Clinical Trial of Lipitor Lipitor is the trade name of the drug atorvastatin, which is used to reduce cholesterol in patients. (Until its patent expired in 2011, this was the largest-selling drug in the world, with annual sales of $13 billion.) Adverse reactions have been studied in clinical trials, and the table below summarizes results for infections in patients from different treatment groups (based on data from Parke-Davis). Use a 0.01 significance level to test the claim that getting an infection is independent of the treatment. Does the atorvastatin (Lipitor) treatment appear to have an effect on infections?


Placebo

Atorvastatin 10 mg

Atorvastatin 40 mg

Atorvastatin 80 mg

Infection

27

89

8

7

No Infection

243

774

71

87

Short Answer

Expert verified

There is not enough evidence to conclude thatgetting an infection is not independent of the treatment.

No, atorvastatin treatment does not have an effect on infection because getting infected does not depend on the type of treatment administered.

Step by step solution

01

Given information

A contingency table is constructed that shows that the number of patients that get/not get an infection when different treatments to reduce cholesterol are administered to them.

02

Hypotheses

The null hypothesis is as follows:

Getting an infection is independent of the treatment.

The alternative hypothesis is as follows:

Getting an infection is not independent of the treatment.

It is considered a right-tailed test.

If the test statistic value is greater than the critical value, then the null hypothesis is rejected, otherwise not.

03

Observed and expected frequencies

Observed Frequency:

The frequencies provided in the table are the observed frequencies. They are denoted by\(O\).

The following contingency table shows the observed frequency for each cell:


Placebo

Atorvastatin 10 mg

Atorvastatin 40 mg

Atorvastatin 80 mg

Infection

\({O_1}\)=27

\({O_2}\)=89

\({O_3}\)=8

\({O_4}\)=7

No Infection

\({O_5}\)=243

\({O_6}\)=774

\({O_7}\)=71

\({O_8}\)=87

Expected Frequency:

The formula for computing the expected frequency for each cell is shown below:

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The row total for the first row is computed below:

\(\begin{aligned}{c}Row\;Tota{l_1} = 27 + 89 + 8 + 7\\ = 131\end{aligned}\)

The row total for the second row is computed below:

\(\begin{aligned}{c}Row\;Tota{l_2} = 243 + 774 + 71 + 87\\ = 1175\end{aligned}\)

The column total for the first column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_1} = 27 + 243\\ = 270\end{aligned}\)

The column total for the second column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_2} = 89 + 774\\ = 863\end{aligned}\)

The column total for the third column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_3} = 8 + 71\\ = 79\end{aligned}\)

The column total for the fourth column is computed below:

\(\begin{aligned}{c}Column\;Tota{l_4} = 7 + 87\\ = 94\end{aligned}\)

The grand total can be computed as follows:

\(\begin{aligned}{c}Grand\;Total = \left( {131 + 1175} \right)\\ = \left( {270 + 863 + 79 + 94} \right)\\ = 1306\end{aligned}\)

Thus, the following table shows the expected frequencies for each of the corresponding observed frequencies:


Placebo

Atorvastatin

10 mg

Atorvastatin

40 mg

Atorvastatin

80 mg

Infection

\[\begin{aligned}{c}{E_1} = \frac{{\left( {131} \right)\left( {270} \right)}}{{1306}}\\ = 27.083\end{aligned}\]

\[\begin{aligned}{c}{E_2} = \frac{{\left( {131} \right)\left( {863} \right)}}{{1306}}\\ = 86.564\end{aligned}\]

\[\begin{aligned}{c}{E_3} = \frac{{\left( {131} \right)\left( {79} \right)}}{{1306}}\\ = 7.924\end{aligned}\]

\[\begin{aligned}{c}{E_4} = \frac{{\left( {131} \right)\left( {94} \right)}}{{1306}}\\ = 9.429\end{aligned}\]

No Infection

\[\begin{aligned}{c}{E_5} = \frac{{\left( {1175} \right)\left( {270} \right)}}{{1306}}\\ = 242.917\end{aligned}\]

\[\begin{aligned}{c}{E_6} = \frac{{\left( {1175} \right)\left( {863} \right)}}{{1306}}\\ = 776.436\end{aligned}\]

\[\begin{aligned}{c}{E_7} = \frac{{\left( {1175} \right)\left( {79} \right)}}{{1306}}\\ = 71.076\end{aligned}\]

\[\begin{aligned}{c}{E_8} = \frac{{\left( {1175} \right)\left( {94} \right)}}{{1306}}\\ = 84.571\end{aligned}\]

04

Test Statistic

The test statistic is computed as shown below:

\[\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}\;\;\; \sim } {\chi ^2}_{\left( {r - 1} \right)\left( {c - 1} \right)}\\ = \frac{{{{\left( {27 - 27.083} \right)}^2}}}{{27.083}} + \frac{{{{\left( {89 - 86.564} \right)}^2}}}{{86.564}} + ..... + \frac{{{{\left( {87 - 84.571} \right)}^2}}}{{84.571}}\\ = 0.773\end{aligned}\]

Thus, \({\chi ^2} = 0.773\).

05

Obtain the critical value, p-value and determine the conclusion of the test

Let r denote the number of rows in the contingency table.

Let c denote the number of columns in the contingency table.

The degrees of freedom are computed below:

\(\begin{aligned}{c}df = \left( {r - 1} \right)\left( {c - 1} \right)\\ = \left( {2 - 1} \right)\left( {4 - 1} \right)\\ = 3\end{aligned}\)

The critical value of\({\chi ^2}\)for 3 degrees of freedom at 0.01 level of significance for a right-tailed test is equal to 11.3449.

The corresponding p-value is approximately equal to 0.855914.

Since the value of the test statistic is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to warrant rejection of the claim thatgetting an infection is independent of the treatment.

No, atorvastatin treatment does not have an effect on infection because getting infected does not depend on the type of treatment administered.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Questions 6–10 refer to the sample data in the following table, which describes the fate of the passengers and crew aboard the Titanic when it sank on April 15, 1912. Assume that the data are a sample from a large population and we want to use a 0.05 significance level to test the claim that surviving is independent of whether the person is a man, woman, boy, or girl.


Men

Women

Boys

Girls

Survived

332

318

29

27

Died

1360

104

35

18

Given that the P-value for the hypothesis test is 0.000 when rounded to three decimal places, what do you conclude? What do the results indicate about the rule that women and children should be the first to be saved?

Questions 6–10 refer to the sample data in the following table, which describes the fate of the passengers and crew aboard the Titanic when it sank on April 15, 1912. Assume that the data are a sample from a large population and we want to use a 0.05 significance level to test the claim that surviving is independent of whether the person is a man, woman, boy, or girl.


Men

Women

Boys

Girls

Survived

332

318

29

27

Died

1360

104

35

18

What distribution is used to test the stated claim (normal, t, F, chi-square, uniform)?

The table below shows results since 2006 of challenged referee calls in the U.S. Open. Use a 0.05 significance level to test the claim that the gender of the tennis player is independent of whether the call is overturned. Do players of either gender appear to be better at challenging calls?

Was the Challenge to the Call Successful?


Yes

No

Men

161

376

Women

68

152

In a study of high school students at least 16 years of age, researchers obtained survey results summarized in the accompanying table (based on data from “Texting While Driving and Other Risky Motor Vehicle Behaviors Among U.S. High School Students,” by O’Malley, Shults, and Eaton, Pediatrics,Vol. 131, No. 6). Use a 0.05 significance level to

test the claim of independence between texting while driving and driving when drinking alcohol. Are those two risky behaviors independent of each other?


Drove when drinking Alcohol?


Yes

No

Texted while driving

731

3054

No Texting while driving

156

4564

A randomized controlled trial was designed to compare the effectiveness of splinting versus surgery in the treatment of carpal tunnel syndrome. Results are given in the table below (based on data from “Splinting vs. Surgery in the Treatment of Carpal Tunnel Syndrome,” by Gerritsen et al., Journal of the American Medical Association,Vol. 288,

No. 10). The results are based on evaluations made one year after the treatment. Using a 0.01 significance level, test the claim that success is independent of the type of treatment. What do the results suggest about treating carpal tunnel syndrome?

Successful Treatment

Unsuccessful Treatment

Splint Treatment

60

23

Surgery Treatment

67

6

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free