2.023 x1-x2 Sp^2 t= 9 13.4 23.65385 0.506 d. Do not reject the null hypothesis since t=0.506<2.023 at the 0.025 significance level Question 13 Statistic Men Women Sample mean 24.38 22.36 Sample standard deviation 5.91 4.8 Sample size 33 37 a= df 32 36 68 Decision rule, t_crit -2.650081 or 2.65008 Null hypothesis H_0: U_1=U_2 Research hypothesis H_1: U_1U_2 Reject the null hypothesis if t does not lie between -2.65 and +2.65 x1-x2 2.02 Sp^2 28.6344 78.75 1. t t= 0.042538 0.043 0.01 Question 8 Sales ($)Audit ($) 127 122 137 103 140 127 159 136 136 149 138 120 142 Mean standard deviation n df a 139.5 128.4286 10.55936 15.4149 6 7 5 6 0.01 0.01 2.7180791838 1. Reject H_0 when t> 2.718 2. Pooled Variance 180.2922 180.29 3. Test statistic t= 1.482077 1.482 4. Do not reject the null hypothesis 5. P-value = Question 10 0.083205 0.083 Claibome Klein $4 $5 $3.40 $3.50 $5.60 $3.50 $4.50 $4.10 $4.10 $3.60 11 $5.70 $6.80 $6.60 $3 $4.30 3.9 3.2 5.8 5.1 6.3 Mean standard deviation n df a $3.80 $4.50 $4.60 $4.80 $5.20 5.4 4.3 $4.8133 $4.3667 1.287781 0.591352 15 12 14 11 1. Degrees of freedom t_crit 25 1.316345 2. Reject H_0 when t> 1.316 3. Test statistic t= x1-x2 Sp^2 t= $0.4467 1.08256 21.21429 0.093 4. Do not reject H_0. It is not possible to conclude that Claibone models earn more Question 12 H_0: U_1U_2 H_1: U_1>U_2 Population 1 N Mean Stdev df Population 2 26 15 107 98 9 7 25 14 39 a. Degrees of freedom b. Decision rule Reject H_0 if t> c. Test statistic t_crit 39 2.022691 2.023 x1-x2 Sp^2 t= 9 13.4 23.65385 0.506 d. Do not reject the null hypothesis since t=0.506<2.023 at the 0.025 significance level Question 13 Statistic Men Women Sample mean 24.38 22.36 Sample standard deviation 5.91 4.8 Sample size 33 37 a= df 32 36 68 Decision rule, t_crit -2.650081 or 2.65008 Null hypothesis H_0: U_1=U_2 Research hypothesis H_1: U_1U_2 Reject the null hypothesis if t does not lie between -2.65 and +2.65 x1-x2 2.02 Sp^2 28.6344 78.75 1. t t= 0.042538 0.043 2. The decision is do not reject the null hypothesis that the means are the same. This is because t=0.43 is greater than -2.65 but less than 2.65 3. P-value = 0.965828 0.9658 Question 14 paper-coupe-coupons Mean age 38.8 38.5 Standard deviation 5.6 12.3 Sample size 26 32 df 25 31 56 Decision rule, t_crit -2.666512 or 2.65008 Null hypothesis H_0: U_1=U_2 Research hypothesis H_1: U_1U_2 Reject the null hypothesis if t does not lie between -2.65 and +2.65 x1-x2 0.3 Sp^2 97.74982 71.92 0.01 t= 0.003578 0.004 1. Degrees of freedom 56 2. Reject H_0 if t<2.667 or t>-2.667 3. Test statistic, t =0.004 4. Do not reject H_0. There is no difference in the mean ages of the two coupon clients Question 8 Sales ($) Mean standard deviation n df a Audit ($) 127 137 140 159 136 138 122 103 127 136 149 120 142 139.5 10.559356041 6 5 0.01 128.4285714286 15.4148969383 7 6 0.01 2.7180791838 1. Reject H_0 when t> 2.718 2. Pooled Variance 180.2922077922 180.29 3. Test statistic t= 1.4820771521 1.482 4. Do not reject the null hypothesis 5. P-value = Question 10 0.083205455 0.083 Claibome Klein $4 $5 $3.40 $3.50 $5.60 $3.50 $4.50 $4.10 $4.10 $3.60 11 $5.70 $3.80 $6.80 $4.50 $6.60 $4.60 $3 $4.80 $4.30 $5.20 3.9 5.4 3.2 4.3 5.8 = [( _1^2)/_1 +(_2^2)/_2 ] ^2/( ((_1^2)/_1 ) 5.1 6.3 Mean standard deviation n df a $4.8133 1.2877814071 15 14 $4.3667 0.5913518155 12 11 1. Degrees of freedom t_crit 117 1.2888291988 2. Reject H_0 when t> 1.289 (/ ((_1^2)/_1 ) ^2/(_11) ( ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 3. Test statistic ((/(/ =( _1 () _2)/((_1^2)/_1 +(_2^2)/_2 ) t= x1-x2 S1^2/N1+S2^2/N2 t t= $0.4467 0.1397001443 $1.20 1.093 4. Do not reject H_0. It is not possible to conclude that Claibone models earn more Question 12 H_0: U_1U_2 H_1: U_1>U_2 Population 1 N Mean Stdev df = (/(/ [(_1^2)/_1 Population 2 26 107 9 25 +(_2^2)/_2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11 ( 15 98 7 14 39 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 a. Degrees of freedom b. Decision rule Reject H_0 if t>2.179 c. Test statistic [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 12 (/ ((_1^2)/_1 ) ^2/(_11) ( 2.1788128297 t_crit 2.179 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) (/(/ _1 () _2)/((_1^2)/_1 +(_2^2)/_2 ) x1-x2 S1^2/N1+S2^2/N2 t 9 6.382051282 3.562561454 d. Reject the null hypothesis since t=3.563 which is greater than 2.023 at the 0.025 significance level = (/(/ [(_1^2)/_1 +(_2^2)/_2 ] ^2/( (/ ((_1^2)/_1 ) Question 13 Statistic Men Women Sample mean 24.38 22.36 Sample standard deviation 5.91 4.8 Sample size 33 37 df 32 Decision rule, t_crit -2.738481482 or 2.73848 Null hypothesis H_0: U_1=U_2 (/ ((_1^2)/_1 ) ^2/(_11) ( Research hypothesis H_1: U_1U_2 Reject the null hypothesis if t does not lie between -2.738 and +2.738 x1-x2 2.02 S1^2/N1+S2^2/N2 1.6811299754 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) t 1.557939984 1. t t= 1.558 2. The decision is do not reject the null hypothesis that the means are the same. This is because t=1.558 is greater than -2.65 but less than 2.65 3. P-value = Question 14 0.1290694957 0.1291 = (/(/ [(_1^2)/_1 +(_2^2)/_2 ] ^2/( (/ ((_1^2)/_1 ) paper-coupons Mean age Standard deviation Sample size df Decision rule, t_crit Null hypothesis Research hypothesis e-coupons 38.8 5.6 26 38.5 12.3 32 30 -2.7499956536 or 2.749995654 H_0: U_1=U_2 H_1: U_1U_2 (/ ((_1^2)/_1 ) ^2/(_11) ( Reject the null hypothesis if t does not lie between -2.75 and +2.75 x1-x2 0.3 S1^2/N1+S2^2/N2 5.9339663462 t 0.1231540548 t= 0.123 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 1. Degrees of freedom 30 2. Reject H_0 if t<2.667 or t>-2.667 3. Test statistic, t =0.123 4. Do not reject H_0. There is no difference in the mean ages of the two coupon clients Question 15 Mean Stdev Sample size First Development Second development 165,000 179,000 36,000 25,000 18 28 Assuming population standard deviation is the same Df = N1+N2-2 a = 0.02 44 t_crit = -2.4141343682 or 2.414134368 1..... Reject H0 if t is not between -2.414 and 2.414 2. Test statistic = ((_// ( _1 () _2)/(_ (1/_1 +1/_2 )) _=((_11) _ ( ( ) ( +_1^2+(_21)_2^2)/(_1+_22) _= _ t= 884250000 -0.000052407 -0.0001 0 t=0.00 (2dp) 3. At .02 significance level, do not reject the null hypothesis. Fairfield cannot conclude that the means are different. t=0.00, which is greater than -2.313 and less than 2.414. Assume unequal variances = _(/(/ [(_1^2)/_1 +(_2^2)/_2 ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2) =( ((/(/ 1 () _2)/((_1^2)/_1 +(_2^2)/_2 ) Assume Equal standard deviations/variances =( ((_// _1 () _2)/(_ (1/_1 +1/_2 )) _=((_11) _ ( ( ) ( +_1^2+(_21)_2^2)/(_1+_22) _2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 117.2357 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 0.111407952 (/ ((_1^2)/_1 ) ^2/(_11) ( 0.000873088 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 0.000077202 (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 11.98359841 13.7865 _1 ) ^2/(_11) ( 0.388225 _2 ) ^2/(_ ( 21) 0.762222 _2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 31.59028 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 1.446185929 (/ ((_1^2)/_1 ) ^2/(_11) ( 0.035008384 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 0.010771074 _2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 30.23282 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 23.55836488 (/ ((_1^2)/_1 ) ^2/(_11) ( 0.058192284 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 0.721039066 t the means are different. 2/(_11)+ ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( Question 8 Sales ($) Mean standard deviation n df a Audit ($) 127 137 140 159 136 138 122 103 127 136 149 120 142 139.5 10.559356041 6 5 0.01 128.4285714286 15.4148969383 7 6 0.01 2.7180791838 1. Reject H_0 when t> 2.718 2. Pooled Variance 180.2922077922 180.29 3. Test statistic t= 1.4820771521 1.482 4. Do not reject the null hypothesis 5. P-value = Question 10 0.083205455 0.083 Claibome Klein $4 $5 $3.40 $3.50 $5.60 $3.50 $4.50 $4.10 $4.10 $3.60 11 $5.70 $3.80 $6.80 $4.50 $6.60 $4.60 $3 $4.80 $4.30 $5.20 3.9 5.4 3.2 4.3 5.8 = [( _1^2)/_1 +(_2^2)/_2 ] ^2/( ((_1^2)/_1 ) 5.1 6.3 Mean standard deviation n df a $4.8133 1.2877814071 15 14 $4.3667 0.5913518155 12 11 1. Degrees of freedom t_crit 117 1.2888291988 2. Reject H_0 when t> 1.289 (/ ((_1^2)/_1 ) ^2/(_11) ( ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 3. Test statistic ((/(/ =( _1 () _2)/((_1^2)/_1 +(_2^2)/_2 ) t= x1-x2 S1^2/N1+S2^2/N2 t t= $0.4467 0.1397001443 $1.20 1.093 4. Do not reject H_0. It is not possible to conclude that Claibone models earn more Question 12 H_0: U_1U_2 H_1: U_1>U_2 Population 1 N Mean Stdev df = (/(/ [(_1^2)/_1 Population 2 26 107 9 25 +(_2^2)/_2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11 ( 15 98 7 14 39 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 a. Degrees of freedom b. Decision rule Reject H_0 if t>2.179 c. Test statistic [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 12 (/ ((_1^2)/_1 ) ^2/(_11) ( 2.1788128297 t_crit 2.179 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) (/(/ _1 () _2)/((_1^2)/_1 +(_2^2)/_2 ) x1-x2 S1^2/N1+S2^2/N2 t 9 6.382051282 3.562561454 d. Reject the null hypothesis since t=3.563 which is greater than 2.023 at the 0.025 significance level = (/(/ [(_1^2)/_1 +(_2^2)/_2 ] ^2/( (/ ((_1^2)/_1 ) Question 13 Statistic Men Women Sample mean 24.38 22.36 Sample standard deviation 5.91 4.8 Sample size 33 37 df 32 Decision rule, t_crit -2.738481482 or 2.73848 Null hypothesis H_0: U_1=U_2 (/ ((_1^2)/_1 ) ^2/(_11) ( Research hypothesis H_1: U_1U_2 Reject the null hypothesis if t does not lie between -2.738 and +2.738 x1-x2 2.02 S1^2/N1+S2^2/N2 1.6811299754 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) t 1.557939984 1. t t= 1.558 2. The decision is do not reject the null hypothesis that the means are the same. This is because t=1.558 is greater than -2.65 but less than 2.65 3. P-value = Question 14 0.1290694957 0.1291 = (/(/ [(_1^2)/_1 +(_2^2)/_2 ] ^2/( (/ ((_1^2)/_1 ) paper-coupons Mean age Standard deviation Sample size df Decision rule, t_crit Null hypothesis Research hypothesis e-coupons 38.8 5.6 26 38.5 12.3 32 30 -2.7499956536 or 2.749995654 H_0: U_1=U_2 H_1: U_1U_2 (/ ((_1^2)/_1 ) ^2/(_11) ( Reject the null hypothesis if t does not lie between -2.75 and +2.75 x1-x2 0.3 S1^2/N1+S2^2/N2 5.9339663462 t 0.1231540548 t= 0.123 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 1. Degrees of freedom 30 2. Reject H_0 if t<2.667 or t>-2.667 3. Test statistic, t =0.123 4. Do not reject H_0. There is no difference in the mean ages of the two coupon clients Question 15 Mean Stdev Sample size First Development Second development 165,000 179,000 36,000 25,000 18 28 Assuming population standard deviation is the same Df = N1+N2-2 a = 0.02 44 t_crit = -2.4141343682 or 2.414134368 1..... Reject H0 if t is not between -2.414 and 2.414 2. Test statistic = ((_// ( _1 () _2)/(_ (1/_1 +1/_2 )) _=((_11) _ ( ( ) ( +_1^2+(_21)_2^2)/(_1+_22) _= _ t= 884250000 -0.000052407 -0.0001 0 t=0.00 (2dp) 3. At .02 significance level, do not reject the null hypothesis. Fairfield cannot conclude that the means are different. t=0.00, which is greater than -2.313 and less than 2.414. Question 16 Manufacturer A B Unacceptable Acceptable Total C 49 126 175 OBSERVED 22 153 175 EXPECTED |O-E| 49 26.5 22.5 22 26.5 4.5 21 26.5 5.5 14 26.5 12.5 126 106.92 19.08 153 1. Reject H0 if 21 154 175 129.8314285714 23.16857143 154 130.68 23.32 161 136.62 24.38 ^2> 7.815 7.8147279033 DF = (NUMBER OF COLUMNS - 1)(N DF = (4-1)(2-1) 3 ^2 is greater than 7.815. Therefore, reject H0. There is a relationship between quality and manufacturer. Question 17 prime time networks cable stations 30% ABC 10% CBS 110 NBC 110 165 Assume unequal variances = _(/(/ [(_1^2)/_1 +(_2^2)/_2 ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2) =( ((/(/ 1 () _2)/((_1^2)/_1 +(_2^2)/_2 ) Assume Equal standard deviations/variances =( ((_// _1 () _2)/(_ (1/_1 +1/_2 )) _=((_11) _ ( ( ) ( +_1^2+(_21)_2^2)/(_1+_22) _2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 117.2357 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 0.111407952 (/ ((_1^2)/_1 ) ^2/(_11) ( 0.000873088 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 0.000077202 (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 11.98359841 13.7865 _1 ) ^2/(_11) ( 0.388225 _2 ) ^2/(_ ( 21) 0.762222 _2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 31.59028 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 1.446185929 (/ ((_1^2)/_1 ) ^2/(_11) ( 0.035008384 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 0.010771074 _2 ] ^2/( (/ ((_1^2)/_1 ) ^2/(_11)+ ( ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 30.23282 [(_1^2)/ ( _1 + ((_2^2)/_ / 2 ] ^2 23.55836488 (/ ((_1^2)/_1 ) ^2/(_11) ( 0.058192284 ((_ ( 2^2)/_ / 2 ) ^2/(_ ( 21) 0.721039066 t the means are different. Expected values D Total 14 161 175 106 594 700 |O-E| (O-E)2 (o-E)2/E (O-E)^2 (O-E)^2/E 506.25 19.10377 A B C 26.5 26.5 106.92 129.8314 22.5 4.5 506.25 20.25 D 26.5 130.68 5.5 30.25 26.5 136.62 12.5 156.25 20.25 0.764151 30.25 1.141509 5.16E-005 156.25 5.896226 14 23.04 9.04 81.72 364.0464 3.404848 45 36.00 9.00 81 536.7827 4.134459 2 20.64 18.64 347.45 543.8224 4.161481 5 15.36 10.36 107.33 594.3844 53 24.00 29.00 841 53 34.40 18.60 345.96 45 25.60 19.40 376.36 2 40.00 38.00 4.35064 ^2 42.95709 F = (NUMBER OF COLUMNS - 1)(NUMBER OF ROWS - 1) F = (4-1)(2-1) quality and manufacturer. CABLE STA TOTAL 165 550 1444.00 2/(_11)+ ((_ ( 2^2)/_ / 2 ) ^2/(_21)) ( 3.546 2.25 16.83 6.99 35.04 10.06 14.7 36.1 125.516