STAT E-50 - Introduction to Statistics SPSS4 Solutions p. 186: 35, 36 p. 216: 36abcd, 57*, 58 (*also create a scatter diagram and find r to analyze the data) p. 278: 12abce p. 186: 35 a) b) The relationship is linear, negative, and strong, and there don't seem to be any outliers. c) r = -.869 Correlations Highway Gas Horsepower Mileage (mpg) Horsepower Pearson Correlation 1 -.869 ** Sig. (2-tailed).000 N 15 15 Highway Gas Mileage (mpg) Pearson Correlation -.869 ** 1 Sig. (2-tailed).000 N 15 15 **. Correlation is significant at the 0.01 level (2-tailed). d) Cars with higher horsepower tend have lower fuel economy (i.e. fewer mpg).
p. 186: 36 a) b) r =.934 Correlations Marijuana (%) Other Drugs (%) Marijuana (%) Pearson Correlation 1.934 ** Sig. (2-tailed).000 N 11 11 Other Drugs (%) Pearson Correlation.934 ** 1 Sig. (2-tailed).000 N 11 11 **. Correlation is significant at the 0.01 level (2-tailed). c) The relationship is linear, positive, and very strong d) No - there is an association between the percent of teens who have used marijuana and the percent of teen who have used other drugs, but correlation doesn t imply a causal relationship. Page 2
p. 216: 36 a) A regression model is appropriate: both of the variables are quantitative, and the scatterplot shows a fairly constant linear relationship with no outliers. b) The equation of the regression line is Mortgages = -7.775 Interest Rate + 220.893 Coefficients a Unstandardized Coefficients Standardized Coefficients Model B Std. Error Beta t Sig. 1 (Constant) 220.893 9.463 23.344.000 Interest.Rate -7.775 1.025 -.840-7.584.000 a. Dependent Variable: Mortgages (in millions dollars) c) The predicted mortgage amount for an interest rate of 20% is $65.39 million d) This is not an appropriate prediction because the value of 20% is well outside of the range of the given data. Page 3
p. 216: 57 a) %Body Fat =.250 Weight - 27.376 Coefficients a Unstandardized Coefficients Standardized Coefficients Model B Std. Error Beta t Sig. 1 (Constant) -27.376 11.547-2.371.029 Weight.250.061.697 4.120.001 a. Dependent Variable: %Body Fat b) A linear model appears to be appropriate. The scatterplot shows a fairly strong positive linear relationship between the variables. c) The slope of the regression line is.250. This tells us that the body fat will increase by about.25% for every additional pound of weight. Page 4
d) The model may not be likely to make reliable estimates. Since R 2 =.485, only 48.5% of the variability in % body fat can be accounted for by the model. Model Summary Model R R Square Adjusted R Square Std. Error of the Estimate 1.697 a.485.457 7.049 a. Predictors: (Constant), Weight e) The residual for a person who weighs 190 and has 21% body fat is.86% Additional questions: The Model Summary shows that r =.697; the scatter diagram is shown in part b. Page 5
p. 216: 58 The scatterplot with Waist as the predictor variable shows a strong positive linear relationship. The linear model for this relationship is %Body Fat = 2.22 Waist - 62.56 For each additional inch in waist size, the model predicts an increase of 2.22% body fat. The value of R 2 for this model is.787 indicating that 78.7% of the variability in % body fat can be accounted for by the model. This is higher than the value when Weight is the predictor (R 2 =.485), indicating that this model is a more powerful predictor than the model based on weight. Coefficients a Unstandardized Coefficients Standardized Coefficients Model B Std. Error Beta t Sig. 1 (Constant) -62.557 10.158-6.159.000 Waist 2.222.273.887 8.144.000 a. Dependent Variable: %Body Fat Model Summary b Model R R Square Adjusted R Square Std. Error of the Estimate 1.887 a.787.775 4.540 a. Predictors: (Constant), Waist b. Dependent Variable: %Body Fat Page 6
p. 278: 12 a) The correlation between the Friday scores and the Monday scores is r =.473 Correlations Fri Mon Fri Pearson Correlation 1.473 * Sig. (2-tailed).017 N 25 25 Mon Pearson Correlation.473 * 1 Sig. (2-tailed).017 N 25 25 *. Correlation is significant at the 0.05 level (2-tailed). b) The scatter diagram shows a weak positive linear association between the Friday score and the Monday score. Generally, students who scored high on Friday also tended to score high on Monday. c) A student with a positive residual scored higher on Monday's test than the model predicted. Page 7
e) The regression equation is Monday = 14.592 +.536Friday Coefficients a Unstandardized Coefficients Standardized Coefficients Model B Std. Error Beta t Sig. 1 (Constant) 14.592 8.847 1.649.113 Fri.536.208.473 2.573.017 a. Dependent Variable: Mon f) According to the model a student who scored 40 on Friday is expected to have a Monday score of about 36. Page 8