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A statistician demonstrated how to test the efficacy of an HIV vaccine. He repor

ID: 3043252 • Letter: A

Question

A statistician demonstrated how to test the efficacy of an HIV vaccine. He reported the results in the 2 times ×2 table shown below. The trial consisted of 8 AIDS patients vaccinated with the new drug and 31 AIDS patients who were treated with a placebo (no vaccination). The table shows the number of patients who tested positive and negative for the MN strain in the trial follow-up period. Complete parts a through e.

                                    Positive                       Negative          totals

Unvaccinated              24                                9                      33

Vaccinated                  2                                  6                      8

Totals                          26                                15                    41

Conduct a test to determine whether the vaccine is effective in treating the MN strain of HIV. Use

=0.05.

What are the null and alternative hypotheses?

Find the test statistic. 2=

Specify the rejection region. Choose the correct answer below.

A. 2greater than>3.84146

B. 2greater than>16.9190

C. 2greater than>9.48773

D. 2greater than>15.5073

Choose from the following:

A. Reject Upper H 0H0. There is insufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

B. Reject Upper H 0H0. There is sufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

C.Fail to reject Upper H 0H0. There is insufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

Are the assumptions for the test you carried out in part a, satisfied?

Yes

No

. In the case of a 2 times ×2 contingency table, R. A. Fisher developed a procedure for computing the exact p-value for the test. The method utilizes the hypergeometric probability distribution. Consider the hypergeometric probability shown on the right which represents the probability that

2 out of 8 vaccinated AIDS patients test positive and

24 out of 33 unvaccinated patients test

Positive long dash—that

is, the probability of the result shown in table, given that the null hypothesis of independence is true.

Compute this probability.  

StartFraction left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 8 2nd Row 1st Column 2 EndMatrix right parenthesis left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 33 2nd Row 1st Column 24 EndMatrix right parenthesis Over left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 41 2nd Row 1st Column 26 EndMatrix right parenthesis EndFraction

                          (8

                           (2

               (33

              (24

    (41

    (26

The probability is p=

.(Round to four decimal places as needed.)

d. Refer to part c.

Two contingency tables that are more unsupportive of the null hypothesis of independence than the observed table are shown in the accompanying table. Explain why these tables provide more evidence to reject Upper H 0H0 than the original table does.

If vaccine and MN strain are independent, then the proportion of positive results should be

relatively the same

different

for both patient groups.

In the two tables presented, the proportion of positive results for the vaccinated group is

smaller

greater

than the proportion for the original table.

Compute the probability of the first table, using the hypergeometric formula.

The probability of the contingency table is

. (Round to four decimal places as needed.)

Compute the probability of the second table, using the hypergeometric formula.

The probability of the contingency table is

(Round to four decimal places as needed.)

e. The p-value of Fisher's exact test is the probability of observing a result at least as unsupportive of the null hypothesis as is the observed contingency table, given the same marginal totals. Sum the probabilities of parts c and d to obtain the p-value of Fisher's exact test.

The p-value of the test is.

(Round to four decimal places as needed.)

Interpret this value in the context of the vaccine trial. Choose the correct answer below.

A. Since the p-value is greater than alpha, there is insufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

B. Since the p-value is less than alpha, there is insufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

C. Since the p-value is less than alpha, there is sufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05

. In the case of a 2 times ×2 contingency table, R. A. Fisher developed a procedure for computing the exact p-value for the test. The method utilizes the hypergeometric probability distribution. Consider the hypergeometric probability shown on the right which represents the probability that

2 out of 8 vaccinated AIDS patients test positive and

24 out of 33 unvaccinated patients test

Positive long dash—that

is, the probability of the result shown in table, given that the null hypothesis of independence is true.

Compute this probability.  

StartFraction left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 8 2nd Row 1st Column 2 EndMatrix right parenthesis left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 33 2nd Row 1st Column 24 EndMatrix right parenthesis Over left parenthesis Start 2 By 1 Matrix 1st Row 1st Column 41 2nd Row 1st Column 26 EndMatrix right parenthesis EndFraction

                          (8

                           (2

               (33

              (24

    (41

    (26

The probability is p=

.(Round to four decimal places as needed.)

Explanation / Answer

Answer:

A statistician demonstrated how to test the efficacy of an HIV vaccine. He reported the results in the 2 times ×2 table shown below. The trial consisted of 8 AIDS patients vaccinated with the new drug and 31 AIDS patients who were treated with a placebo (no vaccination). The table shows the number of patients who tested positive and negative for the MN strain in the trial follow-up period. Complete parts a through e.

  

                                    Positive                       Negative          totals

Unvaccinated              24                                9                      33

Vaccinated                  2                                  6                      8

Totals                          26                                15                    41

Chi-Square Test

Observed Frequencies

Column variable

Calculations

Row variable

C1

C2

Total

fo-fe

R1

24

9

33

3.0732

-3.0732

R2

2

6

8

-3.0732

3.0732

Total

26

15

41

Expected Frequencies

Column variable

Row variable

C1

C2

Total

(fo-fe)^2/fe

R1

20.9268

12.0732

33

0.4513

0.7823

R2

5.0732

2.9268

8

1.8616

3.2268

Total

26

15

41

Data

Level of Significance

0.05

Number of Rows

2

Number of Columns

2

Degrees of Freedom

1

Results

Critical Value

3.8415

Chi-Square Test Statistic

6.3220

p-Value

0.0119

Reject the null hypothesis

Conduct a test to determine whether the vaccine is effective in treating the MN strain of HIV. Use

=0.05.

What are the null and alternative hypotheses?

Ho: the vaccine is not effective in treating the MN strain of HIV

H1: the vaccine is effective in treating the MN strain of HIV

Find the test statistic. 2=6.3220

Specify the rejection region. Choose the correct answer below.

Answer: A. 2greater than>3.84146

B. 2greater than>16.9190

C. 2greater than>9.48773

D. 2greater than>15.5073

Choose from the following:

A. Reject Upper H 0H0. There is insufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

Answer: B. Reject H0. There is sufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

C.Fail to reject Upper H 0H0. There is insufficient evidence to indicate that the vaccine is effective in treating the MN strain of HIV at =0.05.

Are the assumptions for the test you carried out in part a, satisfied?

Yes

Answer: No

( one expected value less than 5).

Chi-Square Test

Observed Frequencies

Column variable

Calculations

Row variable

C1

C2

Total

fo-fe

R1

24

9

33

3.0732

-3.0732

R2

2

6

8

-3.0732

3.0732

Total

26

15

41

Expected Frequencies

Column variable

Row variable

C1

C2

Total

(fo-fe)^2/fe

R1

20.9268

12.0732

33

0.4513

0.7823

R2

5.0732

2.9268

8

1.8616

3.2268

Total

26

15

41

Data

Level of Significance

0.05

Number of Rows

2

Number of Columns

2

Degrees of Freedom

1

Results

Critical Value

3.8415

Chi-Square Test Statistic

6.3220

p-Value

0.0119

Reject the null hypothesis

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