A sales manager for an advertising agency believes there is a
relationship between the number of contacts that a salesperson
makes and the amount of sales dollars earned.
A regression analysis shows the following results:
Coefficients | Standard Error | t Stat | p value | |
Intercept | −12.201 | 6.560 | −1.860 | 0.100 |
Number of contacts | 2.195 | 0.176 | 12.505 | 0.000 |
ANOVA | |||||
df | SS | MS | F | Significance F | |
Regression | 1.00 | 13555.42 | 13555.42 | 156.38 | 0.00 |
Residual | 8.00 | 693.48 | 86.68 | ||
Total | 9.00 | 14248.90 |
Additional information needed to perform the calculation x⎯⎯=33.4, ∑(x−x⎯⎯)2=2,814.4.x¯=33.4, ∑(x-x¯)2=2,814.4. Calculate a 95% confidence interval for the situation where x = 30 calls. (Round your final answers to one decimal).
Multiple Choice
46.7 and 60.6
31.1 and 76.2
51.4 and 55.9
55.8 and 51.5
n= | 10 |
bo= | -12.201 |
b1= | 2.1950 |
sxy =√MSE= | 9.3102 |
Sxx=(n-1)sx^2= | 2814.4 |
x̅ = | 33.4 |
predcited value at X=30: | 53.6490 | ||||
std error confidence interval= | s*√(1/n+(x0-x̅)2/Sxx) | = | 3.0040 | ||
for 95 % CI value of t= | 2.3060 | ||||
margin of error E=t*std error = | 6.9272 | ||||
lower confidence bound=sample mean-margin of error = | 46.7 | ||||
Upper confidence bound=sample mean+margin of error= | 60.6 |
correct option is : 46.7 and 60.6
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