In: Statistics and Probability
The following results are from data concerning the amount
withdrawn from an ATM machine based on the amount of time spent at
the ATM machine (SECONDS) and the gender, FEMALE (dummy variable =
1 for females and = 0 for males) and an interaction term,
SECONDS*FEMALE
SUMMARY OUTPUT |
||||||
Regression Statistics |
||||||
Multiple R |
0.503 |
|||||
R Square |
||||||
Adjusted R Square |
||||||
Standard Error |
||||||
Observations |
50 |
|||||
ANOVA |
||||||
df |
SS |
MS |
F |
Significance F |
||
Regression |
2728.7 |
5.19 |
0.004 |
|||
Residual |
24161.8 |
525.3 |
||||
Total |
32348 |
|||||
Coefficients |
Standard Error |
t Stat |
P-value |
Lower 95% |
Upper 95% |
|
Intercept |
130.9 |
23.51 |
5.89 |
0 |
91.03 |
185.67 |
Seconds |
-2.07 |
0.636 |
0.002 |
-3.401 |
-0.839 |
|
Female |
-75.7 |
30.8 |
-2.26 |
-137.33 |
-8.07 |
|
Seconds*Female |
2.4 |
0.88 |
2.8 |
0.007 |
0.67 |
Based on the regression results, if a male and female each spend
the same amount of time at the ATM matchine (say 38 seconds), how
much more (or less) will a male withdraw? (if a male withdraws more
then your answer should be a positive number and if a male
withdraws less then your answer should be a negative number?
(please express your answer using 1 decimal places)
Answer:
giventhat
Multiple R=R=0.503
R-Square=R2=R*R=0.503*0.503=0.253
Adjusted R square=1-(1-R2)*(N-1)/(N-p-1)=1-(1-0.253)*49/(50-3-1)=0.2043
Residual df=SS(residual)/MS(residual)=24161.8/525.3=45.99 ( next integer is 46)
Total df=N-1=50-1=49
Regression df=Total df - Residual df=49-46=3
confidence interval is equidistant from respective coefficient in this case due to symmetric nature of t-distribution
so Upper 95%=2.4+(2.4-0.76)=4.04
please check your question it is not matching t-stat =coefficient/SE(coefficient)
t-stat=coefficint/SE(coefficient)
P-value is calcuated using ms-excel command=tdist(t-stat, df,two tailed)
for intercept P-value=tdist(5.89, 46,2)=0.000