In: Statistics and Probability
The following data on price ($) and the overall score for 6 stereo headphones that were tested by Consumer Reports were as follows. Brand Price Score Bose 180 78 Scullcandy 160 76 Koss 85 65 Phillips/O'Neill 70 56 Denon 80 40 JVC 45 27 a. Does the t test indicate a significant relationship between price and the overall score? The test t-Conclusion at α = .05 t = (to 2 decimal places.) p-value is What is your conclusion? Use α = .05. . b. Test for a significant relationship using the F test. p-value is What is your conclusion? Use α = .05. Because p-value is .05, we H0: β1 is . c. Show the ANOVA table for these data. Round your answers to three decimal places, if necessary. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Regression Error Total
Price | Score |
180 | 78 |
160 | 76 |
85 | 65 |
70 | 56 |
80 | 40 |
45 | 27 |
using Excel
data -> data analysis -> regression
SUMMARY OUTPUT | |||||
Regression Statistics | |||||
Multiple R | 0.8650 | ||||
R Square | 0.7482 | ||||
Adjusted R Square | 0.6852 | ||||
Standard Error | 11.3775 | ||||
Observations | 6 | ||||
ANOVA | |||||
df | SS | MS | F | Significance F | |
Regression | 1 | 1538.2094 | 1538.2094 | 11.8829 | 0.0261 |
Residual | 4 | 517.7906 | 129.4476 | ||
Total | 5 | 2056.0000 | |||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | |
Intercept | 23.3245 | 10.8171 | 2.1563 | 0.0973 | -6.7085 |
Price | 0.3259 | 0.0945 | 3.4472 | 0.0261 | 0.0634 |
a)
t = 3.45
p-value = 0.0261
p-value < 0.05
hence we reject the null hypothesis
we conclude that there is significant relationship
b)
F = 11.8829
p-value = 0.0261
p-value < 0.05
hence we reject the null hypothesis
we conclude that there is significant relationship
c)
ANOVA | |||||
df | SS | MS | F | Significance F | |
Regression | 1 | 1538.209 | 1538.209 | 11.883 | 0.026 |
Residual | 4 | 517.791 | 129.448 | ||
Total | 5 | 2056.000 |