Answered>Order 19170
a.
Y’ = 0.476X + 33.272
b.
Y’ = -1.931X + 66.363
c.
Y’ = -0.476X + 33.272
d.
Y’ = -0.432X + 32.856
___ 10. Refer to Case Study 7-1. If an individual exercises 20 minutes daily, his predicted % body fat would be _________.
a.
21.63
b.
27.74
c.
27.88
d.
23.75
____ 11. Refer to Case Study 7-1. The value for the standard error of estimate in predicting % fat from daily exercise is _________.
a.
3.35
b.
4.32
c.
2.14
d.
1.66
e.
none of above
____ 12. The assumption of homoscedasticity is that _________.
a.
the range of the Y scores is the same as the X scores
b.
the X and Y distributions have the same mean values
c.
the variability of Y doesn’t change over the X scores
d.
the variability of the X and Y distributions is the same
____ 13. The higher the standard error of estimate is,
a.
the more accurate the prediction is likely to be
b.
the less accurate is the prediction is likely to be
c.
the less confidence we have in the accuracy of the prediction
d.
the more confidence we have in the accuracy of the prediction
e.
b and c
14. (7 Points) A statistics professor conducts a study to investigate the relationship between the performance of students on his exams and their anxiety. Ten students from his class are selected for the experiment. Just prior to taking the final exam, the 10 students are given an anxiety questionnaire. Here are final exam and anxiety scores for the students.
Anxiety Final Exam
28 82
41 58
35 63
39 89
31 92
42 64
50 55
46 70
45 51
37 72
Put in the missing values below
? X = ? Y = ? XY =
? X2 = ? Y2 =
Pearson r for these two variables is
Write the least squares regression equation using anxiety predicting the final exam.
If anxiety level was 49, what is the predicted final exam score.
d) What percent of the variance in the final exam is accounted for by anxiety?
NOTE: Feel free to use SPSS to answer these questions
"Not answered?"
