Question

Practice Exam 2:

A) Psychologists have found that people are generally reluctant to transmit bad news to their peers. This phenomenon has been termed the "MUM effect." To investigate the cause of the MUM effect, 40 undergraduates at Duke University participated in an experiment. Each subject was asked to administer an IQ test to another student and then provide the test taker with his or her percentile score. Unknown to the issue, the test taker was a bogus student who was working with the researchers. The experimenters manipulated two factors: subject visibility and success of test taker, each at two levels. Subject visibility was either visible or not visible to the test taker. The success of the test taker was either top 20% or bottom 20%. Ten subjects were randomly assigned to each of the 2 x 2 = 4 experimental conditions, then the time (in seconds) between the end of the test and the delivery of the percentile score from the subject to the test taker was measured. (This variable is called the latency to feedback.) The data were subjected to appropriate analyses with the following results. The following is a partially completed 2-way analysis table.

                                                                                                                          

Source                       df                  SS                  MS              F                   

Subject visibility        1                                   1380.24         4.26                 

Test taker success      1             1325.16                                                       

Interaction                  1             3385.80                                                       

Error                                       11,664.00                     

Total                                       17,755.20                                                           

1) What is the sum of squares due to factor subject visibility?

1380.24

2) How many degrees of freedom does 'Error' have?

39-(1+1+1) = 36

3) How many treatment combinations are there in the experiment?

4

4) Perform the F- test on 'Test taker success'. Is the factor 'Test taker success' significant( )? F0.05,1,36= 4.12 ( Even though F critical value is provided here, you must make sure that you can comfortably read F tables for your exam) F= MS('Test taker success')/MSE = 1325.16/324 = 4.09; F=4.09< F0.05,1,36= 4.12; Factor is insignificant.

5) Perform the F- test on 'Interaction'. Is the factor 'Interaction' significant( )?F= MS(Interaction)/MSE = 3385.8/324 = 10.45; F=10.45 > F0.05,1,36= 4.12; Interaction is significant.

B) Consider the following Minitab partial output for the independent variable Promotional Exp(in hundreds) and the demand for a product( in thousands)

Regression Analysis: Demand(in Thousands) vs. Promotional exp(in Hundreds)

The regression equation is
Demand(in thousands) = 1.70 + 1.15 Promotional exp(in hundreds)

Predictor Coef SE Coef   
Constant 1.6986 0.1711
Promotional exp(in hundreds) 1.14640 0.05690  

Analysis of Variance

Source DF SS MS
Regression 1 5.1833 5.1833  
Residual Error 8 0.1022 0.0128
Total 9 5.2855

Predicted Values for New Observations

New
Obs Fit SE Fit 95% CI 95% PI
1 4.3353 0.0510 (4.2176, 4.4530) (4.0494, 4.6212)

Values of Predictors for New Observations

Promotional
New exp(in
Obs hundreds)
1 2.30

Please answer the following questions based on the above Minitab output

1) What is the slope(b1) and what is its interpretation in layman's words? For every one unit increase($100 increase) in promotional expenditure, there will be an estimated average increase in demand in the amount of 1.1464*1,000=1,146.4 units.

2) Perform the test of hypothesis ( t-test) on the following, and state your statistical decision. Use
tSTAT = 1.1464/0.0569 = 20.15 > t 0.025, 8 = 2.3060; hence reject null hypothesis

3) Perform the test of hypothesis (F-test) on the following, and state your statistical decision. Use
FSTAT = 5.1833/0.0128= 404.94 > F 0.05,1, 8 = 5.32; hence reject the null hypothesis

4) What is the coefficient of determination (r2), i.e., the percentage of variability in demand explained by the independent variable promotional expenditure?

r2= 5.1833/5.2855= 98.1%

5) What is the 95% confidence interval estimate for the mean or average demand when the promotional expenditure X is 2.3 (i.e., )?

4.2176, 4.4530

Answer 1

A)

1) What is the sum of squares due to factor subject visibility?

Sum of squares due to factor subject visibility = Total sum of squares - Test taker success sum of squares - Interaction sum of squares - Error sum of squares

= 17755.20 - 11664.00 - 3385.80 - 1325.16

= 1380.24

2) How many degrees of freedom does 'Error' have?

Degree of freedom of total = number of observations - 1 = 40 - 1 = 39

Degree of freedom of error = Degree of freedom of total - Degree of freedom of Test taker success - Degree of freedom of  Interaction - Degree of freedom of subject visibility

= 39-(1+1+1) = 36

3) How many treatment combinations are there in the experiment?

There are 4 treatment combinations as given in the problem.

4) Perform the F- test on 'Test taker success'. Is the factor 'Test taker success' significant( )?

Numerator/Denominator Degree of Freedom of F statistic is 1, 36 (DF for factors, DF for error)

Critical value of F0.05,1,36= 4.12

F= MS('Test taker success')/MSE = 1325.16/324 = 4.09;

As, F=4.09 is less than the critical value (4.12), test taker success Factor is insignificant.

5) Perform the F- test on 'Interaction'. Is the factor 'Interaction' significant( )?

F= MS(Interaction)/MSE = 3385.8/324 = 10.45

As, F=10.45 is greater than the critical value (4.12), Interaction is significant.

B)

1) What is the slope(b1) and what is its interpretation in layman's words?

The slope (b1) of the regression line is 1.1464.

For every one $100 increase in promotional expenditure, there will be an estimated average increase in demand for the product is 1.1464*1,000=1,146.4 units.

2) Perform the test of hypothesis ( t-test) on the following, and state your statistical decision.

tSTAT = 1.1464/0.0569 = 20.15

Degree of freedom of tSTAT is DF for Residual Error = 8.

Critical value of t at significance level 0.025 and df = 8 is 2.3060

As, observed tSTAT is greater than the critical value, we reject null hypothesis and conclude that the Promotional exp is significant variable in the model to estimate Demand.

3) Perform the test of hypothesis (F-test) on the following, and state your statistical decision.

FSTAT = 5.1833/0.0128= 404.94

Critical value is F 0.05,1, 8 = 5.32;

As, observed F is greater than the critical value, we reject null hypothesis and conclude that the estimated model is significant a model to estimate Demand.

4) What is the coefficient of determination (r2), i.e., the percentage of variability in demand explained by the independent variable promotional expenditure?

r2= SS(Regression) / SS(Total) = 5.1833/5.2855= 98.1%

5) What is the 95% confidence interval estimate for the mean or average demand when the promotional expenditure X is 2.3 (i.e., )?

From the minitab output,  95% confidence interval estimate for the mean is 95% CI which is

(4.2176, 4.4530)