Share
Explore BrainMass

Chi-square Test, ANOVA & Correlation Testing

1. A study is being conducted to determine whether there is a relationship between jogging and blood pressure. A random sample of 210 subjects is selected, and they are classified as shown in the following table. At ± = 0.05, test the claim that jogging and blood pressure are not related. (apply chi-square independent test)

Blood Pressure
Jogging Status Low Moderate High
Joggers 34 57 21
Non-joggers 15 63 20

(a) State the null and alternative hypothesis.

(b) State your conclusion about the hypothesis based on the test statistic and critical value.

2. An investigator wishes to compare the average time to relief of headache pain under three distinct medications, call them Drugs A, B, and C. Fifteen patients who suffer from chronic headaches are randomly selected for the investigation, and five subjects are randomly assigned to each treatment. The following data reflect times to relief (in minutes) after taking the assigned drug:

Test if there is a significant difference in the mean times among three treatments. Use ± = 0.05. (apply ANOVA)

Drug A Drug B Drug C
30 25 15
35 20 20
40 30 25
25 25 20
35 30 20

(a) State the null and alternative hypothesis.

(b) State your conclusion about the hypothesis based on the test statistic and critical value.

3. Consider the following data reflecting lengths of stay in the hospital (recorded in days) and the total charge (in $000s) for six patients undergoing a minor surgical procedure:

Length of stay 5 7 9 10 12 15
Total charge 6 5 7.2 8 9.4 7.9

Consider length of stay as the independent and total charge as the dependent variable. Calculate value of the correlation coefficient (r) and the test significance at 5% level. (apply correlation analysis and perform a significance test for the correlation coefficient)

Solution Summary

The solution provides step by step method for the calculation of chi square test for association, one way ANOVA and test statistic for significance of correlation coefficient. Formula for the calculation and Interpretations of the results are also included.

$2.19