Unformatted text preview: s a significant linear relationship between x and y. AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 4
Topic: Significance of the slope 11289 Chapter 01  An Introduction to Business Statistics 139. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance Calculate the t statistic used to test H0:
A. 5.908
B. 4.221
C. 5.908
D. 4.221 1 = 0 versus Ha: 1 ≠ 0 at α = .001. 0.34655/0.05866 = 5.908 compared to t statistic of 4.221 (for = .001, df = 13) so reject
the null hypothesis at .001. There is a significant linear relationship between x and y. AACSB: Analytic
Bloom's: Application
Difficulty: Medium
Learning Objective: 4
Topic: Significance of the slope 11290 Chapter 01  An Introduction to Business Statistics 140. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance What is the predicted value of y when x = 9.00?
A. 7.980
B. 1.743
C. 4.1385
D. 1.743
y = 4.8615  0.34655 (9.00) = 1.743 AACSB: Analytic
Bloom's: Application
Difficulty: Easy
Learning Objective: 1
Topic: Simple Linear Regression 11291 Chapter 01  An Introduction to Business Statistics 141. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance Calculate the MSE.
A. .4862
B. .6973
C. .2364
D. .5201
MSE = s2 = (.4862)2 = 0.2364 AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Learning Objective: 3
Topic: Simple Linear Regression 11292 Chapter 01  An Introduction to Business Statistics 142. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance Calculate the SSE
A. 3.073
B. 6.321
C. 3.310
D. 11.324
SSE = MSE * (n2) = .2364 * (13) = 3.073 AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Learning Objective: 3
Topic: Simple Linear Regression 11293 Chapter 01  An Introduction to Business Statistics 143. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance What is the unexplained variance?
A. 11.324
B. 3.073
C. 6.321
D. 3.310
Unexplained variance = SSE = 3.073 AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Learning Objective: 6
Topic: Simple Linear Regression 11294 Chapter 01  An Introduction to Business Statistics 144. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance What is the explained variance?
A. 11.324
B. 3.073
C. 8.251
D. 6.321
Explained variance = Total variance  Unexplained variance = 11.324  3.073 = 8.251 AACSB: Analytic
Bloom's: Application
Difficulty: Easy
Learning Objective: 6
Topic: Simple Linear Regression 11295 Chapter 01  An Introduction to Business Statistics 145. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance What is the coefficient of determination?
A. .8536
B. .7286
C. .2714
D. .3724
r2 = Explained variance/Total variance = 8.251/11.324 = .7286
Explained variance = Total variance  Unexplained variance = 11.324  3.073 = 8.251 AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Learning Objective: 6
Topic: Coefficient of determination 11296 Chapter 01  An Introduction to Business Statistics 146. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = 0.7286
Analysis of Variance What is the correlation coefficient?
A. .8536
B. .7286
C. .4862
D. .8536
r = √.7286 = 0.8536 (use the sign of the slope) AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Learning Objective: 6
Topic: Correlation 11297 Chapter 01  An Introduction to Business Statistics 147. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance Determine the 95% confidence interval for the mean value of y when x = 9.00. Givens: ∑ x =
129.03 and ∑ x2 = 1178.547
A. (.6572 2.829)
B. (1.467 2.019)
C. (.6718 .2814)
D. (1.471 2.015)
Y = 1.743
Distance value = 1/15 + [(9(129.03/15))2/(1178.547  (129.03)2/15))] = .068975
t statistic = 2.160
y ± (t) (s) (√dv)
1.743 ± (2.160) (.4862) (√.068975) → 1.467 to 2.019 AACSB: Analytic
Bloom's: Application
Difficulty: Hard
Learning Objective: 5
Topic: Confidence & prediction intervals 11298 Chapter 01  An Introduction to Business Statistics 148. Consider the following partial computer output from a simple linear regression analysis. S = 0.4862 RSq = ______
Analysis of Variance Determine the 95% prediction interval for the mean value of y when x = 9.00 Givens: ∑ x =
129.03 and ∑ x2 = 1178.547
A. (.6572 2.829)
B. (1.467 2.019)
C. (.6718 .2814)
D. (1.471 2.015)
Y = 1.743
Distance value = 1/15 + [(9  (129.03/15))2/(1178.547  (129.03)2/15))] = .068975
t statistic = 2....
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 Winter '14
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