Ch. 9 - Inferences from Two Samples

Triola - Elementary Statistics 14th Edition

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Triola 14th Edition

Ch. 9 - Inferences from Two Samples

Problem 9.1.14b

Chapter 9, Problem 9.1.14b

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.

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

Verified step by step guidance

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: Calculate the sample proportions for each group. For the text warning group, the sample proportion is p₁ = 366 / 1078. For the picture warning group, the sample proportion is p₂ = 428 / 1071.

Step 3: Compute the pooled proportion (p̂) using the formula: p̂ = (x₁ + x₂) / (n₁ + n₂), where x₁ and x₂ are the number of successes (smokers who tried to quit) in each group, and n₁ and n₂ are the sample sizes of each group.

Step 4: Calculate the test statistic (z) using the formula: z = (p₁ - p₂) / √[p̂(1 - p̂)(1/n₁ + 1/n₂)]. This formula incorporates the pooled proportion and accounts for the sample sizes of both groups.

Step 5: Compare the calculated z-value to the critical z-value for a one-tailed test at the 0.01 significance level. If the calculated z-value is less than the critical z-value, reject the null hypothesis. Additionally, construct a confidence interval for the difference in proportions using the formula: (p₁ - p₂) ± z*√[p₁(1 - p₁)/n₁ + p₂(1 - p₂)/n₂], where z* corresponds to the critical value for the desired confidence level.

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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. In this context, we formulate a null hypothesis (that the proportion of smokers trying to quit with text warnings is equal to that with picture warnings) and an alternative hypothesis (that the proportion with text warnings is less). We then use sample data to determine whether to reject the null hypothesis at a specified significance level.

Confidence Interval

A confidence interval is a range of values, derived from sample statistics, that is likely to contain the true population parameter with a specified level of confidence. In this case, constructing a confidence interval for the difference in proportions of smokers trying to quit between the two groups will help assess whether the observed difference is statistically significant and provides insight into the effectiveness of the warning types.

Proportion Comparison

Comparing proportions involves analyzing the relative frequencies of outcomes in different groups. Here, we compare the proportions of smokers who attempted to quit in the text warning group (366 out of 1078) and the picture warning group (428 out of 1071). This comparison is essential for understanding the impact of different warning types on smoking cessation efforts.

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Difference in Proportions: Hypothesis Tests Example 1

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)

Color and Creativity Researchers from the University of British Columbia conducted trials to investigate the effects of color on creativity. Subjects with a red background were asked to think of creative uses for a brick; other subjects with a blue background were given the same task. Responses were scored by a panel of judges and results from scores of creativity are given below. Higher scores correspond to more creativity. The researchers make the claim that “blue enhances performance on a creative task.”

b. Construct the 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)?

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Textbook Question

F Test Statistic

b. Can the F test statistic ever be a negative number?

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

Textbook Question

Better Tips by Giving Candy An experiment was conducted to determine whether giving candy to dining parties resulted in greater tips. The mean tip percentages and standard deviations are given below along with the sample sizes (based on data from “Sweetening the Till: The Use of Candy to Increase Restaurant Tipping,” by Strohmetz et al., Journal of Applied Social Psychology, Vol. 32, No. 2).

a. Use a 0.05 significance level to test the claim that giving candy does result in greater tips.

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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.

Heights of Presidents A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (cm) of presidents along with the heights of their main opponents (from Data Set 22 “Presidents” in Appendix B).

a. Use the sample data with a 0.05 significance level to test the claim that for the population of heights of presidents and their main opponents, the differences have a mean greater than 0 cm.

Textbook Question

Are Seat Belts Effective? A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts, 31 were killed. Among 7765 occupants wearing seat belts, 16 were killed (based on data from “Who Wants Airbags?” by Meyer and Finney, Chance, Vol. 18, No. 2). We want to use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities.

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