ࡱ>    bjbj DNA44AAAAAAAA8BdDAopMBa]w]w]w]^^^oooooooQqsXoAb^^bboAAw]w]offfbAw]Aw]ofboff mlnw]`CAbxn*poo0on*Ktd`KtTnKtAn^p_f``p^^^ooLfX^^^obbbbKt^^^^^^^^^4 @: Solutions to Practice Problems for Part VII 1. An aircraft company wanted to predict the number of worker-hours necessary to finish the design of a new plane. Relevant explanatory variables were thought to be the plane's top speed, its weight, and the number of parts it had in common with other models built by the company. A sample of twenty-seven of the company's planes was taken, and the following model estimated:  EMBED Equation.3  where  EMBED Equation.3 = Design effort, in millions of worker-hours EMBED Equation.3 = Plane's top speed, in Mach number EMBED Equation.3 = Plane's weight, in tons EMBED Equation.3 = Percentage number of parts in common with other modelsThe estimated partial regression coefficients were  EMBED Equation.3  = 0.661 EMBED Equation.3  = 0.065 EMBED Equation.3  =-0.018Interpret these estimates. All else equal, an increase in the plane's top speed by one Mach number engenders an expected increase of 0.661 millions of worker-hours in design effort. All else equal, an increase in the plane's weight by one ton results in a 0.065 increase in expected design effort in millions of worker-hours. All else equal, an increase in percentage number of parts in common with other models results in a decrease of 0.018 expected design effort measured in millions of worker-hours. 2. The following model was fitted to a sample of twenty-five students using data obtained at the end of their freshman year in college. The aim was to explain students' weight gains.  EMBED Equation.3  where  EMBED Equation.3 = Weight gained, in pounds, during freshman year EMBED Equation.3 = Average number of meals eaten per week EMBED Equation.3 = Average number of hours exercise per week EMBED Equation.3 = Average number of beers consumed per weekThe estimated partial regression coefficients were  EMBED Equation.3  = 7.35 EMBED Equation.3  = 0.653 EMBED Equation.3  = -1.345 EMBED Equation.3  = 0.613Interpret the estimates  EMBED Equation.3 ,  EMBED Equation.3 , and  EMBED Equation.3 . All else being equal, a one-unit increase in the average number of meals eaten per week will result in a 0.653 increase in weight, during the freshman year. All else being equal, a one-unit increase in the average number of hours of exercise per week will result in a 1.345 pound weight loss. All else being equal, a one-unit increase in the average number of beers consumed will result in a 0.613-pound weight gain. Is it possible to provide a meaningful interpretation of the estimate  EMBED Equation.3 ? We might interpret the estimate of  EMBED Equation.3  to mean that a person who doesnt eat, doesnt exercise, and doesnt drink beer is expected to gain 7.35 pounds. On the other hand, that sounds a little odd. In this extreme situation (which probably did not happen in the data set used to fit the regression line), we would logically expect a weight loss, not a weight gain. 3. In the study of Exercise 1, where the least squares estimates were based on twenty-seven sets of sample observations, the total sum of squares and regression sum of squares were found to be SST = 3.881andSSR = 3.549Find and interpret the coefficient of determination.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 91% of the variability in design effort can be explained through its linear dependence on the plane's top speed, weight and percentage number of parts in common with other models. Find the error sum of squares. SSE = 3.881 - 3.549 = 0.332 Find the corrected coefficient of determination.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Find and interpret the coefficient of multiple correlation.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 This is the sample correlation between observed and predicted values of design effort:  EMBED Equation.3  4. A study was conducted to determine whether certain features could be used to explain variability in the prices of air conditioners. For a sample of nineteen air conditioners, the following regression was estimated: y= -68.236+ 0.0023x1+ 19.729x2+ 7.653x3R2 = 0.84(0.005)(8.992)(3.082)where y= Price (in dollars)x1= Rating of air conditioner, in BTU per hourx2= Energy efficiency ratiox3= Number of settingsThe figures in parentheses beneath the coefficient estimates are the corresponding estimated standard errors. Find a 95% confidence interval for the expected increase in price resulting from an additional setting when the values of the rating and the energy efficiency ratio remain fixed. Note that:n= 19 EMBED Equation.3 = 7.653 EMBED Equation.3 = 3.082t(15, 0.025)= 2.131 95% C.I.: EMBED Equation.3 Or (1.0853, 14.2207) Test the null hypothesis that, all else being equal, the energy efficiency ratio of air conditioners does not affect their price against the alternative that the higher the energy efficiency ratio, the higher the price. H0: (2 = 0, HA: (2 > 0  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Note that t(15, 0.025) = 2.131, and t(15, 0.01) = 2.947. We can reject H0 at the 2.5% level but not at the 1% level. (This last analysis is not required in the problem; it is offered here simply to illustrate the level of significance at which this null hypothesis could be rejected. Although we can't get the p-value of this test directly from our table, we can say that the p-value is somewhere between 0.02 and 0.05. Using the Excel function =TDIST(2.194,15,1) we can get the specific p-value for this test: 0.0222.) 5. In a study of foreign holdings in U.S. banks, the following sample regression was obtained, based on fourteen annual observations: y= -3.248+ 0.101x1- 0.244x2+ 0.057x3R2 = 0.93(0.023)(0.080)(0.00925)where y= Year-end share of assets in U.S. bank subsidiaries held by foreigners, as a percentage of total assetsx1= Annual change, in billions of dollars, in foreign direct investment in the U.S. (excluding finance, insurance, and real estate)x2= Bank price-earnings ratiox3= Index of the exchange value of the dollarThe figures in parentheses beneath the coefficient estimates are the corresponding estimated standard errors. Find a 90% confidence interval for (1 and interpret your result. Note that:n= 14 EMBED Equation.3 = 0.101 EMBED Equation.3 = 0.023t(10, 0.05)= 1.812 90% C.I.:= 0.101 ( 1.812(0.023)Or (0.0593, 0.1424)Test the null hypothesis that (2 is zero, against the alternative that it is negative, and interpret your result. H0: (2 = 0, HA: (2 < 0  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Note that -t(10, 0.01) = -2.764, and -t(10, 0.005) = -3.169. We can reject H0 at the 1% level but not at the 0.5% level. The p-value for this test (obtained using Excel), is 0.006126. Test the null hypothesis that (3 is zero against the alternative that it is positive and interpret your result. H0: (3 = 0, HA: (3 > 0  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Note that our t table with 10 degrees of freedom doesn't go beyond 3.169 (where ( = 0.005). We can certainly reject H0 at the 0.5% level, and beyond. 6. A survey research group conducts regular studies of households through mail questionnaires and is concerned about the factors influencing the response rate. In an experiment, thirty sets of questionnaires were mailed to potential respondents. The regression model fitted to the resulting data set was  EMBED Equation.3  where  EMBED Equation.3 = Percentage of responses received EMBED Equation.3 = Number of questions asked EMBED Equation.3 = Length of questionnaire, in number of wordsPart of the SAS computer output from the estimated regression is shown here. R-Square0.637 Parameter Estimatet For H0: Parameter = 0Std. Error Of EstimateIntercept74.3652xl-1.8345-2.890.6349x2-0.0162-1.780.0091Interpret the estimated partial regression coefficients. All else being equal, an increase of one question to the questionnaire results in a decrease of 1.834 in expected percentage of responses received. All else being equal, an increase in one word in length of the questionnaire results in a decrease of 0.016 in expected percentage of responses received. Interpret the coefficient of determination. 63.7% of the variability in percentage responses received is explained by its linear relationship with the number of questions asked and the number of words. Test at the 1% significance level the null hypothesis that taken together, the two independent variables do not linearly influence the response rate. H0: (1 = (2 = 0, HA: At least one (i ( 0 (i = 1, 2) OK, no fair. This one requires an alternative formula for the F statistic, which doesn't appear in Levine, but is on the Regression Formula sheet:  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Now we look at the 0.01 F table (page E-10 in Levine), and see that the F(2,27,0.01) = 5.49. Since our observed F is greater than this critical value, we can reject H0 at the 1% level. Find and interpret a 99% confidence interval for (1. Note that t(27, 0.005) = 2.771 99% C.I.:= -1.8345 + 2.771(0.6349)Or (-3.5938, -0.0752)Test the null hypothesis H0: (2 = 0 against the alternative HA. (2 < 0 and interpret your findings. H0: (2 = 0, HA: (2 < 0  EMBED Equation.3  EMBED Equation.3 Note that -t(27, 0.05) = -1.703, and -t(27, 0.025) = -2.052. We can reject H0 at the 5% level but not at the 2.5% level. 7. Based on data on 2,679 college basketball players, the following model was fitted:  EMBED Equation.3 , where: Y= Minutes played in seasonx1= Field goal percentagex2= Free throw percentagex3= Rebounds per minutex4= Points per minutex5= Fouls per minutex6 = Steals per minutex7 = Blocked shots per minutex8= Turnovers per minutex9= Assists per minuteThe least squares parameter estimates (with standard errors in parentheses):  EMBED Equation.3 = 358.848(44.695) EMBED Equation.3 = -3923.5(120.6) EMBED Equation.3 = 0.6742(0.0639) EMBED Equation.3 = 480.04(224.9) EMBED Equation.3 = 0.2855(0.0388) EMBED Equation.3 = 1350.3(212.3) EMBED Equation.3 = 303.81(77.73) EMBED Equation.3 = -891.67(180.87) EMBED Equation.3 = 504.95(43.26) EMBED Equation.3 = 722.95(110.98)The coefficient of determination was R2 = 0.5239 Find and interpret a 90% confidence interval for  EMBED Equation.3 . Note that our model is based on a sufficiently large data set to use z as an approximation of t. Specifically, t(2669, 0.05) ( 1.645 90% CI:= 480.04 ( 1.645(224.9)Or (110.08, 850.00)We are 90% confident that playing time will increase by somewhere between 110.08 and 850 minutes per season for every increase of 1 steal per minute. Find and interpret a 99% confidence interval for EMBED Equation.3 . t(2669, 0.005) ( 2.576 99% CI: = 1350.3 ( 2.576(212.3)Or (803.4152, 1897.1848)Test against the alternative that it is negative the null hypothesis that  EMBED Equation.3  is 0. Interpret your result. H0: (8 = 0, HA: (8 < 0  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Note that -t(2669, 0.005) ( -2.576. We can reject H0 at any reasonable level of (. Test against the alternative that it is positive the null hypothesis that  EMBED Equation.3  is 0. Interpret your result. H0: (9 = 0, HA: (9 > 0  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 We can reject H0 at any reasonable level of (. Interpret the coefficient of determination. 52.39% of the variability in minutes played can be explained through its linear dependence on the independent variables. Find and interpret the coefficient of multiple correlation.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 R is the correlation between observed and predicted minutes played:  EMBED Equation.3  8. A marketing analyst for a major shoe manufacturer is considering the development of a new brand of running shoes. The marketing analyst wishes to determine which variables can be used in predicting durability (or the effect of long-term impact). Two independent variables are to be considered, X1 (FOREIMP), a measurement of the forefoot shock-absorbing capability, and X2 (MIDSOLE), a measurement of the change in impact properties over time, along with the dependent variable Y (LTIMP), which is a measure of the long-term ability to absorb shock after a repeated impact test. A random sample of 15 types of currently manufactured running shoes was selected for testing. Using Microsoft Excel, we provide the following (partial) output: ANOVADFSSMSFSIGNIFICANCE FRegression212.610206.3051097.690.0001Residual120.774530.06454Total1413.38473 VARIABLECOEFFICIENTSSTANDARD ERRORt STATp-VALUEIntercept-0.02686.06905-0.39Foreimp0.79116.0629512.570.0000Midsole0.60484.071748.430.0000Assuming that each independent variable is linearly related to long-term impact, state the multiple regression equation.  EMBED Equation.3  EMBED Equation.3  Interpret the meaning of the slopes in this problem. For every one-unit increase in FOREIMP, we expect to see a 0.79116-unit increase in LTIMP. For every one-unit increase in MIDSOLE, we expect to see a 0.60484-unit increase in LTIMP. Compute the coefficient of multiple determination  EMBED Equation.3  and interpret its meaning.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 94.21% of the variation in the long-term ability of a shoe to absorb shock can be explained by variation in forefoot absorbing capability and variation in midsole impact. Compute the adjusted R-square ( EMBED Equation.3 ).  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  9. Suppose a large consumer products company wants to measure the effectiveness of different types of advertising media in the promotion of its products. Specifically, two types of advertising media are to be considered: radio and television advertising and newspaper advertising (including the cost of discount coupons). A sample of 22 cities with approximately equal populations is selected for study during a test period of 1 month. Each is allocated a specific expenditure level for both radio and television advertising and newspaper advertising. The sales of the product (in thousands of dollars) and also the levels of media expenditure during the test month are recorded with the following results: CitySALES ($000)RADIO AND TELEVISION ADVERTISING ($000)NEWSPAPER ADVERTISING ($000)197304021,119040387525254625252559103030697130307931353581,177353598824025109824025111,6284545121,5774545131,04450014914500151,3295525161,3305525171,4056030181,4366030191,5216535201,7416535211,8667040221,7177040 Excel output: Regression StatisticsMultiple R0.8993R Square0.8087Adjusted R Square0.7886Standard Error158.9041Observations22 dfSSMSFSignificance FRegression22028032.68961014016.344840.15820.0000Residual19479759.901425250.5211Total212507792.5909 CoefficientsStandard Errort StatP-valueIntercept156.4304126.75791.23410.2322RADIO&TV13.08071.75947.43490.0000NEWSPAPER16.79532.96345.66760.0000 On the basis of the results obtained: State the multiple regression equation.  EMBED Equation.3  EMBED Equation.3 Interpret the meaning of the slopes in this problem. For every one-unit increase in Radio/TV advertising, we expect to see a 13.08-unit increase in Sales. For every one-unit increase in Newspaper advertising, we expect to see a 16.80-unit increase in Sales. Predict the average sales for a city in which radio and television advertising is $20,000 and newspaper advertising is $20,000.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 (or $753,950.00) Compute the coefficient of multiple determination  EMBED Equation.3  and interpret its meaning.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 80.87% of the variation in Sales is explained by variation in Radio/TV advertising and Newspaper advertising. Compute the  EMBED Equation.3 .  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  10. The director of broadcasting operations for a television station wants to study the issue of "standby hours," hours in which unionized graphic artists at the station are paid but are not actually involved in any activity. The variables to be considered are: Standby hours EMBED Equation.3 the total number of standby hours per weekTotal staff present EMBED Equation.3 the weekly total of people-days over a 7-day weekRemote hours EMBED Equation.3 the total number of hours worked by employees at locations away from the central plantThe results for a period of 26 weeks are shown as follows: WEEKSTANDBY HOURSTOTAL STAFF PRESENTREMOTE HOURSWEEKSTANDBY HOURSTOTAL STAFF PRESENTREMOTE HOURS1245338414141613074022177333598152743221513271358656162453352284211372631172013502715196339528181833394406135289409192373274757195334382201753283478118293399211523194499116325343221883253361014731133823188322267111543043532419731723512146312289252613151641311528338826232331270 Excel output: Regression StatisticsMultiple R0.6999R Square0.4899Adjusted R Square0.4456Standard Error35.3873Observations26 dfSSMSFSignificance FRegression227662.542913831.271411.04500.0004Residual2328802.07251252.2640Total2556464.6154 CoefficientsStandard Errort StatP-valueIntercept-330.6748116.4802-2.83890.0093STAFF1.76490.37904.65620.0001REMOTE-0.13900.0588-2.36350.0269 On the basis of the results obtained: State the multiple regression model.  EMBED Equation.3  EMBED Equation.3 Interpret the meaning of the slopes in this problem. For every one-unit increase in the weekly total of people-days over a 7-day week, we expect to see a 1.765-unit increase in the total number of standby hours per week (all other factors held constant). For every one-unit increase in the total number of hours worked by employees at locations away from the central plant, we expect to see a 0.139-unit decrease in the total number of standby hours per week (all other factors held constant). Predict the average standby hours for a week in which the total staff present is 310 people-days and the remote hours are 400.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  Compute  EMBED Equation.3  and interpret its meaning.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 48.99% of the variation in standby hours per week is explained by variation in people-days and in hours worked by employees at locations away from the central plant. Compute  EMBED Equation.3 .  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  11. In Problem 8 above the durability of a brand of running shoe was predicted based on a measurement of the forefoot shock-absorbing capability and a measurement of the change in impact properties over time. The following analysis of variance table was obtained: ANOVADFSSMSFSignificance FRegression212.610206.3051097.690.0001Residual120.774530.06454Total1413.38473Determine whether there is a significant relationship between long-term impact and the two explanatory variables at the 0.05 level of significance. First, we determine the critical cut-off value for F, taking into account our desired level of significance and the degrees of freedom. Looking in the F-table in the back of the book, we can see that (from the F table)  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Decision rule: We will reject the null hypothesis (that is, there is no significant relationship between long-term impact and the two explanatory variables) if our observed F is greater than 3.89.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we can reject the null hypothesis and conclude that there is a significant relationship between long-term impact and the two explanatory variables at the 0.05 level of significance. (This whole procedure is done for you in the Excel output; all you need to do is look at the p-value associated with the F statistic.) Interpret the meaning of the p-value. The p-value for the F statistic (called Significance F in Excel output) is the smallest value of alpha at which we could reject the null hypothesis in the F test. In this case, the p-value of 0.0001 is sufficiently small to allow us to reject the null hypothesis at any reasonable level of alpha. 12. In Problem 9 above the amount of radio and television advertising and newspaper advertising was used to predict sales. Using the computer output you obtained to solve that problem, Determine whether there is a significant relationship between sales and the two explanatory variables (radio and television advertising and newspaper advertising) at the 0.05 level of significance. Here is the ANOVA table from our previous regression analysis: dfSSMSFSignificance FRegression22028032.68961014016.344840.15820.0000Residual19479759.901425250.5211Total212507792.5909Looking in the F-table in the back of the book, we can see that (from the F table)  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Decision rule: We will reject the null hypothesis (that is, there is no significant relationship between long-term impact and the two explanatory variables) if our observed F is greater than 3.522.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we can reject the null hypothesis and conclude that there is a significant relationship between long-term impact and the two explanatory variables at the 0.05 level of significance. (This whole procedure is done for you in the Excel output; all you need to do is look at the p-value associated with the F statistic.) Interpret the meaning of the p-value. The p-value for the F statistic is the smallest value of alpha at which we could reject the null hypothesis in the F test. In this case, the p-value of approximately 0.0000 is sufficiently small to allow us to reject the null hypothesis at any reasonable level of alpha. 13. In Problem 8 above the durability of a brand of running shoe was predicted based on a measurement of the forefoot shock-absorbing capability and a measurement of the change in impact properties over time for a sample of 15 pairs of shoes. Use the following computer output: VARIABLECOEFFICIENTSSTANDARD ERRORt STATp-VALUEIntercept-0.02686.06905-0.39Foreimp0.79116.0629512.570.0000Midsole0.60484.071748.430.0000Set up a 95% confidence interval estimate of the population slope between long-term impact and forefoot impact.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Or, EMBED Equation.3 At the 0.05 level of significance, determine whether each explanatory variable makes a significant contribution to the regression model. On the basis of these results, indicate the independent variables that should be included in this model. In each case, we are testing the null hypothesis that the true slope (either  EMBED Equation.3  or  EMBED Equation.3 ) is in fact zero. In either case we will use a decision rule based on the critical value of t where alpha is 0.05 and there are  EMBED Equation.3  degrees of freedom.  EMBED Equation.3  EMBED Equation.3 For the Foreimp variable, our test statistic is  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we reject the null hypothesis and conclude that there is a relationship between durability and forefoot shock-absorbing capability. This is also evidenced by the very small p-value in the output associated with the forefoot shock-absorbing capability independent variable (approximately 0.0000). For the Midsole variable, our test statistic is  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we reject the null hypothesis and conclude that there is a relationship between durability and the change in impact properties over time. This is also confirmed by the very small p-value associated with the "change in impact properties over time" independent variable (approximately 0.0000). We conclude that both of these independent variables are useful, and ought to remain in the model. (Note that these hypothesis tests are done for you in the Excel output; all you need to do is look at the p-value associated with each estimated slope coefficient.) 14. In Problem 9 above the amount of radio and television advertising and newspaper advertising was used to predict sales. Use the computer output you obtained to solve that problem. Here is a subset of the Excel output produced before: CoefficientsStandard Errort StatP-valueIntercept156.4304126.75791.23410.2322RADIO&TV13.08071.75947.43490.0000NEWSPAPER16.79532.96345.66760.0000Set up a 95% confidence interval estimate of the population slope between sales and radio and television advertising.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Or, EMBED Equation.3 At the 0.05 level of significance, determine whether each explanatory variable makes a significant contribution to the regression model. On the basis of these results, indicate the independent variables that should be included in this model. In each case, we are testing the null hypothesis that the true slope (either  EMBED Equation.3  or  EMBED Equation.3 ) is in fact zero. In either case we will use a decision rule based on the critical value of t where alpha is 0.05 and there are  EMBED Equation.3  degrees of freedom.  EMBED Equation.3  EMBED Equation.3 For the Radio & TV variable, our test statistic is  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we reject the null hypothesis and conclude that there is a relationship between Radio and TV advertising and Sales. This is also evidenced by the very small p-value in the Excel output associated with the Radio & TV independent variable (approximately 0.0000). For the Newspaper variable, our test statistic is  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we reject the null hypothesis and conclude that there is a relationship between Newspaper advertising and Sales. This is also evidenced by the very small p-value associated with the "Newspaper" independent variable (approximately 0.0000). We conclude that both of these independent variables are useful, and ought to remain in the model. (Note that these hypothesis tests are done for you in the Excel output; all you need to do is look at the p-value associated with each estimated slope coefficient.) 15. A real estate association in a suburban community would like to study the relationship between the size of a single-family house (as measured by the number of rooms) and the selling price of the house. The study is to be carried out in two different neighborhoods, one on the east side of the community and the other on the west side. A random sample of 20 houses was selected with the following results: SELLING PRICENUMBER OF ROOMSNEIGHBORHOODSELLING PRICENUMBER OF ROOMSNEIGHBORHOOD109.67East108.56West107.48East181.313West140.39East137.410West146.512East146.210West98.26East142.49West137.89East123.78West124.110East129.68West113.28East143.69West127.89East160.711West125.38East148.39West Excel output: Regression StatisticsMultiple R0.9311R Square0.8670Adjusted R Square0.8513Standard Error7.7393Observations20 dfSSMSFSignificance FRegression26635.48083317.740455.39080.0000Residual171018.248759.8970Total197653.7295 CoefficientsStandard Errort StatP-valueIntercept56.43399.88335.71000.0000ROOMS9.21891.02968.95370.0000EAST-12.69673.5354-3.59130.0023 Using Microsoft Excel: State the multiple regression equation.  EMBED Equation.3  EMBED Equation.3  Interpret the meaning of the slopes in this problem. For every one-unit increase in the number of Rooms, we expect to see a 9.2189-unit increase in Selling Price (an expected increase of $9,218.90 for each additional room). For every one-unit increase in East, we expect to see a 12.6967-unit decrease in Selling Price (all else being equal, a house on the East side is expected to sell for $12,696.70 less than a house on the West side). Predict the average selling price for a house with nine rooms that is located on the east side of the community.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 (or $126,707.30) Perform a residual analysis on your results and determine the adequacy of the fit of the model. A scatter diagram of the residuals doesnt reveal any obvious pattern:  EMBED Excel.Sheet.8  A histogram of the errors seems reasonably bell-shaped, given that we only have 20 data:  EMBED Excel.Sheet.8  Without doing any sophisticated hypothesis tests, it seems reasonable to use the model on the assumption that there is no problem with heteroscedasticity or non-normality of the residuals. Determine whether there is a significant relationship between selling price and the two explanatory variables (rooms and neighborhood) at the 0.05 level of significance. First, we determine the critical cut-off value for F, taking into account our desired level of significance and the degrees of freedom. Looking in the F-table in the back of the book, we can see that (from the F table)  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 You can also get the critical value of F using the Excel function: =FINV(0.05, 2, 17). Therefore, we will reject the null hypothesis (that there is no significant relationship between long-term impact and the two explanatory variables) if our observed F is greater than 3.592.  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we can reject the null hypothesis and conclude that there is a significant relationship between long-term impact and the two explanatory variables at the 0.05 level of significance. At the 0.05 level of significance, determine whether each explanatory variable makes a contribution to the regression model. On the basis of these results, indicate the regression model that should be used in this problem. In each case, we are testing the null hypothesis that the true slope (either  EMBED Equation.3  or  EMBED Equation.3 ) is in fact zero. In either case we will use a decision rule based on the critical value of t where alpha is 0.05 and there are  EMBED Equation.3  degrees of freedom.  EMBED Equation.3  EMBED Equation.3 For the Rooms variable, our test statistic is  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we reject the null hypothesis and conclude that there is a relationship between the number of Rooms and Selling Price. This is also evidenced by the very small p-value in the Excel output associated with the Rooms independent variable (approximately 0.0000). For the East variable, our test statistic is  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Therefore, we reject the null hypothesis and conclude that there is a relationship between Neighborhood and Selling Price. This is also evidenced by the very small p-value associated with the "East" independent variable (approximately 0.0023). We conclude that both of these independent variables are useful, and ought to remain in the model. Set up 95% confidence interval estimates of the population slope for the relationship between selling price and number of rooms and between selling price and neighborhood. For Rooms:  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Or, EMBED Equation.3 For Neighborhood:  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 Or, EMBED Equation.3 Interpret the meaning of the coefficient of multiple determination,  EMBED Equation.3 . 86.7% of the variation in selling prices is explained by the combined effects of the number of rooms and the neighborhood. Compute  EMBED Equation.3 .  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3  EMBED Equation.3 What assumption about the slope of selling price with number of rooms must be made in this problem? We are implicitly assuming that the effect of the number of rooms on selling price is the same, regardless of which part of town a house happens to be in. We could avoid making this assumption by having a separate model for each of the two sides of town.     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R Square? Adjusted R Square<;? Standard Erroroc@  Observations~ 6@ B6XU""""""""""""""""""""""+275>;!T0 "#$% & '()* ./!ANOVA!" "df "SS "MS "F"Significance F# Regression~ #@#d>A#d.A#FAD@#҄N&>$Residual~ $3@$Ե?HA$Z@ $%Total~ %5@%K"CA %&'' Coefficients'Standard Error't Stat'P-value'( Intercept(ƍc@(_@(vӾ?(kK?()RADIO&TV)uO)*@)GYe&?)fPI@)?Gg>)* NEWSPAPER*=X0@*Fz@*/@*~82>* ..FbYp@/ D) ./1/3Ϙ@/D(D)D*//w%.}t@. D* /&`mUBohghA>@111   dMbP?_*+%MZ\\STERNTREE\MVPTR1.MVILLE.DEPTS|sXXLetter!p G'''' e"dX??U} I} }  }  } } m /T0     mmm0 m WEEK STANDBY HOURSSTAFFREMOTE?n@ u@y@@ f@t@@@p@`v@@@`j@@w@@@h@0u@@@`@r@y@@`h@t@w@ @]@Pr@x@ "@]@Pt@pu@ $@`b@ps@ u@ &@@c@s@v@ (@@b@s@r@ *@\@q@@x@,@ d@0s@ y@.@ q@ t@b@0@n@t@l@1@ i@u@p@2@f@0u@{@3@m@pt@}@4@e@t@u@5@c@s@|@6@g@Pt@u@7@g@ t@p@8@h@s@`m@9@Pp@s@d@:@m@t@p@SUMMARY OUTPUT Regression Statistics  Multiple RC_e? BXP""""""""""""""""""""""""""+2 T0!"# % &'() * +,-.  R Square 3/Z? !Adjusted R Square!c_h?!"Standard Error"qA@"# Observations~ #:@#%ANOVA%& &df &SS &MS &F&Significance F' Regression~ '@'h`@'h`@' &@'k͖x@.+.  "phe10 d??ES3` 4#` 4#` 4#` 4# 3d 3Q: ResidualQ ;Q ;Q3_ M MM   4E4D $% M 3O&Q4$% M 3O&Q4FA` 3O! 3*T@#M43*#M! M4% @M3O& Q BPredicted Selling Price ($1000)'4% CQMZ3O{& Q 0Residual Error ($1000)'4523   43d" 3_ M MM  MM<444% SLM03Oa&Q "Scatter Diagram'44e B[@1E_]@H_@t Kc@F>S?X@H_@*`@1E_]@H_@ 1E_]@ [@ Wf@ 4b@ 4b@-la@&E`@&E`@-la@qGBc@-la@eH?a-$@ @T  5Rh@ Nr& 3`b4u'@ o g`@)@*!@e> *   dMbP?_*+%MZ\\STERNTREE\MVPTR1.MVILLE.DEPTS|sXXLetter!p G'''' e"dX??U} m} m }  }  } } m /T0  G   wGwwT0T0TT PRICEROOMS EAST  PredictedResidualh@@? EastQ B[@;D,D-DD.DEast  B!H? DD@ @? EastQ1E_]@;D,D-DD.DEast  B!a-$ DDg@"@? EastQH_@;D,D-DD.DEast  B!S?X@;D,D-DD.DEast  B!8Co9 DD@"@? EastQH_@;D,D-DD.DEast  B!@  D D   [@@ West Q [@ ;D,D-D D.D East  B! @T   D D  @*@ West Q Wf@ ;D,D-D D.D East  B! 5Rh@  D D  @$@ West Q 4b@;D,D-D D.D East  B! Nr& D D @$@ WestQ4b@;D,D-DD.DEast  B!3`b DD@"@ WestQ-la@;D,D-DD.DEast  B!4u'@ DD)@ @ WestQ&E`@;D,D-DD.DEast  B!  DDP@ @ WestQ&E`@;D,D-DD.DEast  B!o DD @"@ WestQ-la@;D,D-DD.DEast  B! g`@ DDc@&@ WestQqGBc@;D,D-DD.DEast  B!)@ DD@"@ WestQ-la@;D,D-DD.DEast  B!*!@ DDSUMMARY OUTPUTRegression Statistics Multiple Rr"?6n18 T0!"# % &'() * +G,-.w  R Square #? !Adjusted R Square!t=?!"Standard Error"@"# Observations~ #4@#%ANOVA%& &df &SS &MS &F&Significance F' Regression~ '@'ӸT{@'ӸT{@')K@'q7c>(Residual~ (1@(riя@(>cM@ ()Total~ )3@)p@ )*++ Coefficients+Standard Error+t Stat+P-value+, Intercept,P@7L@,qA#@, @,/>,-ROOMS-.p"@-k+,Vy?-B$K!@-~n_wt>- .EAST."ʋd).HqH @.G;T .>pb?. I9B?9&`mUBohd>@   dMbP?_*+%"??csU>@ SummaryInformation(%DocumentSummaryInformation8_1004796122-( F`#pOle H\l  David JurandjuranuMicrosoft Excel@kt,@( ,՜.+,D՜.+,` PX|  Columbia Business SchoolD1 5b91015Sheet6 15 Scatter  WorksheetsCharts 6> _PID_GUIDAN{39A4AA80-97EA-11D2-85CA-00A0248058E0} FMicrosoft Excel ChartBiff8Excel.Sheet.89qOh+'0@CompObj'*bObjInfoWorkbook)+$cSummaryInformation(, Ba= < ='t"8X1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial1ZArial"$"#,##0_);\("$"#,##0\)!"$"#,##0_);[Red]\("$"#,##0\)""$"#,##0.00_);\("$"#,##0.00\)'""$"#,##0.00_);[Red]\("$"#,##0.00\)7*2_("$"* #,##0_);_("$"* \(#,##0\);_("$"* "-"_);_(@_).))_(* #,##0_);_(* \(#,##0\);_(* "-"_);_(@_)?,:_("$"* #,##0.00_);_("$"* \(#,##0.00\);_("$"* "-"??_);_(@_)6+1_(* #,##0.00_);_(* \(#,##0.00\);_(* "-"??_);_(@_) 0.0000000000 0.000000000 0.00000000 0.0000000 0.000000 0.00000 0.00000.0 0.000 0.000000000000.000000000000                + ) , *     p  p "x@ &x@  t  t !x@  t  5b P9 #10 215O 15 Scatter[Z15 HistbSheet6 u.df = 10 df = 1000z check with zt1-tail2-tail+3.05 standard deviations away from the mean10 d.f.City SALES ($000) NEWSPAPERRADIO&TVSUMMARY OUTPUTRegression Statistics Multiple RR SquareAdjusted R SquareStandard Error ObservationsANOVA RegressionResidualTotal InterceptdfSSMSFSignificance F Coefficientst StatP-valueWEEK STANDBY HOURSSTAFFREMOTEEastWestPRICEROOMSEAST PredictedMore Frequency* đ,y;v4 MH ͐ .P10P@ mԣ01.04Н0(`T04$0k0dip0mm;v0@4T0v0@:-0h mT00t002 zm00r E000rrt߿0rmtt0pmPercent0]P~0~0Pm  0\y0{T0T00{Tye0yUmp00y LmyyUm00 yy kT0e0{T0{L0{{T0LkT00 /0 أ{ڬp08ם0p00 mfm0mfmS]0mPJmylgm%0a0oyo.   dMbP?_*+%MHP LaserJet 4L@g,,@MSUDOHP LaserJet 4L?\ad "d,,??U T0   4+3.05 standard deviations away from the mean t t check with z1-tail(my?ithffffff@ A-(=S?ffffff@A-&pR?ffffff@A&2-tail(m?ffffff@ A-(=c?ffffff@A-+pb?ffffff@A&df = 10 df = 1000 z10 d.f.  1-tail 2-tail ~ ?! եZ @ D  AL% I @ D  AL ~ @! 4@  D  AL% եZ @ D  AL @LSV(@l>@   dMbP?_*+%MZ\\STERNTREE\MVPTR1.MVILLE.DEPTS|sXXLetter!p G'''' e"dX??U} m}  }  } $ } } m 0T0     0k0mT0m m0m  City SALES ($000)RADIO&TV NEWSPAPER?h@D@@|@D@@X@9@9@@@9@9@@p@>@>@@X@>@>@@@A@A@ @d@A@A@ "@@D@9@ $@@D@9@ &@p@F@F@ (@@F@F@ *@P@I@,@@I@.@Ĕ@K@9@0@Ȕ@K@9@1@@N@>@2@p@N@>@3@ė@@P@A@4@4@@P@A@5@(@Q@D@6@Ԛ@Q@D@SUMMARY OUTPUT Regression Statistics  Multiple R(? R Square? Adjusted R Square<;? Standard Erroroc@  Observations~ 6@ B6XU""""""""""""""""""""""+275>;!T0 "#$% & '()* ./!ANOVA!" "df "SS "MS "F"Significance F# Regression~ #@#d>A#d.A#FAD@#҄N&>$Residual~ $3@$Ե?HA$Z@ $%Total~ %5@%K"CA %&'' Coefficients'Standard Error't Stat'P-value'( Intercept(ƍc@(_@(vӾ?(kK?()RADIO&TV)uO)*@)GYe&?)fPI@)?Gg>)* NEWSPAPER*=X0@*Fz@*/@*~82>* ..FbYp@/ D) ./1/3Ϙ@/D(D)D*//w%.}t@. D* /&`mUBohghA>@111   dMbP?_*+%MZ\\STERNTREE\MVPTR1.MVILLE.DEPTS|sXXLetter!p G'''' e"dX??U} I} }  }  } } m /T0     0k0mT0mm0 m WEEK STANDBY HOURSSTAFFREMOTE?n@ u@y@@ f@t@@@p@`v@@@`j@@w@@@h@0u@@@`@r@y@@`h@t@w@ @]@Pr@x@ "@]@Pt@pu@ $@`b@ps@ u@ &@@c@s@v@ (@@b@s@r@ *@\@q@@x@,@ d@0s@ y@.@ q@ t@b@0@n@t@l@1@ i@u@p@2@f@0u@{@3@m@pt@}@4@e@t@u@5@c@s@|@6@g@Pt@u@7@g@ t@p@8@h@s@`m@9@Pp@s@d@:@m@t@p@SUMMARY OUTPUT Regression Statistics  Multiple RC_e? BXP""""""""""""""""""""""""""+2 T0!"# % &'() * +,-.  R Square 3/Z? !Adjusted R Square!c_h?!"Standard Error"qA@"# Observations~ #:@#%ANOVA%& &df &SS &MS &F&Significance F' Regression~ '@'h`@'h`@' &@'k͖x@.+.   dMbP?_*+%MZ\\STERNTREE\MVPTR1.MVILLE.DEPTS|sXXLetter!p G'''' e"dX??U} m} m }  }  } } m }  Column GX/  T0                0k 0 m   T0  m    mPRICEROOMS EAST  PredictedResidual     Frequency=@$@? EastQ*`@;D,D-DD.DEast  B!u,' DD~ 0  . @$@ WestQ4b@;D,D-DD.DEast  B!Nr& DD~ +  (@ @ @? EastQ1E_]@;D,D-DD.DEast  B!a-$ DD~ %  "@ Pb@(@? EastQt Kc@;D,D-DD.DEast  B!a}"u DD~   ? )@ @ WestQ&E`@;D,D-DD.DEast  B!  DD~   @ @ @? EastQ1E_]@;D,D-DD.DEast  B!PPv' DD~   @  [@@ WestQ[@;D,D-DD.DEast  B!@T  DD~ ?  @@ @$@ WestQ4b@;D,D-DD.DEast  B!3`b  DD~ @  @?  .@@? East Q F>S?X@ ;D,D-D D.D East  B! 8Co9  D D  ~ @  "@@  P@ @ West Q &E`@ ;D,D-D D.D East  B! o  D D  ~ %@  (@?  @"@? East Q H_@ ;D,D-D D.D East  B! gz? D D  ~ +@  .@?  h@@? East Q  B[@ ;D,D-D D.D East  B! H? D D  ~ 0@  2@  c@&@ West Q qGBc@ ;D,D-D D.D East  B! )@ D D  ~ 3@  5@ @"@ WestQ-la@;D,D-DD.DEast  B!4u'@ DD~ 6@  8@  @"@ WestQ-la@;D,D-DD.DEast  B! g`@  DD    More @*@ WestQWf@;D,D-DD.DEast  B!5Rh@ DDy@ @? EastQ1E_]@;D,D-DD.DEast  B!>@ DD@"@ WestQ-la@;D,D-DD.DEast  B!*!@ DD@"@? EastQH_@;D,D-DD.DEast  B!(Residual~ (1@(riя@(>cM@ ()Total~ )3@)p@ )*++ Coefficients+Standard Error+t Stat+P-value+, Intercept,P@7L@,qA#@, @,/>,-ROOMS-.p"@-k+,Vy?-B$K!@-~n_wt>- .EAST."ʋd).HqH @.G;T .>pb?. I9B?9&`mUBohd>@   MZ\\STERNTREE\MVPTR1.MVILLE.DEPTS|sXXLetter!p G'''' e"dX??3` 4#` 4#` 4#` 4# 3d 3Q: ResidualQ ;Q ;Q3_ M MM   4E4D $% M 3O&Q4$% M 3O&Q4FA` 3O! 3*T@#M43*#M! M4% @M3O& Q BPredicted Selling Price ($1000)'4% CQMZ3O{& Q 0Residual Error ($1000)'4523   43d" 3_ M MM  MM<444% SLM03Oa&Q "Scatter Diagram'44e B[@1E_]@H_@t Kc@F>S?X@H_@*`@1E_]@H_@ 1E_]@ [@ Wf@ 4b@ 4b@-la@&E`@&E`@-la@qGBc@-la@eH?a-$@ @T  5Rh@ Nr& 3`b4u'@ o g`@)@*!@e> *  MZ\\STERNTREE\MVPTR1.MVILLE.DEPTS|sXXLetter!p G'''' e"dX??3` 4# ` 4# ` 4# ` 4# 3d  3QQ ; Q ; Q3_4E4D $% M 3O& Q4$% M 3O& Q4FAY  3O! 3 b#M43*#M! M4% cM3OR& Q  Residual Error'4% CMZ3O>& Q  Frequency'4523 O  43" 44% S~M03O?& Q  Histogram'44 e.("@@"@ (@ .@ 2@e@@?@@@?@ ? ? e>    dMbP?_*+%"??U>@ H\l  David JurandjuranuMicrosoft Excel@kt,@( ,՜.+,D՜.+,l( PX|  Columbia Business SchoolD1 5b91015Sheet6 15 Scatter15 Hist  DocumentSummaryInformation8_1004796478/FppOle CompObj.0fWorksheetsCharts 6> _PID_GUIDAN{39A4AA80-97EA-11D2-85CA-00A0248058E0} FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo1Equation Native _10047964964FppOle Xl` F d.f.numerator, d.f.denominator() FMicrosoft Equation 3.0 DS Equation Equation.39q9 =F 2,CompObj35fObjInfo6Equation Native U_10047965579Fpp17() FMicrosoft Equation 3.0 DS Equation Equation.39q!n =3.592 FMicrosoft Equation 3.0 DS EqOle CompObj8:fObjInfo;Equation Native =_1004940548>Fp[ZOle CompObj=?fObjInfo@uation Equation.39qX =SSRkSSEn"k"1 FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native t_1004940553<PCF[Z[ZOle CompObjBDfObjInfoEEquation Native _1004940556HF[Z[ZOle d =6635.521018.217 FMicrosoft Equation 3.0 DS Equation Equation.39q =55.4CompObjGIfObjInfoJEquation Native 9_1004940632AZMF[Z[ZOle CompObjLNfObjInfoOEquation Native H FMicrosoft Equation 3.0 DS Equation Equation.39q,II n"k"1=17 FMicrosoft Equation 3.0 DS Equation Equation.39q_1004940619FURF[Z[ZOle CompObjQSfObjInfoTE\ t 17,0.025() FMicrosoft Equation 3.0 DS Equation Equation.39q ITJ =2.11Equation Native a_1004940627WF[Z[ZOle CompObjVXfObjInfoYEquation Native <_1004940655dn\F[Z[ZOle  FMicrosoft Equation 3.0 DS Equation Equation.39q, JTJ t 0 FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj[]fObjInfo^Equation Native 8_1004940644aF[Z[ZOle CompObj`bfObjInfocEquation Native tX\JTJ ="  1 "0s "  1  FMicrosoft Equation 3.0 DS Equation Equation.39qP, JDI =9.218_1004940648_ifF[Z[ZOle CompObjegfObjInfohEquation Native l_1004940652kF[Z[ZOle CompObjjlf9"01.0296 FMicrosoft Equation 3.0 DS Equation Equation.39q , JDI =8.95 FMicrosoft Equation 3.0 DS EqObjInfomEquation Native <_1004940668pF[Z[ZOle CompObjoqfObjInforEquation Native t_1004940671uF[Z[Z  "%(+.1458;<?BCFILORW\adehkloruyz{|}~uation Equation.39qX, JTJ ="  2 "0s "  2  FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjtvfObjInfowEquation Native tX, JI ="12.6967"03.5354_ FMicrosoft Equation 3.0 DS Equation Equation.39q$, J 5J ="3.59_1004940677zF[Z[ZOle  CompObjy{ fObjInfo| Equation Native @_1004940732xF[Z[ZOle CompObj~f FMicrosoft Equation 3.0 DS Equation Equation.39q, JI "  1 t n"k"1,)2() s  1  FMicrosoft Equation 3.0 DS EqObjInfoEquation Native _1004940736F[Z[ZOle CompObjfObjInfoEquation Native _1004940739}F[Z[Zuation Equation.39q, JJ =9.2189t 17,0.025() 1.0296() FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfo Equation Native !|`, JI =9.21892.111.0296() FMicrosoft Equation 3.0 DS Equation Equation.39q<, JI =9.219_1004940743F[ZOle #CompObj$fObjInfo&Equation Native 'X_1004940746FOle )CompObj*f2.172 FMicrosoft Equation 3.0 DS Equation Equation.39q@, JL@J 7.05,11.39()ObjInfo,Equation Native -\_10048578447FOle /CompObj0fObjInfo2Equation Native 3_1004940750F FMicrosoft Equation 3.0 DS Equation Equation.39q܈l~II "  2 t n"k"1,)2() s  2  FMicrosoft Equation 3.0 DS EqOle 6CompObj7fObjInfo9Equation Native :uation Equation.39qܘoII ="12.6967t 17,0.025() 3.5354() FMicrosoft Equation 3.0 DS Equation Equation.39q_1004940754FOle =CompObj>fObjInfo@hII ="12.69672.113.5354() FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native A_1004940757FOle DCompObjEfObjInfoGEquation Native Hh_1004940760FOle JLl~II ="12.69677.4597 FMicrosoft Equation 3.0 DS Equation Equation.39qHuI|I "20.16,"5.24()CompObjKfObjInfoMEquation Native Nd_1004940784F0c0c 1F&U0Eǹ2 PVԦޑxtk^A7Y/DVAXgK@+s T[`]ʃٴ2mմoNQ9Ky$;{[HY'᜛I^[yx$9TM[%}P{;S7|WMZ`qENVw3nPVnvf?HwdN>19݋ ԺÇ w po3!"Q^$!f=Jm$$If!vh5n5 #vn#v :V l,5n5 4ytC;0TDd B  S A? 2T]ȑ}_ 0'`!(]ȑ}_ `Pxcdd`` @bD"L1JE `x0&Yjl R A@ V UXRYvo`0L`A)??[1$4! ]6J`b9bi40i`}@2U ?=I9PߊԖ S.ph5S?F&&\SS^rc'8!5C^h\̂t8%!k7?W2J o[1ݤSMQF,8Gst&'tHlҟb +rDd @B  S A? 2!L8Ӣ!`!!L8Ӣ4% ~xcdd`` @b1##X`=F !#T57L ! 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