Term Test #2 (March 19, 2009) (Answer Key)
1. Analysis of Variance for Age
Sum of Squares
(a) Factorial design. Type of loan and marital status. 9 treatment combinations.
(b) The three requirements are:
1. The applicant’s age in all 9 combinations of type of loan and marital status are independent.
2. The distribution of applicant’s age in each of the 9 combinations is normally distributed.
3. The variances of applicant’s age in all 9 combinations are equal.
(c) Answer: As in the ANOVA table above.
(d) From the p-value associated with the type, which is 0.0127, which are less than 0.05; therefore, we have enough evidence to conclude that different type of loan has different applicant’s age. From the p-value associated with the marital status, which is 0.4255, which are a lot more than 0.05; therefore, we don’t have enough evidence to conclude that different marital status has different applicant’s age.
(e) We have three pairs for comparison, as: with each is the actual average age and 1 represents installment loan, 2 represents real estate loan, and 3 represents commercial loan.
We need to find two Tukey Kramer’s CR because we have different sample sizes, 19 and 25. From the above ANOVA, we have MSE=117.8150, and the Q3,60 is 3.40; hence
Compare 1 & 2: ; 1 & 3: ; 3 & 2:
Therefore the average applicant’s age of installment loan is different than that of real estate loan and of the commercial loan. (Actually you don’t need to compare three pairs, just the 1&3 and 3&2 are good enough, remember why?)
(f) From the p-value associated with the interaction, it is 0.0749, which are just a little more than 0.05; therefore, we don’t have enough evidence to conclude than there is interaction effect of the type and marital status to applicant’s age. That means, we don’t have enough evidence to conclude the applicant’ age with a certain marital status with a certain type of loan is different than that of the applicants with a different martial status with another type of loan.
(a) 1. One-way ANOVA with 3 sections, i.e. 3 treatments; 2. F-statistic; and 3. F distribution with 2 and 29 degress of freedom. Because there are 3 sections, therefore 2 df for the numerator and (10+12+10)-1-2=29 df for the denominator. From the table in the textbook, the decision point is 3.33.…