(Questions In A manager at o loent bank analyzed the relationship between monthl
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(Questions In A manager at o loent bank analyzed the relationship between monthly salary and three independe variablen length of Service (monorel in montis), Gender (0 female, I - mole) and Job type (0clericut, Teethnical). The following ANOVA sumiturizes the regression results ANOVA | source of Variation Siin of Squarcs Mean Square F Represion TIM 316.771 334782 257 596 Residian! 161134.596 | 56197 4844S | Total 29 3465 481.367 Steixlard Emor 32225 244 S.D | Intercept | Service | Gender Jols Coefficients 78492 9.19 223,78| 28.21|| I Stat p-value 0.02 2.87 D.01 2.50 O 02 40.31 0.76 89.00 89,61| 1. The results for the variable gender' show that A) Males avenge $222.78 more than females in monthly salary. B) Females average $222.78 more than males in monthly salary. C) Gender is not significantly related to Imonthly salary. D) Gender and months (length of service) are correlated. E) Gender and job type are correlated. 2. In the regression model, which of the following are dumny variables? A) Intercept B) Length of service C) Length of service and gender D) Gender and job type E) Length of service, gender, and job type 3. Based on the hypothesis tests for the individual regression coefficients, A) All the regression coefficients are equal to zero. B) "Job" is the only non-significant variable in the model. C) The intercept is the only significant variable in the model. D) "Service" is the only significant variable in the model. E) "Service", "Gender", and "Job" are all significant variables in the model. Show your solution 4. Based on the ANOVA, the multiple coefficient of determination is A) 5957% B) 59.3% C) 40.7% D) 85.6% E) Cannot be computedExplanation / Answer
1. (A) Males average $222.78 more than females in monthly salary, since the coefficient of gender is positive and gender takes a value of 1 for males.
2. (D) Gender and Job-type , self-explanatory
3. (B) Job is the only non significant variable in the model because it has a p-value greater than 0.05
4. (C) R2 = 1 - (RSS/TSS) = 1 - (1462134.596/2465483.367) = 0.407 (40.7%)
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