Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Testing Effectiveness of Nicotine Patches In one study of smokers who tried to quit smoking with nicotine patch therapy, 39 were smoking one year after the treatment and 32 were not smoking one year after the treatment (based on data from “High-Dose Nicotine Patch Therapy,” by Dale et al., Journal of the American Medical Association, Vol. 274, No. 17). Use a 0.05 significance level to test the claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking one year after the treatment. Do these results suggest that the nicotine patch therapy is not effective?

Short Answer

Expert verified

Null hypothesis: The proportion of smokers who are smoking one year after the treatment is equal to 0.5.

Alternative hypothesis: The proportion of smokers who are smoking one year after the treatment is greater than 0.5.

Test statistic: 0.826

Critical value: 1.645

P-value: 0.2044

The null hypothesis is failed to reject.

There is not enough evidence to support the claim that the proportion of smokers who start smoking after one year of the treatment is greater than 50%.

The nicotine patch therapy is not very effective in helping smokers quit smoking because the sample results show that 54.9% of smokers start smoking after one year of the treatment, which is quite high.

Step by step solution

01

Given information

It is given that 39 smokers who have under gone the nicotine patch therapy are smoking after one year of the treatment, while 32 are not smoking one year after the treatment. It is claimed that most of the smokers who take the treatment have started smoking after one year of the treatment.

02

Hypotheses

The null hypothesis is written as follows.

The proportion of smokers who start smoking after one year of the treatment is equal to 0.5.

H0:p=0.5

The alternative hypothesis is written as follows.

The proportion of smokers who start smoking after one year of the treatment is more than 0.5.

H1:p>0.5

The test is right-tailed.

03

Sample size, sample proportion, and population proportion

The sample size is calculated below.

n=39+32=71

The sample proportion of smokers who start smoking after one year of the treatment is computed below.

p^=3971=0.549

The population proportion of smokers who start smoking after one year of the treatment is equal to 0.5.

04

Test statistic

The value of the test statistic is computed below.

z=p^-ppqn=0.549-0.50.51-0.571=0.826

Thus, z=0.826.

05

Critical value and p-value

Referring to the standard normal table, the critical value of z at for a right-tailed test is equal to 1.645.

Referring to the standard normal table, the p-value for the test statistic value of 0.826 is equal to 0.2044.

As the p-value is less than 0.05, the null hypothesis is not rejected.

06

Conclusion of the test

There is not enough evidence to support the claim that the proportion of smokers who start smoking after one year of the treatment is greater than 50%.

Although the claim is rejected, it can be seen that the sample percentage of smokers who start smoking after one year of the treatment is equal to 54.9% (greater than 50%).

Thus, it can be said that nicotine patch therapy is not very effective in helping smokers quit smoking.

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

Hypothesis Test with Known σ How do the results from Exercise 13 “Course Evaluations” change if σis known to be 0.53? Does the knowledge of σ have much of an effect?

Final Conclusions. In Exercises 25–28, use a significance level of α = 0.05 and use the given information for the following:

a. State a conclusion about the null hypothesis. (Reject H0or fail to reject H0.)

b. Without using technical terms or symbols, state a final conclusion that addresses the original claim.

Original claim: The standard deviation of pulse rates of adult males is more than 11 bpm. The hypothesis test results in a P-value of 0.3045.

Lead in Medicine Listed below are the lead concentrations (in ) measured in different Ayurveda medicines. Ayurveda is a traditional medical system commonly used in India. The lead concentrations listed here are from medicines manufactured in the United States (based on data from “Lead, Mercury, and Arsenic in US and Indian Manufactured Ayurvedic Medicines Sold via the Internet,” by Saper et al., Journal of the American Medical Association,Vol. 300, No. 8). Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 14 μg/g.

3.0 6.5 6.0 5.5 20.5 7.5 12.0 20.5 11.5 17.5

Interpreting Power Chantix (varenicline) tablets are used as an aid to help people stop smoking. In a clinical trial, 129 subjects were treated with Chantix twice a day for 12 weeks, and 16 subjects experienced abdominal pain (based on data from Pfizer, Inc.). If someone claims that more than 8% of Chantix users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.18 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.

Cans of coke: Confidence interval if we use the data given in exercise 1, we get this 95% confidence interval estimate of the standard deviation of volumes with the new filling process: 0.1846<σ<0.4900 oz. what does this confidence interval tell us about the new filling process?

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