An Introduction to Statistical Methods and Data Analysis
7th Edition
ISBN: 9781305269477
Author: R. Lyman Ott, Micheal T. Longnecker
Publisher: Cengage Learning
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The Occupational Safety and Health Administration (OSHA) mandates certain regulations that have to be adopted by corporations. Prior to the implementation of the OSHA program, a company found that for a sample of 40 randomly selected months, the mean employee time lost due to job-related accidents was 45 hours. After implementation of the OSHA program, for a random sample of 45 months, the mean employee time lost due to job-related accidents was 39 hours. It can be assumed that the variability of time lost due to accidents is about the same before and after implementation of the OSHA program (with a standard deviation being 3.5 hours).
Find a 90% confidence interval for the difference in the mean time lost due to accidents.
Test the hypothesis that implementation of the OSHA program has reduced the mean employee lost time. Use a level of significance of 0.10.
In a study of the effectiveness of a new device that supports a weak or damaged heart, 150 people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or the experimental treatment that consisted of drugs and surgery to install the new device. After two years, 60% of the 67 patients who received the device had improved and 40% of the patients who received the standard treatment had improved. Does this data provide convincing evidence that the proportion of patients who recieved the experimental treatment who improved is higher than for the patients who received the standard treatment? Test the relevant hypotheses using a significance level of 0.05. (Use
pexperimental − pstandard
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H0: pe − ps = 0, HA: pe − ps ≠ 0
H0: pe − ps = 0, HA: pe − ps < 0
H0: pe − ps = 0, HA: pe − ps > 0
B) Find the test statistic.z = C) What is the p-value?P-Value = D) State your conclusion in terms of…
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Chapter 7 Solutions
An Introduction to Statistical Methods and Data Analysis
Ch. 7.7 - A leading researcher in the study of interstate...Ch. 7.7 - Baseballs vary somewhat in their rebounding...Ch. 7.7 - Random samples of sizes n1 = 25 and n2 = 20 were...Ch. 7.7 - A soft-drink firm is evaluating an investment in a...Ch. 7.7 - A wildlife biologist was interested in determining...Ch. 7.7 - Prob. 30SE
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- A low-level CDC bureaucrat wants to please his boss by gathering evidence thatthe current government-mandated shutdown of society is not causing people’s mentalhealth to deteriorate, so that it can safely be continued for several years if anyexpert says it’s necessary.He polls a random sample of 1600 citizens, gathering data on such items asincome loss, weight gain, access to toilet paper, hours spent binge-watchingNetflix, and number of injuries caused by household fights, and compiles all thisinto a scientifically-weighted “misery index”.The mean misery index from the sample is 99.2; it seems reasonable to use apopulation standard deviation σ = 19.1.a) Does this information provide significant evidence (at the 5% level) that thenationwide mean misery index is less than 100? Set up appropriate null andalternative hypotheses, calculate the appropriate test statistic, find the P-value,and state your conclusion. (10)b) A CDC press release publishing the results of this study claims that…arrow_forwardA low-level CDC bureaucrat wants to please his boss by gathering evidence thatthe current government-mandated shutdown of society is not causing people’s mentalhealth to deteriorate, so that it can safely be continued for several years if anyexpert says it’s necessary.He polls a random sample of 1600 citizens, gathering data on such items asincome loss, weight gain, access to toilet paper, hours spent binge-watchingNetflix, and number of injuries caused by household fights, and compiles all thisinto a scientifically-weighted “misery index”.The mean misery index from the sample is 99.2; it seems reasonable to use apopulation standard deviation σ = 19.1.a) Does this information provide significant evidence (at the 5% level) that thenationwide mean misery index is less than 100? Set up appropriate null andalternative hypotheses, calculate the appropriate test statistic, find the P-value,and state your conclusion. b) A CDC press release publishing the results of this study claims that…arrow_forwardA snack food manufacturer estimates that the variance of the number of grams of carbohydrates in servings of its tortilla chips is 1.25. A dietician is asked to test this claim and finds that a random sample of 22 servings has a variance of 1.35. At ⍶ = 0.05 is there enough evidence to reject the manufacturer’s claim?arrow_forward
- A juice factory buys apple concentrate from two different suppliers. The mean concentration is the same for both suppliers, but it is suspected that the variability of the concentration could be different for the two suppliers. The standard deviation of the concentration in a random sample of n1 = 12 lots produced by Supplier 1 is s1 = 6.4 g / L, and for Supplier 2, n2 = s2 = 4.8 g / L in a random sample of 12 lots. Is there sufficient evidence to conclude that the variance of the two populations is different by calculating the appropriate test statistic? (α = 0.05)arrow_forwardA consumer advocacy group wanted to study whether different airline carriers differed in terms of their delayed flights. In particular, the researchers were interested in the relationship between p1, the proportion of Alpha Airlines flights that were delayed at least 15 minutes, and p2, the proportion of Beta Airlines flights that were delayed at least 15 minutes. A random sample of 1,000 Alpha flights and a separate random sample of 1,000 Beta flights found that 67 of the Alpha fights and 160 of the Beta flights were delayed at least 15 minutes. The conditions for inference were checked and verified. Does this set of samples provide strong evidence that Alpha Airlines has a smaller proportion of flights that are delayed at least 15 minutes than Beta Airlines, at the α = 0.05 significance level? Find the z-table here. A. The test statistic is z = –6.56 and the P-value ≈ 0. Since the P-value ≈ 0 < 0.05, there is not sufficient evidence that Alpha Airlines has fewer delayed…arrow_forwardThe one-way ANOVA partitions the total variability into into a. Sswithin and MSbetween b.Sstotal and Ssbetween c. Sswithin and ssbetween d. Sswithin and sstotalarrow_forward
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