ࡱ>  }xyz{|D :bjbj ǡ}|-d C[ [ [ [ 222A>A>A>A>A>A$gF ITbA21"222bA[ [ KC3<3<3<2[ [ >[ @,=<5R>(AaC0C>]I;@]I>>4]I,@223<22222bAbA3<222C2222]I222222222 : 88. Refer to the Baseball 2005 data, which reports information on the 30 major league teams for the 2005 baseball season. a. Select the variable team salary and find the mean, median, and the standard deviation. b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the year in which the stadium was built from the current year to find the stadium age and work with that variable.) Find the mean, median, and the standard deviation. c. Select the variable that refers to the seating capacity of the stadium. Find the mean, median, and the standard deviation. Solution: a. Select the variable team salary and find the mean, median, and the standard deviation. Team Salary( in Million Dollars)Boston123.5New York Yankees208.3Oakland55.4Baltimore73.9Los Angeles Angels97.7Cleveland41.5Chicago White Sox75.2Toronto45.7Minnesota56.2Tampa Bay29.7Texas55.8Detroit69.1Seattle87.8Kansas City36.9Atlanta86.5Arizona62.3Houston76.8Cincinnati61.9New York Mets101.3Pittsburgh38.1Los Angeles Dodgers83San Diego63.3Washington48.6San Francisco90.2St. Louis92.1Florida60.4Philadelphia95.5Milwaukee39.9Chicago Cubs87Colorado48.2 Mean $73.0600Median66.2Standard Deviation34.23285 b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the year in which the stadium was built from the current year to find the stadium age and work with that variable.) Find the mean, median, and the standard deviation. Team Built AgeAge (2009-Built Age)Boston191297New York Yankees192386Oakland196643Baltimore199217Los Angeles Angels196643Cleveland199415Chicago White Sox199118Toronto198920Minnesota198227Tampa Bay199019Texas199415Detroit20009Seattle199910Kansas City197336Atlanta199316Arizona199811Houston20009Cincinnati20036New York Mets196445Pittsburgh20018Los Angeles Dodgers196247San Diego20045Washington196148San Francisco20009St. Louis196643Florida198722Philadelphia20045Milwaukee20018Chicago Cubs191495Colorado199514 Mean 28.2Median17.5Standard Deviation25.94477 c. Select the variable that refers to the seating capacity of the stadium. Find the mean, median, and the standard deviation. Team Seating Capacity of the StadiumBoston33871New York Yankees57746Oakland43662Baltimore48262Los Angeles Angels45050Cleveland43368Chicago White Sox44321Toronto50516Minnesota48678Tampa Bay44027Texas52000Detroit40000Seattle45611Kansas City40529Atlanta50062Arizona49075Houston42000Cincinnati42059New York Mets55775Pittsburgh38127Los Angeles Dodgers56000San Diego42445Washington56000San Francisco40800St. Louis49625Florida42531Philadelphia43500Milwaukee42400Chicago Cubs38957Colorado50381 Mean 45912.60Median44174.00Standard Deviation5894.20 56. Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study. a. What is the likelihood all four of the selected flights arrived within 15 minutes of the scheduled time? b. What is the likelihood that none of the selected flights arrived within 15 minutes of the scheduled time? c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes of the scheduled time? Solution: Let X be the sample parameter for any flight on Northwest Airlines arrives within 15 minutes of the scheduled time. Sample of size, n = 4 p = 0.9. q= 0.1 a. P(X=4) = 0.94 = 0.6561 b. P(X=0) = (1-0.9)4 = 0.0001 c. P (Xd"3) = 1 - 0.6561 = 0.3439 since P (Xd"3) = 1- P(X=4) 64. An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two emails per hour. Assume the arrival of these emails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 1 email between 4 P.M. and 5 P.M. yesterday? b. What is the probability she received 5 or more email during the same period? c. What is the probability she did not receive any email during the period? Solution:  where e is the HYPERLINK "http://en.wikipedia.org/wiki/E_%28mathematical_constant%29" \o "E (mathematical constant)"base of the natural logarithm (e = 2.71828...) k is the number of occurrences of an event - the probability of which is given by the function k! is the HYPERLINK "http://en.wikipedia.org/wiki/Factorial" \o "Factorial"factorial of k  is a positive HYPERLINK "http://en.wikipedia.org/wiki/Real_number" \o "Real number"real number, equal to the expected number of occurrences that occur during the given interval. For instance, if the events occur on average 4 times per HYPERLINK "http://en.wikipedia.org/wiki/Minute" \o "Minute"minute, and you are interested in the number of events occurring in a 10 minute interval, you would use as your model a Poisson distribution with =104=40. (Source: HYPERLINK "http://en.wikipedia.org/wiki/Poisson_distribution"http://en.wikipedia.org/wiki/Poisson_distribution) (a) P(received exactly 1 email between 4 P.M. and 5 P.M. yesterday)=f(1,2) = 2e-2 = 0.2707 (b)P(received 5 or more email during the same period) =1-k=0"k=4f(k;) = 1-[(e-2+2e-2+2e-2+(4/3)e-2+(2/3)e-2] = 0.05265 (c) P(she did not receive anyemail during period) = f(0,2)= e-2 = 0.13534 50. Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation 2,000 miles. a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more? b. What percent of the trucks logged more than 57,060 but less than 58,280 miles? c. What percent of the Fords traveled 62,000 miles or less during the year? d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles? Explain. Solution: a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more? Here, sample mean is 65,200 miles and population mean is 60,000 miles. the standard deviation is 2,000 miles. (a) Calculation of test-statistics, z = (X - ) / x Where X is a normal random variable,  is the mean, and x is the standard deviation. Hence, z= 65,200-60,000/2,000 = 5,200/2000 = 2.6 P (the Ford Super Duty F-750s logged 65,200 miles or more) = P(Z>2.6)=1-P(Z<2.6)=1-0.9953=0.0047 Hence answer is 0.47% b. What percent of the trucks logged more than 57,060 but less than 58,280 miles? For X = 57,060, Z = 57,060-60,000/2,000= -1.47 For X=58,280 Z= 58,250-60,000/2,000 = -0.875 P(the trucks logged more than 57,060 but less than 58,280 miles) = P(-1.475) -> 0 Hence, it is not reasonable to assume that any of the trucks were driven more than 70,000 miles. 38. The mean amount purchased by a typical customer at Churchills Grocery Store is $23.50 with a standard deviation of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions. a. What is the likelihood the sample mean is at least $25.00? b. What is the likelihood the sample mean is greater than $22.50 but less than $25.00? c. Within what limits will 90 percent of the sample means occur? Solution: Here, population mean is $23.50 Standard Deviation  = 5 n= size of sample = 50 Calculation of test-statistics, z = (X - ) / x Where X is a normal random variable,  is the mean, and  is the standard deviation. x = /"n x = /sqrt(n) = 5/"50 = 0.7071 (a) What is the likelihood the sample mean is at least $25.00? Calculation of test-statistics, z= (25-23.5)/(5/"50) = 2.12 P(the sample mean is at least $25.00) = P(Z e" 2.12) = 0.017 (b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00? for sample mean = $22.50, z = 22.5-23.5/(5/"50 = -1.41 P(the sample mean is greater than $22.50 but less than $25.00) = P(-1.41 < Z < 2.12) = 0.9037 (c) Within what limits will 90 percent of the sample means occur? 90% Confidence Interval can be determined by  - z*/"(n) to  + z*/"(n). Since it is 90% Confidence Interval, significant level will be 0.10. Critical vaue of z for significant level 0.10 = 1.645 So, (23.5 - 1.645*5/"(50), 23.5 + 1.645*5/"(50)) = (22.33,24.67) 54. Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000. a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium. b. How large a sample is needed to find the population mean within $250 at 99 percent confidence? Solution: Here, population mean is $10,979 n= size of sample = 20 Calculation of test-statistics, z = (X - ) / x Where X is a normal random variable,  is the mean, and  is the standard deviation. x = /sqrt(n) a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium. 90% Confidence Interval can be determined by  - z*/sqrt(n) to  + z*/sqrt(n). Since it is 90% Confidence Interval, significant level will be 0.10. Critical value of z for significant level 0.10 = 1.645 So, (10979-1.645*1000/"20, 10979+1.645*1000/"20) = (10611.12,11346.81) b. How large a sample is needed to find the population mean within $250 at 99 percent confidence? Since it is 99% Confidence Interval, significant level will be 0.01. Critical value of z for significant level 0.01 = 2.5758 z = (X - ) / /sqrt(n) Now, z < (X - ) / /sqrt(n) Then z*/sqrt(n) < 250 => 2.5758 * 1000/sqrt(n) < 250 => 2.5758 * 1000/250 < sqrt(n) => 10.312 < sqrt(n) => 106.33 < n This gives the sample size of 107. 42. During recent seasons, Major League Baseball has been criticized for the length of the games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample of 17 games revealed the following times to completion. (Note that the minutes have been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes is reported at 2.40 hours.)  Solution: Size of sample, n = 17 Degree of freedom = n-1 = 17-1 = 16 Sum of sample = i=1"n=17xi = (2.98 + 2.40 + 2.70 + 2.25 + 3.23 + 3.17 + 2.93 + 3.18 +2.80 +2.38+ 3.75+ 3.20+ 3.27+ 2.52+ 2.58+ 4.45+ 2.45) = 50.24 Time (in Hours)SumMean(xm)(xi- xm)2Standard Deviation12.9850.24Mean = Sum/n = 50.24/17 = 2.95520.000605 = "( i=1"n=25(xi-xm)2/(n-1)) = "(5.009623/16) = "0.3131 =0.559522.40.30846232.70.06522642.250.49758153.230.07540863.170.04605672.930.00064583.180.05044892.80.024147102.380.331078113.750.631399123.20.059832133.270.098977142.520.189568152.580.140921164.452.233847172.450.255423 Here X = 2.9552  = 0.5595 We can define one-tailed statistics for above observations as follows: Null Hypothesis: H0: ( =3.5 vs. Ha: ( < 3.5 Rejection region: t < -t( Here significance level is 0.05, So, t0.05 = - 2.120 (Using Statistical Ratio Calculator From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating t with 0.05 significance and DF =16) I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in sample. Now, t = (X - ) / (s/"n) Where X is a normal random variable,  is the mean, and s is the standard deviation. and n is the sample size. Calculating t-test statistics: t = 2.9552-3.5/(0.5595/"17) = -0.5448* "17/0.5595= -4.0147 Calculating p-value: Degree of freedom = DF = 17-1 = 16 P(t16 < -4.0147) = 0.000501 (Using http://www.graphpad.com/quickcalcs/PValue1.cfm with t=-4.0147and DF = 16) The p-value of 0.0501% is less than significant level of 5%. Hence, we can reject the null hypothesis and conclude that the mean time for a game is less than 3.50 hours. 58. The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, five years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than five years ago?  Solution: Let us define, Total housing costs as a percent of monthly income five years ago = C-5 Current total housing costs as a percent of monthly income five years ago = C0 So, differenec between housing costs as a percent of monthly income five years ago and current value = "C = C-5 - C0 Null Hypothesis: H0: "C = 0 vs H1: "C > 0 Rejection region: t > t( Here significance level is 0.05, So, t0.05 = 1.729 (Using Statistical Ratio Calculator From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating t with 0.05 significance and DF =19) Table for Paired t-Test Calculation: Homeowner 5 Years AgoNow"C 11710722039-1932937-844327165361224643413745242181926-79492821104926231135323121632-1613232121433122115444041644422172822618291910193935420221210 Mean X =6.35 Standard Deviation (xs) = 12.4778 I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in sample. Now, t = (X - ) / (s/"n) Where X is a normal random variable,  is the mean, and s is the standard deviation. n is the sample size. Calculating t-test statistics: t = (6.35 -0) / (12.4778 / "20) = 2.275 Conclusion: Since calculated value of t is greater than critical value of t i.e.2.275 > 1.729. We reject the null hypothesis and conclude that the percent is less now than 5 years ago. 42. Martin Motors has in stock three cars of the same make and model. The president would like to compare the gas consumption of the three cars (labeled car A, car B, and car C) using four different types of gasoline. For each trial, a gallon of gasoline was added to an empty tank, and the car was driven until it ran out of gas. The following table shows the number of miles driven in each trial.  Solution: Results of ANOVASummaryCount SumAverageVarianceRegular364.721.570.64Super Regular357.119.033.52Unleaded360.620.21.00Premium Unleaded359.319.771.02Car A478.919.735.06Car B47919.750.92Car C483.820.950.24ANOVASource of VariationSSDFMSFP-ValueFCritRows10.2133.402.410.164.76Columns3.9221.961.390.325.14Errors8.4661.41Hence applying 0.05 level of significance,a.F = 2.41, no difference as F < FCritb.F = 1.39, no difference as F < FCrit 37. A regional commuter airline selected a random sample of 25 flights and found that the correlation between the number of passengers and the total weight, in pounds, of luggage stored in the luggage compartment is 0.94. Using the .05 significance level, can we conclude that there is a positive association between the two variables? Solution: Null Hypothesis: H0: r = 0 ( There is no positive association the number of passengers and the total weight) vs. Ha: r > 0 ( There is positive association between the number of passengers and the total weight) Degre of Freedom=n-2=23Significant level = 0.05(Using http://www.danielsoper.com/statcalc/calc10.aspx with probability value = 0.05 and df= 23)one-tailed test, critical value of t =1.713872Correlation Coefficientr=0.94Test-Statistics 13.21Since 13.21 > 1.31, we can conclude that there is a positive association between no of passengers and total weight of luggages 40. A suburban hotel derives its gross income from its hotel and restaurant operations. The owners are interested in the relationship between the number of rooms occupied on a nightly basis and the revenue per day in the restaurant. Below is a sample of 25 days (Monday through Thursday) from last year showing the restaurant income and number of rooms occupied.  Solution : Attached excel sheet has solution.  EMBED Excel.Sheet.8  17. The district manager of Jasons, a large discount electronics chain, is investigating why certain stores in her region are performing better than others. She believes that three factors are related to total sales: the number of competitors in the region, the population in the surrounding area, and the amount spent on advertising. From her district, consisting of several hundred stores, she selects a random sample of 30 stores. For each store she gathered the following information. Y _ total sales last year (in $ thousands). X1 _ number of competitors in the region. X2 _ population of the region (in millions). X3 _ advertising expense (in $ thousands).  Multiple Regression and Correlation Analysis 549 a. What are the estimated sales for the Bryne store, which has four competitors, a regional population of 0.4 (400,000), and advertising expense of 30 ($30,000)? b. Compute the R2 value. c. Compute the multiple standard error of estimate. d. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not equal to zero. Use the .05 level of significance. e. Conduct tests of hypotheses to determine which of the independent variables have significant regression coefficients. Which variables would you consider eliminating? Use the .05 significance level. Solution: Y = total sales last year (in $ thousands).X1 = number of competitors in the region.X2 =population of the region (in millions).X3 = advertising expense (in $ thousands). Analysis of varianceSOURCEDFSSMSRegression33050.001016.67Error262200.0084.62Total295250.00PredictorCoefStDevt-ratioConstant14.007.002.00X1-1.000.70-1.43X230.005.205.77X30.200.082.50 a. What are the estimated sales for the Bryne store, which has four competitors, a regional population of 0.4 (400,000), and advertising expense of 30 ($30,000)? Solution: The regression equation for sales, Y = -1.00 * X1 + 30 * X2 + 0.20 * X3 + 14 = - X1 + 30 X2 + 0.20 X3 + 14 So, the estimated sales for Bryne store, Y = -4+30*0.4+0.20*30+14 = -4+12+6+14 = $28 thousands b. Compute the value of R2 Solution: R = SSR/SST = 3050.00/5250.00 = 0.58 c. Compute the multiple standard error of estimate. Solution: Multiple Standard error of the estimate = ( MSE =(0.58 = 0.76 d. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not equal to zero. Use the .05 level of significance. Solution: Null Hypothesis: Ho: the regression coefficients are equal to zero. Ha: the regression coefficients are not equal to zero. Significant level:  = 0.05 Critical Value of F: DFn = 3 DFd = 26 Critical Table value of F at 0.05 significant level, Probability = 0.05 DFn=3 DFd=26 Fcrit= 2.9752 5.00% of a F distribution has values greater than 2.975. (Using  HYPERLINK "http://www.graphpad.com/quickcalcs/StatRatio2.cfm" http://www.graphpad.com/quickcalcs/StatRatio2.cfm ) Calculating F-ratio: F = MSR/MSE = 1016.67/84.62 = 12.01 Conclusion: Since F > Fcrit , we can reject the null hypothesis and conclude that the regression coefficients are not equal to zero. e. Conduct tests of hypotheses to determine which of the independent variables have significant regression coefficients. Which variables would you consider eliminating? Use the .05 significance level. Solution: PredictorCoefStDevt-ratioConstant14.007.002.00X1-1.000.70-1.43X230.005.205.77X30.200.082.50 Null Hypothesis: Ho: Independent variables do not affect the sales price Ha: Independent variables affect the sales price Significant level:  = 0.05 Decision Rule: Reject the null hypothesis if t > tcrit Critical value of t: Probability = 0.0499999156603 df=29 Two tailed: tcrit= 2.0452 5.000% of a t distribution has values greater than 2.045 or less than -2.0452. (Using  HYPERLINK "http://www.graphpad.com/quickcalcs/StatRatio2.cfm" http://www.graphpad.com/quickcalcs/StatRatio2.cfm ) Conclusion: Since 5.77 > 2.045 for X2, we reject the null hypothesis. Since calculated value of t is less that tcrit for X1 , X3 , we fail to reject the null hypothesis. Hence, we can consider eliminating X1 ,and X3 , 18. Suppose that the sales manager of a large automotive parts distributor wants to estimate as early as April the total annual sales of a region. On the basis of regional sales, the total sales for the company can also be estimated. If, based on past experience, it is found that the April estimates of annual sales are reasonably accurate, then in future years the April forecast could be used to revise production schedules and maintain the correct inventory at the retail outlets. Several factors appear to be related to sales, including the number of retail outlets in the region stocking the companys parts, the number of automobiles in the region registered as of April 1, and the total personal income for the first quarter of the year. Five independent variables were finally selected as being the most important (according to the sales manager). Then the data were gathered for a recent year. The total annual sales for that year for each region were also recorded. Note in the following table that for region 1 there were 1,739 retail outlets stocking the companys automotive parts, there were 9,270,000 registered automobiles in the region as of April 1 and so on. The sales for that year were $37,702,000.  a. Consider the following correlation matrix. Which single variable has the strongest correlation with the dependent variable? The correlations between the independent variables outlets and income and between cars and outlets are fairly strong. Could this be a problem? What is this condition called?  b. The output for all five variables is on the following page. What percent of the variation is explained by the regression equation?  c. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not zero. Use the .05 significance level. d. Conduct a test of hypothesis on each of the independent variables. Would you consider eliminating outlets and bosses? Use the .05 significance level. e. The regression has been rerun below with outlets and bosses eliminated. Compute the coefficient of determination. How much has R2 changed from the previous analysis?  f. Following is a histogram and a stem-and-leaf chart of the residuals. Does the normality assumption appear reasonable?  g. Following is a plot of the fitted values of Y (i.e., Y) and the residuals. Do you see any violations of the assumptions? Multiple Regression and Correlation Analysis  Solution: Number ofAverageAnnualNumber ofAutomobilesPersonalAge ofSalesRetailRegisteredIncomeAutomobilesNumber of($ millions),Outlets,(millions),($ billions),(years)SupervisorsYX1X2X3X4X5 37,702  1,739 9.2785.43.59 24,196  1,221 5.8660.755 32,055  1,846 8.8168.14.47 3,611 120 3.8120.245 17,625  1,096 10.3133.83.57 45,919  2,290 11.6295.14.113 29,600  1,687 8.9669.34.115 8,114 241 6.2816.35.911 20,116  649  7.7734.95.516 12,994  1,427  10.92 15.1 4.1 10 a. Consider the following correlation matrix. Which single variable has the strongest correlation with the dependent variable? The correlations between the independent variables outlets and income and between cars and outlets are fairly strong. Could this be a problem? 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This condition is called MultiCollinearity. Yes, it could be a problem since the Multiple Standard Error of Estimate (MSE) can be distorted by multicollinearity which can affect the result of experiment by changing the degree of statistical significance of independent variables. b. The output for all five variables is on the following page. What percent of the variation is explained by the regression equation? The regression equation is  sales = -19.7 - 0.00063 outlets + 1.74 cars + 0.410 income + 2.04 age - 0.034 bosses  Predictor CoefStDevt-ratio Constant -19.6725.422-3.63 outlets -0.0006290.002638-0.24 cars 1.73990.5533.15 income 0.409940.043859.35 age 2.03570.87792.32 bosses -0.03440.188-0.18Analysis of varianceSOURCEDFSSMSRegression51593.81318.76Error49.082.27Total91602.89 Solution: The percent of the variation = SSR/SST = 1593.81/1602.89 = 0.9943. c. Conduct a global test of hypothesis to determine whether any of the regression coefficients are not zero. Use the .05 significance level. Solution: Null Hypothesis: Ho: the regression coefficients are equal to zero. Ha: the regression coefficients are not equal to zero. Significant level:  = 0.05 Critical Value of F: DFn = 5 DFd = 4 Critical Table value of F at 0.05 significant level, Probability = 0.05 DFn=5 DFd=4 Fcrit = 6.2561 5.00% of a F distribution has values greater than 6.256. (Using  HYPERLINK "http://www.graphpad.com/quickcalcs/StatRatio2.cfm" http://www.graphpad.com/quickcalcs/StatRatio2.cfm ) Calculating F-ratio: F = MSR/MSE = 318.76/2.27 = 140.42 Conclusion: Since F > Fcrit , we can reject the null hypothesis and conclude that the regression coefficients are not equal to zero. d. Conduct a test of hypothesis on each of the independent variables. Would you consider eliminating outlets and bosses? Use the .05 significance level. Solution: Null Hypothesis: Ho: Independent variables do not affect the sales price Ha: Independent variables affect the sales price Significant level:  = 0.05 Decision Rule: Reject the null hypothesis if t > tcrit Critical value of t: degree of freedom = n-(k+1) = 4. Probability = 0.0499999581466 df=4 Two tailed: t= 2.7764 5.000% of a t distribution has values greater than 2.776 or less than -2.7764. (Using  HYPERLINK "http://www.graphpad.com/quickcalcs/StatRatio2.cfm" http://www.graphpad.com/quickcalcs/StatRatio2.cfm ) Conclusion: Since t > tcrit for constant, cars and income, we reject the null hypothesis. Since calculated value of t is less that tcrit for outlets,and bosses, we cannot reject the null hypothesis. For age, we fail to reject the null hypothesis. Hence, we can definitely consider eliminating outlets and bosses. e. The regression has been rerun below with outlets and bosses eliminated. Compute the coefficient of determination. How much has R2 changed from the previous analysis? The regression equation is  sales = -18.9 + 1.61 cars + 0.400 income + 1.96 age  Predictor CoefStDevt-ratio Constant -18.9243.636-5.20 cars 1.61290.19798.15 income 0.400310.0156925.52 age 1.96370.58463.36Analysis of varianceSOURCEDFSSMSRegression31593.66531.22Error69.231.54Total91602.89 Solution: Current coefficient of determination R2 = 1593.66/1602.89 = 0.9942 Current coefficient of determination R2 = 1593.81/1602.89 = 0.9943 Value of R2 changed from the previous analysis = 0.9943 0.9942 = 0.0001. There is very small change in R2 due to difference in only two variables Outlets and Bosses which we have found in previous hypothesis that they can be eliminated. f. Following is a histogram and a stem-and-leaf chart of the residuals. Does the normality assumption appear reasonable?  Solution: Yes, even distribution of data in histogram and stem-and-leaf chart of the residuals makes the assumption of normality reasonable. It looks to follow the bell-curve with symmetrical shape. g. Following is a plot of the fitted values of Y (i.e., Y) and the residuals. Do you see any violations of the assumptions? Multiple Regression and Correlation Analysis  Solution: The plot of the fitted values of Y (i.e., Y) and the residuals reflects that Y and residuals are not related systematically. Hence, the homoescedacity assumption of regression model is not violated. 22. Banner Mattress and Furniture Company wishes to study the number of credit applications received per day for the last 300 days. The information is reported on the next page. Nonparametric Methods: Chi-Square Applications  To interpret, there were 50 days on which no credit applications were received, 77 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 2.0? Use the .05 significance level. Hint: To find the expected frequencies use the Poisson distribution with a mean of 2.0. Find the probability of exactly one success given a Poisson distribution with a mean of 2.0. Multiply this probability by 300 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. Solution: No of Credit ApplicationsObserved Frequency (fo)Expected Frequency (fe) = f(x)*300Chi-square statistics = ( fo - fe)2/ fe p-value05040.6012.17617781.2010.21728181.2010.00034854.1340.69543127.0670.5715+1315.7960.495Total:3003004.1550.5272  (Source:  HYPERLINK "http://infinity.cos.edu/faculty/woodbury/Stats/Tutorial/Pois_Form.htm" J $7$8$H$Ifgd& 7$8$H$gd&[]8** $7$8$H$Ifgd&kd$$Iflֈw`!'%ffffff t$644 lBabp<yt M]^`abcdefhikw|Ff $7$8$H$Ifgd&gkd6$$Ifl0`!'%f f t$644 lBabpyt MFfZFfFf6 $7$8$H$Ifgd&  !#$&,3:?@ACDFOWFfFf~Ff  $7$8$H$Ifgd&W]cdeghjklmnoprsuFf4Ff $7$8$H$Ifgd&8*** $7$8$H$Ifgd&kd!$$Iflֈw`!'%fFrFFff t$644 lBabp<yt M $7$8$H$Ifgd& kd$$Ifl֞w`!'%fF fffff t$644 lBabpFyt M $7$8$H$Ifgd& kd$$Ifl֞w`!'%fF fffff t$644 lBabpFyt M $7$8$H$Ifgd& kd$$Ifl֞w`!'%fF fffff t$644 lBabpFyt M $7$8$H$Ifgd& kd-!$$Ifl֞w`!'%fF fffff t$644 lBabpFyt M V2i}<L\I 7$8$H$gd&Ff]$ $7$8$H$Ifgd&@BPREFI^        R v      $ d f < = D E           C H vopq̿̿h&B*CJaJph# &jLh&h&B*CJUaJph# h&h&5B*CJaJph# h&h&0JCJaJ&jh&h&B*CJUaJph# h&h&B*CJaJph# h&h&B*CJH*aJph# 8I^      R z    $ f  <      $7$8$H$Ifgd& 7$8$H$gd&*kdJ&$$IflֈviR!%ffffff t$644 lBabp<yt M $7$8$H$Ifgd&Zkd'$$Ifl\R!%fsfff t$644 lBabp(yt M $7$8$H$Ifgd& !"$05;CDEGHJU]cijkmnpw~Ff1Ff .Ffy* $7$8$H$Ifgd&FfS<Ff8Ff/5 $7$8$H$Ifgd&Ff? $7$8$H$Ifgd&  8*** $7$8$H$Ifgd&kdA$$IflֈiR!%fFdFFff t$644 lBabp<yt M   $7$8$H$Ifgd& kd=C$$Ifl֞viR!%fFq fffff t$644 lBabpFyt M "*124 $7$8$H$Ifgd&45 kdD$$Ifl֞viR!%fFq fffff t$644 lBabpFyt M57=?DIJL $7$8$H$Ifgd&LM kdSF$$Ifl֞viR!%fFq fffff t$644 lBabpFyt MMOUW_`ac $7$8$H$Ifgd&cd kdG$$Ifl֞viR!%fFq fffff t$644 lBabpFyt MdfhjlnprtuvQQoqr|}:;< 7$8$H$gdtN 7$8$H$gd&FfK $7$8$H$Ifgd&qr|}9:?Bnpwx"#yĵđđĵ}lđđĵZKZhg LhdzB*CJaJph# #hg Lhdz5B*CJ\aJph# h&h&5B*CJaJph# &jrh&h&B*CJUaJph# #h&h&6B*CJ]aJph# #h&h&5B*CJ\aJph# hBo5B*CJ\aJph# h&h&B*CJaJph# h&htNB*CJaJph# h&htN5B*CJaJph# htNB*CJaJph# <=>?"xy 7$8$H$gdP 7$8$H$gd&BCDNOjkho깧~v~ov~~j~co[htNhBo5 h?IhtN hBoH* ho}hBoho}hBoH* hBoH* h?IhBo hBo5hn}ohBo5hBohdzB*CJaJph# #hg Lhdz6B*CJ]aJph# hg LhdzB*CJaJph# Cj h)h-|5B*CJOJQJU\^JaJmHnHph# u)hdz5B*CJOJQJ\^JaJph# #b hCDOQkt~$&P#$/IfgdBogdBo 7$8$H$gdP<,,,$&P#$/IfgdBokd$$Iflr qfCfffk t 6P0644 lBaytBo,kd{$$Iflr qfCffffk t 6P0644 lBaytBo$&P#$/IfgdBo$&P#$/IfgdBo<,,,$&P#$/IfgdBokdf$$Iflr qfCffffk t 6P0644 lBaytBo,kdQ$$Iflr qfCffffk t 6P0644 lBaytBo$&P#$/IfgdBo"()$&P#$/IfgdBo)*,/6<,,,$&P#$/IfgdBokd<$$Iflr qfCffffk t 6P0644 lBaytBo6<=>A,kd'$$Iflr qfCffffk t 6P0644 lBaytBo$&P#$/IfgdBoADKQR$&P#$/IfgdBoRSZ^b<,,,$&P#$/IfgdBokd$$Iflr qfCffffk t 6P0644 lBaytBobhop,kd$$Iflr qfCffffk t 6P0644 lBaytBo$&P#$/IfgdBoopqrs{|}EFHIKjp9:ƽhg LhBoB*CJaJph# h?IhtNhtNhhBo0JUjhBoU hn}ohBohBohn}ohBo5jhBo5UhBoB*CJaJph# h?IhBopqsIJK9:gdtNgdBo 7$8$H$gdPhttp://infinity.cos.edu/faculty/woodbury/Stats/Tutorial/Pois_Form.htm ) Since calculated value of p is 0.5272 which is greater than significance level 0.05. 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