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Testing Claims About Proportions. In Exercises 7–22, 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.

Clinical Trials of OxyContin OxyContin (oxycodone) is a drug used to treat pain, butit is well known for its addictiveness and danger. In a clinical trial, among subjects treatedwith OxyContin, 52 developed nausea and 175 did not develop nausea. Among other subjectsgiven placebos, 5 developed nausea and 40 did not develop nausea (based on data from PurduePharma L.P.). Use a 0.05 significance level to test for a difference between the rates of nauseafor those treated with OxyContin and those given a placebo.

a. Use a hypothesis test.

b. Use an appropriate confidence interval.

c. Does nausea appear to be an adverse reaction resulting from OxyContin?

Short Answer

Expert verified

a.There is not sufficient evidence to concludethat there is a significant difference between the rates of nausea for subjects who were treated with OxyContin and subjects who were given a placebo.

b. The 95% confidence interval is equal to (0.011, 0.225), andsince 0 is not included in the interval, it is concluded that there is a difference between the rates of nausea for subjects treated with OxyContin and subjects given a placebo.

c. Nausea appears to be an adverse reaction resulting from OxyContin.

Step by step solution

01

Given information

In a clinical trial, 52 subjects developed nausea, and 175 did not develop nausea among the subjects who were treated with OxyContin.

Also, among the subjects who were given placebos, five developed nausea, and 40 did not develop nausea.

02

Describe the hypothesis to be tested.

Null hypothesis:There is no difference between the rates of nausea for subjects who were treated with OxyContin and subjects who were given a placebo.

H0:p1=p2

Alternate hypothesis:There is a significant difference between the rates of nausea for subjects who were treated with OxyContin and subjects who were given a placebo.

H1:p1p2

03

Calculate the sample statistics

Let n1and n2be the number of subjects treated with OxyContin and the number of subjects given a placebo, respectively.

The sample size n1 is computed below:

n1=52+175=227

The sample size n2 is computed below:

n2=5+40=45

Assume that x1 and x2 are the number ofsubjects who developed nausea and were given OxyContin and placebo, respectively.

Letp^1 be the sample proportionof subjects who developed nausea and were treated with OxyContin.

p^1=x1n1=52227=0.229

q^1=1-p^1=1-0.229=0.771

Letp^2be the sample proportion ofsubjects who developed nausea and were given a placebo.

p^2=x2n2=545=0.111

q^2=1-p^2=1-0.111=0.889

The value of the pooled sample proportion is equal to:

p¯=x1+x2n1+n2=52+5227+45=0.210

Hence,

q¯=1-p¯=1-0.210=0.790

04

Compute the value of the test statistic

The test statistic is equal to:

z=p^1-p^2-p1-p2p¯q¯n1+p¯q¯n2=0.229-0.111-00.2100.790227+0.2100.79045=1.776

Referring to the standard normal distribution table, the critical values of z corresponding to α=0.05 for a two-tailed test are -1.96 and 1.96.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0757.

Here, the value of the test statistic lies between the two critical values, and the p-value is greater than 0.05.

Therefore, the null hypothesis is failed to reject.

05

Conclusion of the Test

a.

There is not sufficient evidence to concludethat there is a significant difference between the rates of nausea for subjects who were treated with OxyContin and subjects who were given a placebo.

06

Describe the confidence interval

If the level of significance for a two-tailed test is equal to 0.05, then the corresponding confidence level to construct the confidence interval is equal to 95%.

The expression for computing the confidence interval is given below:

p^1-p^2-E<p1-p2<p^1-p^2+E

07

Calculate the margin of error

E is the margin of error and has the following formula:

E=zα2p^1q^1n1+p^2q^2n2=1.96×0.2290.771227+0.1110.88945=0.1069

08

Construct the confidence interval

b.

Substituting the required values, the following confidence interval is obtained:

p^1-p^2-E<p1-p2<p^1-p^2+E(0.229-0.111)-0.1069<p1-p2<(0.229-0.111)+0.10690.011<p1-p2<0.225

Thus, the 95% confidence interval is equal to (0.011, 0.225).

This confidence interval does not contain zero, which means that there is a significant difference between the two proportions of subjects who developed nausea.

Therefore, there is sufficient evidence to conclude thatthere is a significant difference between the rates of nausea for subjects who were treated with OxyContin and subjects who were given a placebo.

09

Examining nausea as an adverse reaction

c.

The sample results show that 22.9% of the subjects who were treated with OxyContin developed nausea.

Since the proportion is high, it can be said that nausea is an adverse reaction of OxyContin.

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Most popular questions from this chapter

Hypothesis Tests and Confidence Intervals for Hemoglobin

a. Exercise 2 includes a confidence interval. If you use the P-value method or the critical value method from Part 1 of this section to test the claim that women and men have the same mean hemoglobin levels, will the hypothesis tests and the confidence interval result in the same conclusion?

b. In general, if you conduct a hypothesis test using the methods of Part 1 of this section, will the P-value method, the critical value method, and the confidence interval method result in the same conclusion?

c. Assume that you want to use a 0.01 significance level to test the claim that the mean haemoglobin level in women is lessthan the mean hemoglobin level in men. What confidence level should be used if you want to test that claim using a confidence interval?

Degrees of Freedom

For Example 1 on page 431, we used df\( = \)smaller of\({n_1} - 1\)and\({n_2} - 1\), we got\(df = 11\), and the corresponding critical values are\(t = \pm 2.201.\)If we calculate df using Formula 9-1, we get\(df = 19.063\), and the corresponding critical values are\( \pm 2.093\). How is using the critical values of\(t = \pm 2.201\)more “conservative” than using the critical values of\( \pm 2.093\).

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1−1 and n2−1.)Color and Cognition Researchers from the University of British Columbia conducted a study to investigate the effects of color on cognitive tasks. Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Higher scores correspond to greater word recall.

a. Use a 0.05 significance level to test the claim that the samples are from populations with the same mean.

b. Construct a confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)?

c. Does the background color appear to have an effect on word recall scores? If so, which color appears to be associated with higher word memory recall scores?

Red Background n = 35, x = 15.89, s = 5.90

Blue Background n = 36, x = 12.31, s = 5.48

Testing Claims About Proportions. In Exercises 7–22, 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.

Cell Phones and Handedness A study was conducted to investigate the association between cell phone use and hemispheric brain dominance. Among 216 subjects who prefer to use their left ear for cell phones, 166 were right-handed. Among 452 subjects who prefer to use their right ear for cell phones, 436 were right-handed (based on data from “Hemi- spheric Dominance and Cell Phone Use,” by Seidman et al., JAMA Otolaryngology—Head & Neck Surgery, Vol. 139, No. 5). We want to use a 0.01 significance level to test the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones. (Try not to get too confused here.)

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

Braking Reaction Times: Normal? The accompanying normal quantile plot is obtained by using the braking reaction times of females listed in Exercise 6. Interpret this graph.

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