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.)

Coke and Pepsi Data Set 26 “Cola Weights and Volumes” in Appendix B includes volumes of the contents of cans of regular Coke (n = 36, x = 12.19 oz, s = 0.11 oz) and volumes of the contents of cans of regular Pepsi (n = 36, x = 12.29 oz, s = 0.09 oz).

a. Use a 0.05 significance level to test the claim that cans of regular Coke and regular Pepsi have the same mean volume.

b. Construct the confidence interval appropriate for the hypothesis test in part (a).

c. What do you conclude? Does there appear to be a difference? Is there practical significance?

Short Answer

Expert verified

a. There is sufficient evidence that there is a difference between the volumes of the contents of cans of Coke and Pepsi.

b. The confidence interval for the difference between the means of sample is (-0.052 oz,0.148 oz).

c. The result is statistically significant but has no practical significance.

Step by step solution

01

Given information

The given problem is based on the data of Coke and Pepsi. This data set contains information about the volume of the contents in cans of regular Coke and regular Pepsi, summarized as follows:

\(\begin{array}{l}{{\rm{1}}^{{\rm{st}}}}\;{\rm{sample}}\;{\rm{:Regular}}\;{\rm{coke}}\\{n_1}\; = 36,\\{s_1} = 0.11\;oz\;\\{{\bar x}_1} = 12.19\;oz\end{array}\)

\(\begin{array}{l}{{\rm{2}}^{{\rm{nd}}}}\;{\rm{sample}}\;:{\rm{Regular}}\;{\rm{Pepsi}}\;\\{n_2} = 20\;\\{s_2} = 0.09\;oz\;\\{{\bar x}_2} = 12.29\;oz\end{array}\)

02

State the hypothesis

a.

The claim is states that the mean volume of a can of Coke and a can of Pepsi is the same.

\(\begin{array}{l}{H_{0\;}}:\;{\mu _1} = {\mu _2}\\{H_1}\;:\;{\mu _1} \ne \;{\mu _2}\end{array}\)

Here,\({\mu _1},{\mu _2}\)are the population mean volume contents for Coke and Pepsi, respectively.

The samples are independent with unknown and unequal population standard deviations.

03

Compute the test statistic

The formula for t-statistic is given below.

\({t_{stat}} = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }};\;({\rm{here}},\;\left( {{\mu _1} - {\mu _2}} \right)\;{\rm{is}}\;{\rm{supposed}}\;{\rm{to}}\;{\rm{be}}\;0\,)\)

04

Find critical values

For t-distribution, find the degrees of freedom as follows:

\(\begin{array}{c}df = \min \left( {\left( {{n_1} - 1} \right),\left( {{n_2} - 1} \right)} \right)\\ = \min \left( {\left( {36 - 1} \right),\left( {36 - 1} \right)} \right)\\ = 35\end{array}\)

The critical values are obtained as follows:

\(\begin{array}{c}P\left( {t > {t_{\frac{\alpha }{2}}}} \right) = \frac{\alpha }{2}\\P\left( {t > {t_{\frac{{0.05}}{2}}}} \right) = \frac{{0.05}}{2}\\P\left( {t > {t_{0.025}}} \right) = 0.025\end{array}\)

Thus, the critical value obtained from the t-table for 35 degrees of freedom is 2.0301.

05

Compute test statistic

The test statistic of the means of populations is as follows:

\(\begin{array}{c}{t_{stat}} = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {12.19 - 12.29} \right)}}{{\sqrt {\frac{{{{\left( {0.11} \right)}^2}}}{{36}} + \frac{{{{\left( {0.09} \right)}^2}}}{{36}}} }}\\ = - 4.22159\end{array}\)

The test statistic is \({t_{stat}} = - 4.22\).

06

Decision rule using the critical value

The decision criterion for this problem statement is given below.

If\(\left| {{t_{stat}}} \right|\; > \;{t_{crit}}\); Reject a null hypothesis at \(\alpha \)level of significance

If \(\left| {{t_{stat}}} \right|\; < {t_{crit}}\) ; Fail to accept null hypothesis at \(\alpha \)level of significance

In this case, \(\left| {{t_{stat}} = - 4.222} \right|\; > \;{t_{crit}} = 2.0301\).

Thus, the null hypothesis is rejected. It shows that there is not enough evidence to support the claim that the volumes of cans of Pepsi and Coke are the same.

07

Confidence interval for the difference of means of population

b.

For the 0.05 significance test, concerning a two-tailed test, the most appropriate confidence level is 95%.

The formula for the confidence interval of the means of population is given by

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\).

Here, E is the margin of error that is computed as follows:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}} \times \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \\ = {t_{\frac{{0.05}}{2}}} \times \sqrt {\frac{{{{0.11}^2}}}{{36}} + \frac{{{{0.09}^2}}}{{36}}} \\ = 2.0301 \times 0.024\\ = 0.0481\end{array}\)

Substitute the values in the formula.

\(\begin{array}{c}{\rm{C}}{\rm{.I}} = \left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\\ = \left( {12.19 - 12.29} \right) - 0.59 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {12.19 - 12.29} \right) + 0.59\\ = - 0.052 < \left( {{\mu _1} - {\mu _2}} \right) < 0.148\end{array}\)

The 95% confidence interval includes 0, which implies that there is not enough evidence that the mean of the contents for Pepsi and Coke is the same.

There is a significant difference between the volumes of the cans.

08

Conclude the results

c.

Thus, it is concluded that there is a statistical significance for the difference in mean of the contents in Pepsi and Coke.

There does not appear to be any practical significance as the difference in the sample means is 0.10 oz, which is very small.

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

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.

Smoking Cessation Programs Among 198 smokers who underwent a “sustained care” program, 51 were no longer smoking after six months. Among 199 smokers who underwent a “standard care” program, 30 were no longer smoking after six months (based on data from “Sustained Care Intervention and Postdischarge Smoking Cessation Among Hospitalized Adults,” by Rigotti et al., Journal of the American Medical Association, Vol. 312, No. 7). We want to use a 0.01 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program.

a. Test the claim using a hypothesis test.

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

c. Does the difference between the two programs have practical significance?

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.

Tennis Challenges Since the Hawk-Eye instant replay system for tennis was introduced at the U.S. Open in 2006, men challenged 2441 referee calls, with the result that 1027 of the calls were overturned. Women challenged 1273 referee calls, and 509 of the calls were overturned. We want to use a 0.05 significance level to test the claim that men and women have equal success in challenging calls.

a. Test the claim using a hypothesis test.

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

c. Based on the results, does it appear that men and women have equal success in challenging calls?

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.

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.

Before/After Treatment Results Captopril is a drug designed to lower systolic blood pressure. When subjects were treated with this drug, their systolic blood pressure readings (in mm Hg) were measured before and after the drug was taken. Results are given in the accompanying table (based on data from “Essential Hypertension: Effect of an Oral Inhibitor of Angiotensin-Converting Enzyme,” by MacGregor et al., British Medical Journal, Vol. 2). Using a 0.01 significance level, is there sufficient evidence to support the claim that captopril is effective in lowering systolic blood pressure?

Subject

A

B

C

D

E

F

G

H

I

J

K

L

Before

200

174

198

170

179

182

193

209

185

155

169

210

After

191

170

177

167

159

151

176

183

159

145

146

177

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