STAT3220FA23Unit2ClassworkSolutions

STAT 3220 Unit 2 Classwork Solutions Your name 1. Which insect repellents protect best against mosquitoes? Consumer Reports (June 2000) tested 14 products that all claim to be an effective mosquito repellent. Each product was classified as either lotion/cream or aerosol/spray. The cost of the product (in dollars) was divided by the amount of the repellent needed to cover exposed areas of the skin (about 1/3 ounce) to obtain a cost-per-use value. Effectiveness was measured as the maximum number of hours of protection (in half-hour increments) provided when human testers exposed their arms to 200 mosquitoes. The data from the report are listed in the object ‘repellent“. a) Consider the relationship between the * type of repellent and the maximum hours of protection . Create a visualization of this relationship and comment on it. #Write code for appropriate visualization boxplot(Hours~Type,data=repellent) Aerosol/Spray Lotion/Cream 0 5 10 15 20 Type Hours We 1

do not see a strong relationship here because the means are very close b) Suppose you want to use repellent type to model the maximum number of hours of protection (y). Create the appropriate number of dummy variables for repellent type and write the model. Use “Aerosol/Spray” as the base level. E ( Hours ) = β 0 + β 1 TypeLotion where TypeLotion= 1 if Type= Lotion/Cream c) Fit a model to the data that includes the repellent type and cost. Include just the regression output #Fit the model with Type and Cost as predictors repc<-lm(Hours~Type+Cost,data=repellent) summary(repc) Call: lm(formula = Hours ~ Type + Cost, data = repellent) Residuals: Min 1Q Median 3Q Max -6.3283 -1.2664 -0.1718 2.4606 4.6804 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.252 1.517 1.485 0.165700 TypeLotion/Cream -2.391 1.854 -1.290 0.223635 Cost 6.830 1.175 5.812 0.000117 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 3.425 on 11 degrees of freedom Multiple R-squared: 0.7586, Adjusted R-squared: 0.7147 F-statistic: 17.29 on 2 and 11 DF, p-value: 0.0004025 d) Test whether cost is a useful predictor of maximum number of hours (y). Use α = . 10 . Include the hypotheses, test statistic, p-value, and conclusion in context. • Hypotheses: H 0 : β 2 = 0 vs H a : β 2 = 0 2

2. Earnings of Mexican Street Vendors. Detailed interviews were conducted with over 1,000 street vendors in the city of Puebla, Mexico, in order to study the factors influencing vendors’ incomes. Vendors were defined as individuals working in the street and included vendors with carts and stands on wheels and excluded beggars, drug dealers, and prostitutes. The researchers collected data on gender, age, hours worked per day, annual earnings, and education level. We have a sample of 15 vendors from this list and we will examine the following: • Annual Earnings (response, y) • Age (explanatory, x1) • Hours Worked Per Day (Explanatory, x2) a) Consider the interaction model E ( y ) = β 0 + β 1 X 1 + β 2 X 2 + β 3 X 1 X 2 . Use the street vendor data to fit the model. #Fit the model vena<-lm(Earnings~Age*Hours,data=streentven) summary(vena) Call: lm(formula = Earnings ~ Age * Hours, data = streentven) Residuals: Min 1Q Median 3Q Max -936.5 -281.3 -117.6 255.6 787.4 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1041.894 1303.593 0.799 0.441 Age -13.238 29.234 -0.453 0.659 Hours 103.306 162.014 0.638 0.537 Age:Hours 3.621 3.840 0.943 0.366 Residual standard error: 550.3 on 11 degrees of freedom Multiple R-squared: 0.6135, Adjusted R-squared: 0.5081 F-statistic: 5.82 on 3 and 11 DF, p-value: 0.0124 4

b) What is the estimated slope relating annual earnings (y) to age (x1) when number of hours worked (x2) is 10? Interpret the result. The estimated slope for age is (-13.238+3.621xHours). Therefore, for a one year increase in age the expected earnings will increase by -13.238+3.621x10 =$22.972 when hours is held constant at 10. c) What is the estimated slope relating annual earnings (y ) to hours worked (x2 ) when age (x1 ) is 40? Interpret the result. The estimated slope for age is (103.306+3.621xAge). Therefore, for a one hour increase in hours worked the expected earnings will increase by 103.306+3.621x40 =$248.15 when age is held constant at 40. d) Test whether the interaction between age and hours is a useful predictor of earnings (y). Use α = . 10 . Include the hypotheses, test statistic, p-value, and conclusion in context. • Hypotheses: H 0 : β 3 = 0 vs H a : β 3 = 0 • Test Statistic: 0.943 • Pvalue: 0.366 • With a pvalue this large, we fail to reject the null hypothesis. There is not enough evidence to conclude the interaction between age and hours is a useful predictor of earnings. 5

Residuals: Min 1Q Median 3Q Max -1242.2 -313.8 -97.7 292.2 1863.4 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.441e+04 1.313e+03 10.977 6.97e-16 *** RPM 1.212e-01 1.145e-01 1.059 0.29387 CPRATIO -4.211e+02 1.240e+02 -3.395 0.00123 ** I(RPM^2) -2.277e-07 2.016e-06 -0.113 0.91042 I(CPRATIO^2) 7.111e+00 2.547e+00 2.791 0.00706 ** recodeENGINEAdvanced -1.088e+02 3.284e+02 -0.331 0.74161 recodeENGINETraditional 2.358e+02 2.899e+02 0.813 0.41930 RPM:CPRATIO 9.100e-04 6.045e-03 0.151 0.88086 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 558.3 on 59 degrees of freedom Multiple R-squared: 0.8905, Adjusted R-squared: 0.8775 F-statistic: 68.53 on 7 and 59 DF, p-value: < 2.2e-16 round(coef(gasa),5) (Intercept) RPM CPRATIO 14413.84263 0.12122 -421.12001 I(RPM^2) I(CPRATIO^2) recodeENGINEAdvanced 0.00000 7.11101 -108.80654 recodeENGINETraditional RPM:CPRATIO 235.77999 0.00091 • New Model: heatrate = β 0 + β 1 RPM + β 2 CPRatio + β 3 RPM * CPRatio + β 4 RPM 2 + β 5 CPRatio 2 + β 6 X 1 + β 7 X 2 + • Prediction Equation: \ heatrate = 14413 . 84 + 0 . 121 RPM - 421 . 12 CPRatio + 0 . 0009 RPM * CPRatio + 0 . 000 RPM 2 + 7 . 11 CPRatio 2 + 235 . 77 X 1 - 109 . 80 X 2 • where X1={1 if traditional, 0 otherwise} and X2= {1 if advanced, 0 otherwise} 7

b) Conduct the test to determine if engine type is important for predicting heat rate. #Fit the model redgas<-lm(HEATRATE~RPM*CPRATIO+I(RPMˆ2)+I(CPRATIOˆ2),data=gasturbine) summary(redgas) Call: lm(formula = HEATRATE ~ RPM * CPRATIO + I(RPM^2) + I(CPRATIO^2), data = gasturbine) Residuals: Min 1Q Median 3Q Max -1196.10 -281.46 -34.99 302.94 1896.08 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.558e+04 1.143e+03 13.635 < 2e-16 *** RPM 7.823e-02 1.104e-01 0.708 0.48144 CPRATIO -5.231e+02 1.034e+02 -5.061 4.11e-06 *** I(RPM^2) -1.806e-07 1.969e-06 -0.092 0.92724 I(CPRATIO^2) 8.840e+00 2.163e+00 4.087 0.00013 *** RPM:CPRATIO 4.452e-03 5.582e-03 0.798 0.42821 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 563.5 on 61 degrees of freedom Multiple R-squared: 0.8846, Adjusted R-squared: 0.8752 F-statistic: 93.55 on 5 and 61 DF, p-value: < 2.2e-16 anova(redgas,gasa) Analysis of Variance Table Model 1: HEATRATE ~ RPM * CPRATIO + I(RPM^2) + I(CPRATIO^2) Model 2: HEATRATE ~ RPM * CPRATIO + I(RPM^2) + I(CPRATIO^2) + recodeENGINE Res.Df RSS Df Sum of Sq F Pr(>F) 1 61 19370350 8