Statistics: Statistical Hypothesis Testing and Critical Value
Lind, Chapter 13, Exercise 42
A sample of 12 homes sold last week in St. Paul, Minnesota, is selected. Can we conclude that as the size of the home (reported below in thousands of square feet) increases, the selling price (reported in $ thousands) also increases?
a. Compute the coefficient of correlation. Coefficients Standard Error t Stat P-value
Intercept 59.95717345 28.65750326 2.092198085 0.062896401
Home Size 31.69164882 24.44661008 1.296361692 0.223968044
The formula for the coefficient of correlation, . = 0.076.
a. Compute the coefficient of correlation is .308
b. Determine the coefficient of determination.
Regression Statistics
Multiple R 0.37931016
R Square 0.143876197
Adjusted R Square 0.058263817
Standard Error 15.25058359
Observations 12
The coefficient of determination is 14%.
b. Determine the coefficient of determination is .095
c. Can we conclude that there is a positive association between the size of the home and the selling price? Use the .05 significance level. Yes, there is a positive correlation between the variables.
c. Can we conclude that there is a positive association between the size of the home and the selling price? Use the .05 significance level.
Step 1:
Step 2:
The alternative hypothesis is which indicates a one tail test. The degrees of freedom (df) = n-2=12-2=10. The t value at the significance level of .05 and df of 8 and from appendix F is 1.812. The decision rule is to reject the null hypothesis because the test statistic is greater then 1.812.
Step 3:
According to Appendix F, we can assume that P 1.744), there is insufficient evidence to reject the null hypothesis.
P-value Approach: Since the P-value is less than the level of significance of the test, namely (p-value > 0.05), there is insufficient evidence to reject the null hypothesis.
Step 5: Interpret the results.
At the level of significance, there is no evidence that the insurance quotes differ. Lind,
A sample of 12 homes sold last week in St. Paul, Minnesota, is selected. Can we conclude that as the size of the home (reported below in thousands of square feet) increases, the selling price (reported in $ thousands) also increases?
a. Compute the coefficient of correlation. Coefficients Standard Error t Stat P-value
Intercept 59.95717345 28.65750326 2.092198085 0.062896401
Home Size 31.69164882 24.44661008 1.296361692 0.223968044
The formula for the coefficient of correlation, . = 0.076.
a. Compute the coefficient of correlation is .308
b. Determine the coefficient of determination.
Regression Statistics
Multiple R 0.37931016
R Square 0.143876197
Adjusted R Square 0.058263817
Standard Error 15.25058359
Observations 12
The coefficient of determination is 14%.
b. Determine the coefficient of determination is .095
c. Can we conclude that there is a positive association between the size of the home and the selling price? Use the .05 significance level. Yes, there is a positive correlation between the variables.
c. Can we conclude that there is a positive association between the size of the home and the selling price? Use the .05 significance level.
Step 1:
Step 2:
The alternative hypothesis is which indicates a one tail test. The degrees of freedom (df) = n-2=12-2=10. The t value at the significance level of .05 and df of 8 and from appendix F is 1.812. The decision rule is to reject the null hypothesis because the test statistic is greater then 1.812.
Step 3:
According to Appendix F, we can assume that P 1.744), there is insufficient evidence to reject the null hypothesis.
P-value Approach: Since the P-value is less than the level of significance of the test, namely (p-value > 0.05), there is insufficient evidence to reject the null hypothesis.
Step 5: Interpret the results.
At the level of significance, there is no evidence that the insurance quotes differ. Lind,