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Ch. 9 - Inferences from Two Samples
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 9, Problem 9.1.14a

Cigarette Pack Warnings A study was conducted to find the effects of cigarette pack warnings that consisted of text or pictures. Among 1078 smokers given cigarette packs with text warnings, 366 tried to quit smoking. Among 1071 smokers given cigarette packs with warning pictures, 428 tried to quit smoking. (Results are based on data from “Effect of Pictorial Cigarette Pack Warnings on Changes in Smoking Behavior,” by Brewer et al., Journal of the American Medical Association.) Use a 0.01 significance level to test the claim that the proportion of smokers who tried to quit in the text warning group is less than the proportion in the picture warning group.


a. Test the claim using a hypothesis test.

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Step 1: Define the null and alternative hypotheses. The null hypothesis (H₀) states that the proportion of smokers who tried to quit in the text warning group is greater than or equal to the proportion in the picture warning group: H₀: p₁ ≥ p₂. The alternative hypothesis (H₁) states that the proportion of smokers who tried to quit in the text warning group is less than the proportion in the picture warning group: H₁: p₁ < p₂.
Step 2: Identify the sample proportions and sample sizes. For the text warning group, the sample size is n₁ = 1078 and the number of successes (smokers who tried to quit) is x₁ = 366. The sample proportion is calculated as p̂₁ = x₁ / n₁. For the picture warning group, the sample size is n₂ = 1071 and the number of successes is x₂ = 428. The sample proportion is calculated as p̂₂ = x₂ / n₂.
Step 3: Calculate the pooled proportion. The pooled proportion (p̂) is used when comparing two proportions and is calculated as p̂ = (x₁ + x₂) / (n₁ + n₂). This represents the overall proportion of successes across both groups.
Step 4: Compute the test statistic. The test statistic for comparing two proportions is given by the formula: z = (p̂₁ - p̂₂) / √[p̂(1 - p̂)(1/n₁ + 1/n₂)]. Substitute the values of p̂₁, p̂₂, p̂, n₁, and n₂ into the formula to calculate the z-score.
Step 5: Determine the critical value and make a decision. Using a significance level of α = 0.01, find the critical value for a one-tailed test from the standard normal distribution table. Compare the calculated z-score to the critical value. If z < critical value, reject the null hypothesis (H₀). Otherwise, fail to reject the null hypothesis.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about population parameters based on sample data. It involves formulating a null hypothesis (H0) and an alternative hypothesis (H1), then using sample data to determine whether to reject H0. In this case, the null hypothesis would state that the proportion of smokers trying to quit is the same for both warning types, while the alternative hypothesis would suggest that the proportion is lower for the text warning group.
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Step 1: Write Hypotheses

Significance Level

The significance level, often denoted as alpha (α), is the threshold for determining whether the results of a hypothesis test are statistically significant. In this scenario, a significance level of 0.01 indicates that there is a 1% risk of rejecting the null hypothesis when it is actually true. This means that if the p-value obtained from the test is less than 0.01, the evidence is strong enough to support the claim that the text warning group has a lower proportion of smokers trying to quit.
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Step 4: State Conclusion Example 4

Proportions and Sample Size

Proportions represent the fraction of a whole, often expressed as a percentage, and are crucial in comparing groups in statistical studies. In this case, the proportions of smokers trying to quit in both groups (text and picture warnings) are calculated from their respective sample sizes. Understanding how to compute and compare these proportions is essential for conducting the hypothesis test and interpreting the results accurately.
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Sampling Distribution of Sample Proportion
Related Practice
Textbook Question

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)


Queues Listed on the next page are waiting times (seconds) of observed cars at a Delaware inspection station. The data from two waiting lines are real observations, and the data from the single waiting line are modeled from those real observations. These data are from Data Set 30 “Queues” in Appendix B. The data were collected by the author.


a. Use a 0.01 significance level to test the claim that cars in two queues have a mean waiting time equal to that of cars in a single queue.


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Textbook Question

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)


Bicycle Commuting A researcher used two different bicycles to commute to work. One bicycle was steel and weighed 30.0 lb; the other was carbon and weighed 20.9 lb. The commuting times (minutes) were recorded with the results shown below (based on data from “Bicycle Weights and Commuting Time,” by Jeremy Groves, British Medical Journal).


a. Use a 0.05 significance level to test the claim that the mean commuting time with the heavier bicycle is the same as the mean commuting time with the lighter bicycle.


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Textbook Question

Second-Hand Smoke Samples from Data Set 15 “Passive and Active Smoke” include cotinine levels measured in a group of smokers ( n = 40, x_bar = 172.48 ng/mL, 119.50 ng/mL ) and a group of nonsmokers not exposed to tobacco smoke ( n = 40, x_bar = 16.35 ng/mL, 62.53 ng/mL ). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced.


a. Use a 0.05 significance level to test the claim that the variation of cotinine in smokers is greater than the variation of cotinine in nonsmokers not exposed to tobacco smoke.

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Textbook Question

In Exercises 5–16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.


Do Men Talk Less than Women? Listed below are word counts of males and females in couple relationships (from Data Set 14 “Word Counts” in Appendix B).


a. Use a 0.05 significance level to test the claim that men talk less than women.


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Textbook Question

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)


Readability of Font On a Computer Screen The statistics shown below were obtained from a standard test of readability of fonts on a computer screen (based on data from “Reading on the Computer Screen: Does Font Type Have Effects on Web Text Readability?” by Ali et al., International Education Studies, Vol. 6, No. 3). Reading speed and accuracy were combined into a readability performance score (x), where a higher score represents better font readability.


a. Use a 0.05 significance level to test the claim that there is no significant difference in readability between Roman and Arial fonts.


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Textbook Question

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)


Magnet Treatment of Pain People spend around \$5 billion annually for the purchase of magnets used to treat a wide variety of pains. Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using the visual analog scale, and the results given below are among the results obtained in the study (based on data from “Bipolar Permanent Magnets for the Treatment of Chronic Lower Back Pain: A Pilot Study,” by Collacott, Zimmerman, White, and Rindone, Journal of the American Medical Association, Vol. 283, No. 10). Higher scores correspond to greater pain levels.


a. Use a 0.05 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment (similar to a placebo).


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