Question

7. Psychological and social factors can influence the survival of patients with serious diseases. One study compared the relationship between survival of patients with coronary heart disease and pet ownership. Patients were either classified as having a pet or not and whether they survived for 1 year. Based on the data below, is there an association between owning a pet and survival? No need to do the entire significance test, just the parts below. Pet Ownership Patient Status Yes No Alive 28 50 Dead 11 3 a. State the hypotheses for the appropriate test in the context of this problem. b. If the X test statistic has a value of 8.85 then give the p-value and conclusion of your significance test. 8. A researcher wishes to determine whether the rate of water flow (x) (in liters per second) over an experimental soil bed can be used to predict the amount of soil washed away(y) (in kilograms). The researcher measures the amount of soil washed away for various flow rates, finds that the sample mean flow rate is 6.1 L/s with a standard deviation of 0.5 L/s. Also the sample mean for the amount of eroded soil is 4.2 kg with a standard deviation of 1.2 kg. If the correlation between flow rate (x) and the amount of eroded soil (y) is 0.65 then give a. The equation of the least squares regression line of y on x. b. What is the predicted amount of eroded soil for a flow rate of 10 L/s? 9. The correlation between amount of time spent studying for a final and the final exam grade is 0.9. Explain what this means.

Answer 1

We would be looking at the first question all parts here as:

Q7) a) As we are testing here whether there is an association between the two variables: owing a pet and survival, therefore this is a test for independence for which the null and the alternative hypothesis here would be given as:

H0: There is no association between the two variables: owing a pet and survival

Ha: There is association between the two variables: owing a pet and survival

b) Degrees of freedom here is computed as:
Df = (num of column - 1)(num of rows - 1) = 1

Therefore the p-value here is obtained from the chi square distribution tables as:

p= P(Xỉ > 8.85) = 0.0029

As the p-value here is 0.0029 which is very very low, therefore the test is significant here and we can reject the null hypothesis here, therefore we have sufficient evidence here that there is an association between the two variables: owing a pet and survival