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Case 1
The following is part of an ANOVA table that was obtained from data regarding three treatments and a total of 15 observations.
Source of
Variation
Sum of
Squares
Degrees of
Freedom
Between Treatments
64
Error (Within Treatments)
96
Refer to case 1 . The number of degrees of freedom corresponding to between treatments is
Question 1 options:
12
2
3
4
Refer to case 1 . The mean square between treatments (MSTR) is
Question 2 options:
36
16
8
32
Refer to case 1 . The number of degrees of freedom corresponding to within treatments is
Question 3 options:
12
2
3
15
Refer to case 1. The computed F test statistics is
Question 4 options:
32
8
0.667
4
Refer to Case 1. If at 95% confidence, we want to determine whether or not the means of the populations are equal, using Excel the p-value is
Question 5 options:
TDIST (F, df1,df2)
FDIST(t,df1,df2)
FDIST (F,df1,df2)
None of these alternatives is correct
Refer to Case 1 . If p-value between 0.05 to 0.1
the conclusion of the test is that the means
Question 6 options:
are equal
may be equal
are not equal
None of these alternatives is correct.
Case 2
The following information regarding a dependent variable Y and an independent variable X is provided
Refer to case 2. The total sum of squares (SST) is
Question 7 options:
-156
234
1870
1974
Refer to case 2. The sum of squares due to error (SSE) is
Question 8 options:
-156
234
1870
1974
Refer to case 2. The mean square error (MSE) is
Question 9 options:
1870
13
1974
935
Refer to case 2. The slope of the regression equation is
Question 10 options:
-0.667
0.667
100
-10
Refer to case 2. The Y intercept is
Question 11 options:
-0.667
0.667
100
-100
Refer to case 2. The coefficient of correlation is
Question 12 options:
-0.2295
0.2295
0.0527
-0.0572
Case 3
In order to determine whether or not the number of automobiles sold per day (Y) is related to price (X1 in $1,000), and the number of advertising spots (X2), data were gathered for 7 days. Part of the regression results is shown below.
Coefficient
Standard Error
Intercept
0.8051
X1
0.4977
0.4617
X2
0.4733
0.0387
Analysis of Variance
Source of
Degrees
Sum of
Mean
Variation
of Freedom
Squares
Square
F
Regression
40.700
Error
1.016
Determine the least squares regression function relating Y to X1 and X2.
Question 13 options:
y-estimated=7.017-8.6233X1+0.0858X2
y-estimated=-7.0174+8.6233X1+0.0858X2
y-estimated=7.0174+8.6233X1+0.0858X2
non of these alternatives is correct
Refer to case 3
If the company charges $20,000 for each car and uses 10 advertising spots, how many cars would you expect them to sell in a day?
Question 14 options:
15 (rounded from 15.49)
16
17
14
Refer to case 3
At a = 0.05, test to determine if the equation developed represents a significant relationship between the independent variables and the dependent variable.
The test statistics is
Question 15 options:
F
t
We can use F or t
None of these alternatives is correct.
Refer to test 3
At a = 0.05, test to determine if the fitted equation developed in Part a represents a significant relationship between the independent variables and the dependent variable.
to find p-value we should use excel function
Question 16 options:
p-value is FDIST (F,df1,df2)
p-value is FDIST (t,df1,2)
p-value is FDIST (t,df2,2)
None of these alternatives is correct
Refer to case 3. At a = 0.05, test to determine if the fitted equation developed in Part a represents a significant relationship between the independent variables and the dependent variable.
Conclusion is
Question 17 options:
F = 80.12; p-value > .01; Don’t reject H0
The model is not significant at alpha 0.05
F = 8.12
F = 80.12; p-value < .01 (almost zero); reject H0 and conclude that te model is significant at alpha 0.05 None of these alternatives is correct. Refer to case 3 At 95% confidence, test to see if price is a significant variable. Use test statistics Question 18 options: F t both t and F None of these alternatives is correct. Refer to case 3. At 95% confidence, test to see if price is a significant variable. Conclusion is Question 19 options: t = 0.008; p-value less than alpha . Reject H0; price is a significant variable at alpha 0.05 t = 1.078; p-value is between 0.1 and 0.2; do not reject H0; price is not a significant variable at 95% confidence. t = 1.078; p-value is less than 0.1 ; reject H0 price is a significant variable at 95% confidance None of these alternatives is correct. Refer to case 3 Determine the multiple coefficient of determination and interpret this number Question 20 options: 0.9756. 97.56% of the variation of the number of automobiles sold per day is explained by the variation in price and the number of advertising spots. 0.9756 97.56% of the variation of the number of automobiles sold per day is explained by the variation in price. 0.3467 34.67% of the variation of the number of automobiles sold per day is explained by the variation in price. 0.7654 76.54% of the variation of the number of automobiles sold per day is explained by the variation in the number of advertising spots.

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