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A.For questions 1-7, refer to the following information: Thrombolytic drugs are substances that cause the breakdown of blood clots obstructing the flow of blood through the vessels. They are used by injection during or shortly after a heart attack or stroke to prevent clots from blocking blood flow to the heart muscle or brain. In an experiment, three thrombolytic treatments were studied: streptokinase (SK), accelerated alteplase (AtPA) and streptokinase plus alteplase (SK+tPA). A total of 4082 patients affected by Acute Myocardial Infarction (AMI) participated in the study. Each patient received one of the drugs, randomly assigned, shortly after the infarction. Efficacy was measured by the mortality at 30 to 35 days. The data are summarized in the following table. Using a 0.05 significance level, you need to test the claim that survival is associated with the type of thrombolytic drug by answering the following questions.

1. What are the most appropriate null and alternative hypotheses? 

a. H0: Survival and type of thrombolytic drug follow a uniform distribution

H1: Survival and type of thrombolytic drug do not follow a uniform distribution

b. H0: 25% of those who died and 30% of those who lived took SK+tPA

H1: Dead and alive patients who took SK+tPA differ from 25% and 30%

c. H0: Survival and type of thrombolytic drug are independent

H1: Survival and type of thrombolytic drug are dependent

d. H0: Survival and type of thrombolytic drug are dependent

H1: Survival and type of thrombolytic drug are independent

e. H0: There is no linear correlation between survival and type of thrombolytic drug

H1: There is a linear correlation between survival and type of thrombolytic drug

2. What type of test should we perform to test the null hypothesis? We are asked to test the claim that survival is related to the type of thrombolytic drug. To do so, we use a test of independence known as Pearson’s chi-square test.

a. Goodness-of-fit test for multinomial experiments

b. Fisher’s exact test

c. Pearson’s chi-square test of independence

d. McNemar’s test

e. Test of no significant linear correlation

3. What is the expected frequency for those who used AtPA and died (labeled A in the output table)? 

E = (row total)(column total) / (grand total) = (284)(1034) / 4082 = 71.94

a. 0.05

b. 0.67

c. 65.00

d. 71.94

e. 962.06

4. What is the contribution to the overall test statistic of the cell corresponding to those who used SK and lived (labeled B in the output table)? (1 point)

Let O = observed frequency and E = expected frequency. Then the cell chi-square value is

calculated as: (O – E)

2 / E = (1869 – 1875.7)

2 / 1875.7 = 0.024.

a. 0.024

b. 0.32

c. 71.80

d. 1869.00

e. 1875.70

5. What is the value of the test statistic? (2 points)

χ2 = 0.3238 + 0.6694 + 0.0006 + 0.0242 + 0.0501 + 0.0000 = 1.068

a. 1.07

b. 1.14

c. 10.68

d. 37.42

e. Cannot be determined with the given information

6. How many degrees of freedom does this test statistic have? (1 point)

df = (r – 1)(c – 1) = (2 – 1)(3 – 1) = 2

a. 1

b. 2

c. 5

d. 4081

7. Given the p-value corresponding to the test statistic in (5) is 0.59, what is your conclusion regarding H0? 

a. Because 0.59 > 0.05, we reject H0 and conclude there is not sufficient evidence to support the claim that survival is related to the type of thrombolytic drug.

b. Because 0.59 < 0.05, we reject H0 and conclude there is sufficient evidence to support the claim that survival is related to the type of thrombolytic drug.

c. Because 0.59 > 0.05, we fail to reject H0 and conclude there is not sufficient evidence

to support the claim that survival is related to the type of thrombolytic drug.

d. Because 0.59 < 0.05, we fail to reject H0 and conclude there is sufficient evidence to  support the claim that survival is related to the type of thrombolytic drug.

e. We cannot directly compare 0.59 to 0.05. Instead we need to compare 0.59 to a critical  value.

B.For questions 8-18, refer to the following information: In their 1992 JAMA article “A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich,” Mackowiak, Wasserman and Levine examined whether body temperatures (degrees Fahrenheit) can be predicted by heart rate (beats per minute). Think about it. When you have a fever, do you have a higher heart rate? Could a higher heart rate be a mechanism of the body to generate higher body temperatures? The researchers generated the following results using SAS® software; however, they do not know how to interpret any of the output. Confident of your statistical abilities, they have come to you with their output and a list of questions. It’s now your job to help them interpret their output. Don’t let them down!

8. What type of analysis is this?

This output allows us to describe the linear relationship between two continuous variables using a linear regression equation. This is referred to as a simple linear regression analysis.

a. Two-sample t-test

b. Paired t-test

c. Simple linear regression

d. Multiple linear regression

e. Contingency table analysis

9. What are the independent and dependent variables? 

Researchers want to examine whether heart rate can predict body temperature. The independent variable is the predictor/explanatory variable x, which is heart rate. The dependent variable is the response/outcome variable y, which is body temperature.

a. Dependent variable is unknown, independent variable = body temperature

b. Dependent variable is unknown, independent variable = heart rate

c. Dependent variable = heart rate, independent variable = body temperature

d. Dependent variable = body temperature, independent variable = heart rate

e. Variable order does not matter here, i.e. (x, y) is the same as (y, x). Either body temperature or heart rate can be the independent variable. The other is then the dependent variable

10. What are the most appropriate null and alternative hypotheses for this analysis? (1 point)

The slope of the regression line, β1, tells us whether or not the two variables are significantly related to one another. If the slope is 0, then the variables are not related. This is the null hypothesis. The alternative is the opposite, that the slope is not 0.

a. H0: b0 = 0, H1: b0 ≠ 0

b. H0: b1 = 0, H1: b1 ≠ 0

c. H0: β0 = 0, H1: β0 ≠ 0

d. H0: β1 = 0, H1: β1 ≠ 0

e. H0: ρ = 0, H1: ρ ≠ 0

11. What is the size of our sample? 

In simple linear regression, the total degrees of freedom is n – 1. Therefore, our sample size is n = 129 + 1 = 130.

a. 1

b. 128

c. 129

d. 130

e. Cannot be determined from the given information

C.For questions 12-14 refer to the following: In the analysis of variance table, some wise guy replaced 3 important values with the letters A, B and C. You need to find these missing pieces. 12. Model degrees of freedom (df), labeled A 

The degrees of freedom for the model is equal to the number of independent variables in the model. For simple linear regression, this number is 1. Alternatively, you could have used the rule that model df + error df = total df. Here, A + 128 = 129 yields A = 1.

a. 0

b. 1

c. 2

d. 6

e. 31

13. Total Sum of Squares, labeled B (1 point)

The total sum of squares (SS) represents the total variation in our data, i.e. a comparison of the observed values of the dependent variable relative to the mean of the dependent variable. Remember SStotal = SSmodel + SSerror = 4.46176 + 64.88316 = 69.34492.

a. 0.51

b. 4.46

c. 60.42

d. 64.88

e. 69.34

14. Mean Square Error, labeled C 

The mean square error (MSE) measures the average of the squares of the errors. Simply put, the MSE is the sum of squares error divided by the corresponding degrees of freedom. So, MSE = 64.88316 / 128 = 0.50690.

a. 0.51

b. 4.46

c. 60.42

d. 64.88

e. 69.34

15. What is the value of the test statistic and the p-value that can be used to test the null hypothesis? 

For simple linear regression, the F test and the t test are equivalent ways to test the null

hypothesis that the slope is equal to 0. Using the Analysis of Variance table, F = 8.80 with pvalue = 0.0036. Using the Parameter Estimates table, t = 0.02633 / 0.00888 = 2.97 with pvalue = 0.0036. Remember the relationship F = t

a. t = 0.026, p = 0.0036

b. t = 2.97, p = 0.0036

c. t = 146.43, p < 0.0001

d. F = 8.80, p = 0.0036

e. Both a and d are correct

f. Both b and d are correct

g. Both c and d are correct

16. Using a 0.05 significance level, what decision and conclusion should you make regarding the null hypothesis? 

From (15), our p-value is 0.0036, which is less than our significance level of 0.05. Therefore, we reject H0 of no significant slope. This means that there is a significant linear relationship between heart rate and body temperature, so we can use heart rate to predict body temperature.

a. Because p-value < 0.05, we fail to reject H0 and conclude that heart rate cannot be used topredict body temperature.

b. Because p-value < 0.05, we fail to reject H0 and conclude that heart rate can be used to predict body temperature.

c. Because p-value < 0.05, we reject H0 and conclude that heart rate cannot be used to predict body temperature.

d. Because p-value < 0.05, we reject H0 and conclude that heart rate can be used tonpredict body temperature.

e. Because p-value > 0.05, we fail to reject H0 and conclude that heart rate cannot be used to

predict body temperature.

17. What is the marginal change in body temperature, i.e. by how many degrees Fahrenheit does body temperature change when heart rate increases by 1 beat per second? (1 point) The marginal change in the dependent variable that occurs when the independent variable changes by one unit is represented by the slope b1 in the regression equation. Here, the estimate of the slope parameter is +0.02633.

a. Increases by 0.026 degrees Fahrenheit

b. Decreases by 0.026 degrees Fahrenheit

c. Increases by 96.31 degrees Fahrenheit

d. Decreases by 96.31 degrees Fahrenheit

e. Increases by 25.36%

18. What is the estimated value of the linear correlation coefficient and how do we best interpret this value? 

The coefficient of determination R2 is given in the output as 0.0643. We interpret R2 as the percentage of the variation in the dependent variables that can be explained by the linear relationship between the independent and dependent variables. The linear correlation coefficient r is simply the square root of this value, which is 0.2536.

a. r = 0.0643, so 6.43% of the variation in body temperature can be explained by the linear relationship between body temperature and heart rate.

b. r = 0.2536, so 25.36% of the variation in body temperature can be explained by the linear relationship between body temperature and heart rate.

c. r = 0.2536, so 6.43% of the variation in body temperature can be explained by the linear relationship between body temperature and heart rate.

d. r = 0.0643, so 25.36% of the variation in body temperature can be explained by the linear relationship between body temperature and heart rate.

e. r = 0.0041, so 0.41% of the variation in body temperature can be explained by the linear relationship between body temperature and heart rate.

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