Sarah is the office manager for a group of financial advisors who provide financial services for individual clients. She would like to investigate whether a relationship exists between the number of presentations made to prospective clients in a month and the number of new clients per month. The accompanying table shows the number of presentations and corresponding new clients for a random sample of six employees.Sarah would like to use simple regression analysis to estimate the number of new clients per month based on the number of presentations made by the employee per month. What is the standard error of the slope?
Employee |
Presentations |
New Clients |
---|---|---|
1 |
7 |
2 |
2 |
9 |
3 |
3 |
9 |
4 |
4 |
10 |
3 |
5 |
11 |
5 |
6 |
12 |
3 |
A.
0.4622
B.
0.2526
C.
0.6155
D.
0.5720
Ans:
Employee | Presentations | New Clients | y'=0.3043x+0.3913 | (y-y')^2 | (x-9.667)^2 |
1 | 7 | 2 | 2.5214 | 0.271858 | 7.11128889 |
2 | 9 | 3 | 3.13 | 0.0169 | 0.44448889 |
3 | 9 | 4 | 3.13 | 0.7569 | 0.44448889 |
4 | 10 | 3 | 3.4343 | 0.188616 | 0.11108889 |
5 | 11 | 5 | 3.7386 | 1.59113 | 1.77768889 |
6 | 12 | 3 | 4.0429 | 1.08764 | 5.44428889 |
Total | 46 | 17 | 3.913045 | 15.33333334 | |
Average | 9.67 | 3.33 |
SE of regression slope = sb1 = sqrt [ Σ(yi – ŷi)2 / (n – 2) ] / sqrt [ Σ(xi – x-bar)2 ].
standard error of the slope=SQRT(3.913/(6-2))/SQRT(15.333)=0.2526
Option B is correct.
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