COMM121 UNIVERSITY OF WOLLONGONG (Statistics For Business) FACULTY OF BUSINESS Spring Session 2015 Tutorial 7 Questions (Week 11 Beginning 12 October) Instructions: Please prepare answers to problems with * prior to your tutorial class (questions 1, 3 and 4). Submit the answers to your tutor at the commencement of the class. Chapter 11: ANOVA 1)* In order to examine the differences in ages of teachers among five schools, an educational statistician took random samples of six teachers' ages in each school. Complete the following ANOVA table and conduct hypothesis testing to see if there is a difference in the means of teachers age between the schools (=0.05) . Source SS df F MS Between Groups Within Group 1024.83 Total 2846.96 2) Suppose you are using a completely randomised design to study some phenomenon. There are three treatment levels and a total of 17 people in the study. Complete the following ANOVA table. Use = 0.05 to find out whether the mean of groups are equal. Source SS Between Groups 29.64 Within Groups Total 98.06 df MS F Chapter 12: Simple Linear Regression 3)* A professor of economics wants to study the relationship between income (y in $1000s) and education (x in years). A random sample eight individuals is taken and a simple regression is conducted. The following table shows the regression output: Regression Statistics Multiple R 0.9603 R Square 0.9223 Adjusted R Square 0.9093 Standard Error 2.4356 Observations 8 ANOVA df Regression Residual Total SS MS F 1 6 7 422.2810 35.5940 457.8750 422.2810 5.9323 71.1830 t Stat P-value 2.4384 8.4370 0.0506 0.0002 Coefficients Intercept X Significance F 0.0002 10.6165 2.9098 Standard Error 4.3539 0.3449 Use the following regression output to answer the following questions: a. Write the regression equation b. Determine the coefficient of determination and interpret its meaning. c. Interpret the meaning of the slope b1 in this problem. d. Predict the income if years of eduction is 9 years. e. At the 0.05 level of significance, is there evidence of a linear relationship between income and education? 2 4)* A financier whose speciality is investing in movie production has observed that, in general, movies with \"big-name' stars seem to generate more revenue than those movies whose stars are less well known. To examine his belief he records the gross revenue and the payment (in $ millions) given to highest-paid performers in the movie for ten recently released movies. Using Excel a regression is run for the collected data set. Complete the ANOVA table (find values of A, B, C, D, E, F and G) and then answer the questions based on the table. Regression Statistics Multiple R 0.9954 R Square 0.9908 Adjusted R Square 0.9896 Standard Error 2.0247 Observations 10 ANOVA Regression Residual Total Intercept X df A B C SS D 32.7948 3552.9000 Coefficients Standard Error 4.2254 1.3816 8.2851 0.2827 MS E F F G Significance F 0.0000 t Stat P-value 3.0584 0.0156 29.3036 0.0000 f. Write the regression equation g. Determine the coefficient of determination and interpret its meaning. h. Interpret the meaning of the slope b1 in this problem. i. Predict the the gross revenue of a move whose top two stars earn $5 million. j. At the 0.05 level of significance, is there evidence of a linear relationship between gross revenue and cost of two highest paid performers? 3 Source Between groups Within Group Total SS 2846.961024.83=1822.13 1024.83 2846.96 df 4 5 4+5=9 MS 1822.13/4=455.532 5 1024.83/5=204.48 F 455.5325/204.48=2.227 8 Here the null hypothesis H0: there is no difference in the means of tech age between schools against the alternative hypothesis H1: there is a difference in the means of tech age between schools. From the above anova result we see that F=2.2278, the critical F value at 5% level of significance with 4,5 degree of freedom is F0.05/2, 4, 5=7.38. Since the calculated |F|=2.2278 is less than the critical F=7.38 therefore we accept the null hypothesis that there is no difference in the means of tech age between schools. Source Between Groups Within group Total SS 29.64 98.06-29.64=69.02 98.06 df 3-1=2 17-1=16 2+16=18 MS 29.64/2=14.82 669.02/16=4.314 F 14.82/4.314=3.43 5 Null hypothesis H0: The means of groups are equal against the alternative hypothesis H 1: The means of groups are not equal. From the above anova table we see that calculated F=3.435, the critical F value at 5% level of significance with 2 and 16 degree of freedom is F0.05/2. 2, 16= 4.687. Since the calculated |F|=3.435 is less than critical F=4.687 therefore we accept the null hypothesis that the means of groups are equal. Solution: a) The regression equation is Income (y)=10.6165+2.9098*education. b) The coefficient of determination of the regression equation is 0.9223 which means the regression model explains 92.23% variation of the total variation. c) The slope b1=2.9098, which means if we increase education by one unit then income will increase by 2.9098 units. d) The predicted income when education is 9 years is Income=10.6165+2.9098*9=36.7047. e) Here the null hypothesis H0: there is o evidence about the relationship between education and income against the alternative hypothesis H1: there is a evidence about the relation between education and income. From the above anova table we see that the F=71.1830 and the corresponding P value is 0.0002. since the p value =0.0002 is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that there is a sufficient evidence about the relationship between education and income. Solution: f) The regression equation is Gross Revenue (y)=4.2254+8.2851* Payments. g) The coefficient of determination of the regression is 0.9908 which means the regression model explains 99.08% variation of the total variation. h) The slope b1=8.2851, which means if we increase the payment by one units the gross revenue will be increased by 8.2851 units. i) The predicted Gross revenue when payment is $ 5 million is Gross revenue=4.2254+8.2851* $5=$45.65. h) Here the null hypothesis H0: there is no relation between gross revenue and cost of two highest paid performers against the alternative hypothesis H1: there is a relation between gross revenue and cost of two highest paid performers. From the above anova table we see that the F= 858.699 and the corresponding P value is 0.000. Since the p value =0.000 is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that there is a relation between gross revenue and cost of two highest paid performers. Source Between groups Within Group Total SS 2846.961024.83=1822.13 1024.83 2846.96 df 4 5 4+5=9 MS 1822.13/4=455.532 5 1024.83/5=204.48 F 455.5325/204.48=2.227 8 Here the null hypothesis H0: there is no difference in the means of tech age between schools against the alternative hypothesis H1: there is a difference in the means of tech age between schools. From the above anova result we see that F=2.2278, the critical F value at 5% level of significance with 4,5 degree of freedom is F0.05/2, 4, 5=7.38. Since the calculated |F|=2.2278 is less than the critical F=7.38 therefore we accept the null hypothesis that there is no difference in the means of tech age between schools. Source Between Groups Within group Total SS 29.64 98.06-29.64=69.02 98.06 df 3-1=2 17-1=16 2+16=18 MS 29.64/2=14.82 669.02/16=4.314 F 14.82/4.314=3.43 5 Null hypothesis H0: The means of groups are equal against the alternative hypothesis H 1: The means of groups are not equal. From the above anova table we see that calculated F=3.435, the critical F value at 5% level of significance with 2 and 16 degree of freedom is F0.05/2. 2, 16= 4.687. Since the calculated |F|=3.435 is less than critical F=4.687 therefore we accept the null hypothesis that the means of groups are equal. Solution: a) The regression equation is Income (y)=10.6165+2.9098*education. b) The coefficient of determination of the regression equation is 0.9223 which means the regression model explains 92.23% variation of the total variation. c) The slope b1=2.9098, which means if we increase education by one unit then income will increase by 2.9098 units. d) The predicted income when education is 9 years is Income=10.6165+2.9098*9=36.7047. e) Here the null hypothesis H0: there is o evidence about the relationship between education and income against the alternative hypothesis H1: there is a evidence about the relation between education and income. From the above anova table we see that the F=71.1830 and the corresponding P value is 0.0002. since the p value =0.0002 is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that there is a sufficient evidence about the relationship between education and income. Solution: f) The regression equation is Gross Revenue (y)=4.2254+8.2851* Payments. g) The coefficient of determination of the regression is 0.9908 which means the regression model explains 99.08% variation of the total variation. h) The slope b1=8.2851, which means if we increase the payment by one units the gross revenue will be increased by 8.2851 units. i) The predicted Gross revenue when payment is $ 5 million is Gross revenue=4.2254+8.2851* $5=$45.65. h) Here the null hypothesis H0: there is no relation between gross revenue and cost of two highest paid performers against the alternative hypothesis H1: there is a relation between gross revenue and cost of two highest paid performers. From the above anova table we see that the F= 858.699 and the corresponding P value is 0.000. Since the p value =0.000 is less than 0.05 at 5% level of significance therefore we reject the null hypothesis and accept the alternative hypothesis that there is a relation between gross revenue and cost of two highest paid performers