Statistics: Statistical Hypothesis Testing and Critical Value

1873 Words Nov 8th, 2006 8 Pages
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
…show more content…
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 3.89, we obtained in step 2; we reject the Null Hypothesis, in favor of the Alternate Hypothesis.

Step 5: Interpret the results.
At the level of significance of 0.05, there is a difference in the mean number of hours spent per week on the computer by the banking, retail, and insrance industry.
At the level of significance of 0.05, there is a difference in the mean number of hours worked per week in the Banking, Retail, and Insurance industry for sample of five weeks.

18. There are three hospitals in the Tulsa, Oklahoma, area. The following data show the number of outpatient surgeries performed at each hospital last week. At the .05 significance level, can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week? Solution:
Step 1: State the null and alternative hypotheses.
In words: NUMBER OF SURGERIES
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