Statistics

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, Chapter 12, Exercise 10 and 18

W3-Mini Lecture - One-Way, and Two-Way Anova

10. The manager of a computer software company wishes to study the number of hours senior executives spend at their desktop computers by type of industry. The manager selected a sample of five executives from each of three industries. At the .05 significance level, can she conclude there is a difference in the mean number of hours spent per week by industry?

Solution:

Step 1: State the null and the alternative hypothesis.

: At least one pair of means are different

Step 2: Decision Rule.

Remember, that decision rule states region, or regions for which the null hypothesis is rejected. From the above table, k =3, and N = 15, therefore the dfn = k - 1 = 3 -1 =2, and dfd = N - k = 15 -3 =12. Now, at the level of significance, is given as 0.05. Using the "Critical values of the F distribution at " (Appendix G), the critical value of F= 3.98, is found at the cross section of the appropriate degrees of freedom, by looking down on the first column, that represents the dfdenominator column, and stop at dfd = 12, and cross the top, on the first row, that represents the dfnumerator, stop at dfn = 2. Therefore, we will reject the null, if the computed F-statistic will be greater than 3.98, found from the table.

From the above table, k=3. And N=15, therefore the =k-1=3-1=2, and =N-k=15-3=12. Now, at the level of significance, is given as 0.05. Using the "Critical values of the F distribution at " (Appendix G), the critical value of F= 3.89, is found at the cross section of the appropriate degrees of freedom, by looking down on the first column, that represents the df denominator column, and stop at = 12, and cross the top, on the first row, that represents the df numerator, stop at = 2. Therefore, we will reject the null, if the computed F-statistic will be greater than 3.89, found from the table.

Step 3: compute the test statistic.

The excel output returns the test statistic as 5.73.

The excel output returns the p-value as 0.018.

Anova: Single Factor

SUMMARY

Groups Count Sum Average Variance

Banking 5 54 10.8 1.2

Retail 5 40 8 2

Insurance 5 42 8.4 2.8

ANOVA

Source of Variation SS df MS F P-value

Between Groups 22.93333333 2 11.46666667 5.733333333 0.017880475

Within Groups 24 12 2

Total 46.93333333 14

Step 4: Make Decision.

Since the value of the test statistic is 5.73, is greater then the critical value of the test, 3.98, we reject the Null Hypothesis, in favor of the Alternative Hypothesis.

Since the value of the test statistic 5.733, we computed, in the previous step is greater than the critical value of the test