Question
SPSS OUTPUT BASED QUESTIONS REGRESSION OUTPUT Variables Entered/Removed a Model Variables Entered Variables Removed Method 1 Attitude towards internet, Familiarity with the internet, Gender, Attitude
SPSS OUTPUT BASED QUESTIONS
REGRESSION OUTPUT
Variables Entered/Removeda | |||
Model | Variables Entered | Variables Removed | Method |
1 | Attitude towards internet, Familiarity with the internet, Gender, Attitude towards technologyb | . | Enter |
a. Dependent Variable: Internet usage in hours per week | |||
b. All requested variables entered. |
Model Summary | ||||
Model | R | R Square | Adjusted R Square | Std. Error of the Estimate |
1 | .922a | .852 | .825 | 1.89643 |
a. Predictors: (Constant), Attitude towards internet, Familiarity with the internet, Gender, Attitude towards technology |
ANOVAa | ||||||
Model | Sum of Squares | Df | Mean Square | F | Sig. | |
1 | Regression | 445.289 | 4 | 111.322 | 30.953 | .000b |
Residual | 89.911 | 25 | 3.596 | |||
Total | 535.200 | 29 | ||||
a. Dependent Variable: Internet usage in hours per week | ||||||
b. Predictors: (Constant), Attitude towards internet, Familiarity with the internet, Gender, Attitude towards technology |
Coefficientsa | ||||||
Model | Unstandardized Coefficients | Standardized Coefficients | t | Sig. | ||
B | Std. Error | Beta | ||||
1 | (Constant) | -.457 | 3.112 | -.147 | .884 | |
Attitude towards technology | 1.370 | .406 | .446 | 3.373 | .002 | |
Familiarity with the internet | 1.509 | .272 | .610 | 5.559 | .000 | |
Gender | -1.506 | .917 | -.178 | -1.643 | .030 | |
Attitude towards internet | -.708 | .499 | -.185 | -1.417 | .040 | |
a. Dependent Variable: Internet usage in hours per week |
Questions
Interpret the value of R Square (in the Table above) for the tested model.
- What is the hypothesis being tested in the ANOVA table with respect to the model?
- What will be the final regression equation used for predicting dependent variable from given values of independent variables?
- Write the null and alternate hypotheses for the problem given above.
CROSS-TABS & CHI-SQUARE OUTPUT
The analysis given below is carried out between in-store promotion and normalized sales.
Case Processing Summary | ||||||
Cases | ||||||
Valid | Missing | Total | ||||
N | Percent | N | Percent | N | Percent | |
InStorePromotion * NormalizedSales_Categorical | 60 | 100.0% | 0 | 0.0% | 60 | 100.0% |
InStorePromotion * NormalizedSales_CategoricalCrosstabulation | |||||
NormalizedSales_Categorical | Total | ||||
1.00 | 2.00 | ||||
InStorePromotion | High | Count | 0 | 20 | 20 |
% within InStorePromotion | 0.0% | 100.0% | 100.0% | ||
Medium | Count | 8 | 12 | 20 | |
% within InStorePromotion | 40.0% | 60.0% | 100.0% | ||
Low | Count | 16 | 4 | 20 | |
% within InStorePromotion | 80.0% | 20.0% | 100.0% | ||
Total | Count | 24 | 36 | 60 | |
% within InStorePromotion | 40.0% | 60.0% | 100.0% |
Chi-Square Tests | |||
Value | Df | Asymp. Sig. (2-sided) | |
Pearson Chi-Square | 26.667a | 2 | .000 |
Likelihood Ratio | 33.825 | 2 | .000 |
Linear-by-Linear Association | 26.222 | 1 | .000 |
N of Valid Cases | 60 | ||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 8.00. |
Questions
- How is thePearson Chi-Square valuecalculated in the problem given above?
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