Math Cracker - Solved Statistics Problems

Problem 1: A researcher is interested in the effects of family size on delinquency for a group of offenders and examines families with one to four children. She obtains a sample of 16 families, four of each size, and identifies the number of arrests per child for delinquency. The data is as follows: Group 1 4 children n=4 Group 2 3 children n=4 Group 3 2 children n=4 Group 4 1 child n=4 Family 1 10 8 5 4 Family 2 8 8 6 5 Family 3 9 6 7 2 Family 4 10 9 9 2 a) Calculate the total sum of squares. b) Calculate the mean square (between groups). c) Calculate the F-ratio d) Use the Turkey HSD (alpha=0.05) to test for significance between groups. Which groups differed? e) Based on your results, write a 1-2 paragraph essay that describes your observations obtained from this sample in regard to the effects of family size on delinquency for a group of offenders. Solution: (a) The following table with descriptive statistics is obtained from the information provided Obs. Group 1 http://www.mathcracker.com Group 2 Group 3 Group 4 10 8 5 4 8 8 6 5 9 6 7 2 10 9 9 2 Mean 9.25 7.75 6.75 3.25 St. Dev. 0.957 1.258 1.708 1.5 We need to test H0 :1  2  3  4 H A : Not all the means are equal With the data found in the table above, we can compute the following values, which are needed to construct the ANOVA table. We have: k SS Between   ni  xi  x  i1 from which we get 2 SSBetween  4 9.25 6.75  4 7.75 6.75  4 6.75 6.75  4 3.25  6.75  78 2 2 2 Now we also see that, k SSWithin    ni 1 si2 i1 which implies SSWithin   4  1  0.9572   4  1 1.2582   4  1 1.7082   4  1 1.52  23 Hence, SSTotal = 78+23 = 101 (b) Therefore MS Between  Also, we obtain that SS Between 78   26 k 1 3 2 MSWithin  23 SSWithin   1.917 N  k 1 12 (c) Therefore, the F-statistics is computed as F MSBetween 26   13.5652 MSWithin 1.917 The critical value for   0.05 , df1  3 and df 2  12 is given by FC  3.4903 and the corresponding p-value is p  Pr  F3,12  13.5652   0.000 Observed that the p-value is less than the significance level  0.05 , then we reject H 0 . (d) The HSD difference is computed as follows: HSD  Q * 1.917 MSE  4.20  2.91 4 n The following table is obtained: Post hoc analysis Tukey simultaneous comparison t-values (d.f. = 12) Group 4 Group 3 Group 2 3.3 6.8 7.8 Group 4 3.3 Group 3 6.8 3.58 Group 2 7.8 4.60 1.02 Group 1 9.3 6.13 2.55 1.53 Group 1 9.3 critical values for experimentwise error rate: 0.05 2.97 0.01 3.89 (e) Based on the above results, we have enough evidence to reject the null hypothesis of equal means, at the 0.05 significance level. Summarizing, we have the following ANOVA table: Source SS df MS F p-value Crit. F Between Groups 78 3 26 13.5652 0.000 3.4903 Within Groups 23 12 1.917 Total 101 15 The pairwise differences that are significant are between Group 1 and Group 4, Group 2 and Group, and Group 3 and Group 4. In fact, the mean for Group 4 is significantly lower when compared to the means for groups 1, 2 and 3, respectively. Problem 2: Movie Success. Using the data in Table 7.2, make a scatter diagram for the relationship between production budget and viewer rating of movies. Estimate the correlation coefficient. Based on these data, do you think a large production budget is likely to result in a movie with a high viewer rating? Explain. Solution: The scatter plot is shown below. Scatterplot of Rating vs Budget 9 Rating 8 7 6 5 4 0 50 100 Budget 150 200 It seems like there's a mild negative linear relationship between Budget and Rating. The actual correlation coefficient is computed as As predicted by the visual trend, the correlation is negative, but since it's very small, the relationship is fairly weak. This means that is not certain that a larger budget will produce a higher rating, as it's not certain that a larger budget will produce a lower rating, but there a inclination to have lower rating with higher budgets. Problem 3: Which of these models is a better representation of the relationship between students’ age and starting salary? Explain your decision. Solution: As mentioned in the previous part, the model obtained once the outlier was eliminated is relatively similar to the model with n=25 cases, as the regression coefficients don't change dramatically. But still this relatively small difference in coefficients makes a relatively large difference in R^2. In fact, for the model with n = 25 we get R2 = 0.334, and for the model with n = 25 we get R2 = 0.447. This makes the second model (with n = 24) the preferred one. The preferred model is Starting Salary^ = -67,941.7485 + 3,635.6857* Age Problem 4: Compute an imprisonment rate per 1000 population for 2000. Introduce this incarceration rate as an independent variable into the model run in Part B.  Test the hypothesis that the R squared =0.  Does this model fit the data better than the model in Part B above? Explain.  Does each of the independent variables have a statistically significant effect on homicide? Explain.  How strong is the effect of each of the independent variables? Explain.  Which of the independent variables has the stronger effect on the homicide rate? Explain. Solution: The new variable is computed as ImprPer1000 = Prison20/pop20 (let us recall that pop20 is already given in 1000’s). The following is obtained with Excel: Regression Analysis R² 0.499 Adjusted R² 0.466 n 49 R 0.707 k 3 Std. Error 1.639 Source SS df MS F p-value Regression 120.4481 3 40.1494 14.95 6.87E-07 Residual 120.8453 45 2.6855 Total 241.2935 48 Dep. Var. homrt20 ANOVA table Regression output variables confidence interval coefficients std. error t (df=45) p-value 95% lower 95% upper std. coeff. Intercept -1.9451 1.1721 -1.660 .1040 -4.3059 0.4156 0.000 ImprPer1000 0.2430 0.1707 1.423 .1616 -0.1009 0.5869 0.175 sglmom80 24.3703 5.6975 4.277 .0001 12.8949 35.8458 0.539 unempl20 36.9631 27.2997 1.354 .1825 -18.0215 91.9476 0.150 The model is Homicide Rate in 2000 = -1.9451 + 0.2430* ImprPer1000 + 24.3703* sglmom80 + 36.9631* unempl20 Notice that the model is significant overall, since F(3, 45) = 14.95, p = 0.000000687 < 0.05, so then R2 is significantly greater than zero. This model fits only slightly better than the previous one, since now Adj. R2 = 0.466, which means that in this case the amount of explained variation in the response variable by this model is 46.6%. Notice that in this model, the variable sglmom80 is individually significant, with t = 4.277 and p = 0.0001 < 0.05, but the variable uempl20 is not individually significant, t = 1.354, p = 0.1825 > 0.05. The variable ImprPer1000 is not significant either, since t = 1.423, p = 0.1616 > 0.05. The effect of ImprPer1000 and uempl20 is quite moderate since the standardized coefficients associated to them are less than 0.2 (this is, an increase in one standard deviation in either of the variables brings a change of less than 0.2 standard deviations in the response variable). The variable with the strongest effect is sglmom80, with a standardized coefficient of 0.539. Problem 5: Using the data below, answer the following questions using a table format. Xi 4 yi 5 4 a. x 4 e. b. 2 i f. x 2 -1 7 4 4 i y i 1 4 2 i c. x y i 1 i d. Show that i  4  g.   xi yi   i 1  i i i 1 4 2 i i i1 4 h. Show that x i 1 0 4  x . y   x y i1 4  (x  x )   ( y  y )  i 1 4 4 y i 1 i 1 i. Show that 3 4 i i 1 6 2 i i1 i i  4     xi   i 1  2