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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 9.3.6a

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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Step 1: Formulate the null and alternative hypotheses. The null hypothesis (H₀) states that there is no difference in word counts between men and women, or men talk as much as women. The alternative hypothesis (H₁) states that men talk less than women. Mathematically, H₀: μ₁ = μ₂ and H₁: μ₁ < μ₂, where μ₁ is the mean word count for men and μ₂ is the mean word count for women.
Step 2: Calculate the differences between paired samples (word counts for men and women). For each pair, subtract the word count for men from the word count for women. This will give you the differences (d). For example, for the first pair: d = 21,261 - 13,560.
Step 3: Compute the mean (d̄) and standard deviation (s_d) of the differences. Use the formulas for the mean and standard deviation: d̄ = (Σd) / n and s_d = sqrt((Σ(d - d̄)²) / (n - 1)), where n is the number of pairs.
Step 4: Perform a t-test for paired samples. Calculate the test statistic t using the formula: t = (d̄ - 0) / (s_d / sqrt(n)), where d̄ is the mean of the differences, s_d is the standard deviation of the differences, and n is the number of pairs.
Step 5: Compare the calculated t-value to the critical t-value from the t-distribution table at a significance level of 0.05 and degrees of freedom (df = n - 1). If the calculated t-value is less than the critical t-value, reject the null hypothesis and conclude that men talk less than women. 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.

Paired Sample Data

Paired sample data involves two related groups where each member of one group is matched with a member of another group. In this context, the word counts of men and women in couple relationships are paired, allowing for a direct comparison of their communication levels. This method is essential for analyzing differences while controlling for individual variability.
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Hypothesis Testing

Hypothesis testing is a statistical method used to determine if there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. In this case, the null hypothesis states that men do not talk less than women, while the alternative hypothesis suggests that they do. The significance level of 0.05 indicates the threshold for determining statistical significance in the results.
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Step 1: Write Hypotheses

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. The assumption that the differences in word counts have an approximately normal distribution is crucial for applying certain statistical tests, such as the t-test, which relies on this property to validate the results.
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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)


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

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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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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Textbook Question

Overlap of Confidence Intervals In the article “On Judging the Significance of Differences by Examining the Overlap Between Confidence Intervals,” by Schenker and Gentleman (American Statistician, Vol. 55, No. 3), the authors consider sample data in this statement: “Independent simple random samples, each of size 200, have been drawn, and 112 people in the first sample have the attribute, whereas 88 people in the second sample have the attribute.”


a. Use the methods of this section to construct a 95% confidence interval estimate of the difference p1-p2. What does the result suggest about the equality of p1 and p2

.

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