STAT3220FA23Unit2Classwork

STAT 3220 Unit 2 Classwork 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. tapply(repellent$Hours,repellent$Cost,summary) $‘0.11‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 6 6 6 6 6 6 $‘0.19‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 5.5 5.5 5.5 5.5 5.5 5.5 $‘0.22‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 3 3 3 3 3 3 $‘0.24‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 1

6.5 6.5 6.5 6.5 6.5 6.5 $‘0.27‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 1 1 1 1 1 1 $‘0.36‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 7 7 7 7 7 7 $‘0.46‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 3 3 3 3 3 3 $‘0.67‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 0.500 1.125 1.750 1.750 2.375 3.000 $‘0.75‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 7 7 7 7 7 7 $‘1‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 2 2 2 2 2 2 $‘1.77‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 14 14 14 14 14 14 $‘2.08‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 13.5 13.5 13.5 13.5 13.5 13.5 $‘2.75‘ Min. 1st Qu. Median Mean 3rd Qu. Max. 2

c) Fit a model to the data that includes the repellent type and cost. Include just the regression output repmod1<-lm(Hours~Type+Cost,data=repellent) summary(repmod1) 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. Hypothesis : HO: B1 = 0 (Cost does not contribute for model on maximum number of hours) Ha: B1 does not equal O (Cost does contribute for model on maximum number of hours) Distribution of test statistic : 2 and 11 DF test stat : 5.812 P-value : 0.000117 Decision : 0.000117 < 0.1 -> Reject the null We reject the null hypothesis as we have sufficient evidence to support that cost is a useful predictor in predicting the maximum amount of hours. 4

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 influ- encing 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. streetven1<-lm(Earnings~Age*Hours,data=streentven) summary(streetven1) 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 5

3. During periods of high electricity demand, especially during the hot summer months, the power output from a gas turbine engine can drop dramatically. One way to counter this drop in power is by cooling the inlet air to the gas turbine. An increasingly popular cooling method uses high pressure inlet fogging. The performance of a sample of 67 gas turbines augmented with high pressure inlet fogging was investigated in the Journal of Engineering for Gas Turbines and Power (January 2005). One measure of performance is heat rate (kilojoules per kilowatt per hour). Heat rates for the 67 gas turbines, saved in the gasturbine file. • Engine: traditional, advanced, and aeroderivative • Shafts: 1, 2, 3 • RPM: cycle speed (revolutions per minute) • CPRatio: cycle pressure ratio • Inlet_temp: inlet temperature ( ◦ C) • Exh_temp: exhaust gas temperature ( ◦ C) • Airflow: air mass flow rate (kilograms per second) • Power: horsepower (Hp units) • Heatrate (response): kilojoules per kilowatt per hour A complete second-order model for heat rate (y) as a function of cycle speed, cycle pressure ratio is written as y = β 0 + β 1 RPM + β 2 CPRatio + β 3 RPM ∗ CPRatio + β 4 RPM 2 + β 5 CPRatio 2 + ϵ y = β 0 + β 1 RPM + β 2 CPRatio + β 3 RPM ∗ CPRatio + β 4 RPM 2 + β 5 CPRatio 2 + ϵ a) Now consider the qualitative predictor, engine type, at three levels (traditional, ad- vanced, and aeroderivative). Add the dummy variables such that X1={1 if traditional, 0 otherwise} and X2= {1 if advanced, 0 otherwise} to the model above and write the model. Fit the model. Include your regression output and write your prediction equation. recodeENGINE<-relevel(factor(gasturbine$ENGINE),ref="Aeroderiv") gasturbine1<- lm(HEATRATE~RPM+CPRATIO+RPM*CPRATIO+I(RPMˆ2)+I(CPRATIOˆ2) +recodeENGINE, d summary(gasturbine1) 7

Call: lm(formula = HEATRATE ~ RPM + CPRATIO + RPM * CPRATIO + I(RPM^2) + I(CPRATIO^2) + recodeENGINE, data = gasturbine) 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(gasturbine1) ,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: y = 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+ ϵ Prediction equation: \ heatrate = 14413 . 84+0 . 121 RPM − 421 . 12 CPRatio + 8

Model 2: HEATRATE ~ RPM + CPRATIO + RPM * CPRATIO + I(RPM^2) + I(CPRATIO^2) + recodeENGINE Res.Df RSS Df Sum of Sq F Pr(>F) 1 61 19370350 2 59 18389213 2 981137 1.5739 0.2158 Hypothesis : HO: B6 = B7 = 0 Ha: B6 and/or B7 does not equal 0 test stat : 1.5739 P-value : 0.2158 With a p-value this large, we fail to reject the null hypothesis. There is not enough evidence to conclude the engine type is a useful predictor of heat rate in the model. c) Do you recommend the model original model or the model from part b Why? I would prefer the original model from part 1 as engine type was not significant, hence we do not need it as predictor as the p-value was 0.2158 which causes us to reject the model using engine type as a predictor of heat rate. Although, we may want to continue evaluating the model to make it simpler rather than more complex. 10