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1 Applied Statistics Questions 1. Wall Street securities firms recently paid out record year-end average bonuses of $141,300 per employee. Suppose we drew a sample n = 64 employees at the Jones & Ryan securities firm to test whether their mean year-end bonus is different from the reported mean of $141,300 for the population, and found that the sample mean is $138,000. Assume s = $18,000. (a) At a level of a = .01, please use the six-step hypothesis-testing framework we employed in class, and write out the last three steps in this table. 1. 𝐻𝑜: μ = 141,300 2. 𝐻𝑎: μ ¹ 141,300 3. n = 64, α = .01 4. Using the test statistic Z, we reject the null hypothesis if: 5. Collect data, calculate test statistic, Z: 6. Conclusion: (b) What is the p-value associated with the above sample result? _______________ (2 pts) 2. The U.S. government released a report stating that 55% of adults do not exercise regularly. A Los Angeles-based researcher decided to investigate to see whether this claim was true for California. When she carried out the study, she found that 321 of 535 adults reported that they did not exercise regularly. (a) At a level of a = .10, please use the six-step hypothesis-testing framework we employed in class, and write out the last three steps in this table. (6 pts) 1. 𝐻𝑜: p = .55 2. 𝐻𝑎: p ¹.55 3. n = 535, α = .10 4. Using the test statistic Z, we reject the null hypothesis if: 5. Collect data, calculate test statistic, Z: (b) What is the p-value associated with the above sample result? 2 _______________ 3. Speaking to a group of analysts, a brokerage firm executive recently claimed that at least 75% of investors are currently confident of meeting their investment objectives. A UBS Investor Optimism Survey, conducted during the same period, and involving 346 investors, found that 242 were confident of meeting their investment objectives. (a) At a level of a = .02, please use the six-step hypothesis-testing framework we employed in class, and write out the last three steps in this table. 1. 𝐻𝑜: p ³ .75 2. 𝐻𝑎: p < .75 3. n = 346, α = .02 4. Using the test statistic Z, we reject the null hypothesis if: 5. Collect data, calculate test statistic, Z: 6. Conclusion: (b) What is the p-value associated with the above sample result? _______________ 4. A German manufacturer of industrial machine tools (high-tech, high value-added) purchases inputs and raw materials from suppliers across the EU. One particular component must meet very precise standards in terms of tensile strength. Accordingly, the quality control manager evaluates the quality of each incoming shipment by selecting a sample of n = 81 items for “stress testing.” (The sample mean “strength” is calculated as an estimate of the population mean, μ.) Assume that design specifications require components from the supplier to have a mean “strength” exceeding 1,400. Since the manufacturer is concerned only with whether the strength exceeds 1,400 (but not by how much), an upper-tail test is appropriate. Moreover, because the supplier-in-question has recently had some quality control problems themselves, the null hypothesis assumes they still are having these issues. Accordingly, we are looking for evidence supporting the alternative hypothesis before we accept the most recent shipment. During the most recent inspection, a simple random sample of n = 81 was selected, and the average “strength” was determined to be 1,412. Assume that the standard deviation of the population is  = 105. (a) At the significance level of .05, test to determine whether the most recent 3 shipment of components should be accepted as meeting the design specification. Please use the six-step hypothesis-testing framework we employed in class, and write out the last three steps in this table. 1. 𝐻𝑜: μ ≤ 1,400 2. 𝐻𝑎: μ > 1,400 3. n = 81, α = .05 4. Using the test statistic Z, we reject the null hypothesis if: 5. Collect data, calculate test statistic, Z: 6. Conclusion: (b) What is the p-value associated with the above sample result? _______________ 5. Consumption of alcoholic beverages by young women of drinking age has been increasing in the UK, the US, and Europe. Data (annual consumption in liters) consistent with the findings reported in a recent Wall Street Journal article are shown for a sample of 10 European women: 103, 73, 124, 110, 171, 110, 47, 109, 0, and 160. (a) Please provide a 95% confidence interval estimate of the annual population alcohol consumption (in liters) among European women: ________________ (b) Please provide a 90% confidence interval estimate of the annual population alcohol consumption (in liters) among European women: __________ 6. When a human resources consulting firm conducted a survey of 1,060 employees at medium-sized and large companies to determine how dissatisfied employees were with their jobs, they found that 673 reported that they experienced strong dissatisfaction with their jobs. (a) Please provide a 99% confidence interval estimate of the proportion of the population of employees who strongly dislike their current work experience. __________ (b) Please provide a 95% confidence interval estimate of the proportion of the population of employees who strongly dislike their current work experience. __________ 4 7. The travel-to-work time for residents of the world’s large cities has been studied and reported in many publications. Suppose that a preliminary simple random sample of residents of a certain metropolitan area in North America is used to develop a planning value of 12.75 minutes for the population standard deviation. (a) If we want to estimate the population mean travel-to-work time for this metropolitan area with a margin of error of 3 minutes, what sample size would you recommend? Please assume 99% confidence. __________ (3 pts) (b) If we want to estimate the population mean travel-to-work time for this metropolitan area with a margin of error of 2 minute, what sample size would you recommend? Please assume 90% confidence. __________ 8. A study is being conducted to determine the percentage of people (18 years and older) who smoke, and a preliminary estimate of 0.33 is being used. (a) How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .03? Please use 95% confidence. __________ (b) Assuming this study uses the sample size you recommend in (a), and finds 273 smokers, please develop a 99% confidence interval estimate for the proportion of smokers in the population. Use the new point estimate, not .33. __________ 9. Interpret the results below, and answer the following questions. Suppose we regress the dependent variable y on 4 independent variables x1 , x2 , x3 , and x4 . After running this regression on twenty-five observations we have the following information from the ANOVA table: Regression Sum of Squares is 26,264,566; Residual Sum of Squares is 15,114,983. (a) What is the coefficient of determination? _______________ (b) What is the correlation between the actual dependent variable and the predicted dependent variable? _______________ (c) What is the adjusted coefficient of determination? _______________ (d) Test the overall significance of the regression model at the  = .05 level. Please use the six-step hypothesis-testing framework we employed in class, and write out the last three steps in this table. 1. 𝐻𝑜: r = 0 2 5 2. 𝐻𝑎: r 0 3. n = 25, α = .05 4. Using the test statistic F, we reject the null hypothesis if: 5. Collect data, calculate test statistic, F: 6. Conclusion: (e) The printout results also provides the following information about the partial regression coefficients: Independent variables Unstandardized coefficients Standard error x1 63.73 51.96 x2 0.045 0.012 x3 0.26 0.074 x4 52.41 215.96 Test the significance of each of the four independent variables, one at a time, at the  = .05 level. Please use the six-step hypothesis-testing framework we employed in class, and write out the last three steps in each table. 1. 𝐻𝑜: = 0 2. 𝐻𝑎: 0 3. n = 25, α = .05 4. Using the test statistic t, we reject the null hypothesis if: 5. Collect data, calculate test statistic, t: 6. Conclusion: 1. 𝐻𝑜: = 0 2. 𝐻𝑎: 0 3. n = 25, α = .05 4. Using the test statistic t, we reject the null hypothesis if: 5. Collect data, calculate test statistic, t: 6. Conclusion: 2   1  1   2  2  6 1. 𝐻𝑜: = 0 2. 𝐻𝑎: 0 3. n = 25, α = .05 4. Using the test statistic t, we reject the null hypothesis if: 5. Collect data, calculate test statistic, t: 6. Conclusion: 1. 𝐻𝑜: = 0 2. 𝐻𝑎: 0 3. n = 25, α = .05 4. Using the test statistic t, we reject the null hypothesis if: 5. Collect data, calculate test statistic, t: 6. Conclusion: 10. A federal funding program is available to low-income neighborhoods. To qualify for the funding, a neighborhood must have a mean household income of less than $15,000 per year; neighborhoods with mean annual household income of $15,000 or more do not qualify. Funding decisions are based on a sample of residents in the neighborhood. A hypothesis test with a .02 level of significance is conducted. If the funding guidelines call for a maximum probability of .05 of not funding a neighborhood with a mean annual household income of $14,000, what sample size should be used in the funding decision study? Use σ = $4000 as a planning value. (Note: this is #75, p. 430, Ch. 9) ____________ 11. An automobile fuel efficiency study tested the following hypothesis, where mpg is miles per gallon of gasoline. Hypothesis Conclusion 𝐻𝑜: μ ≥ 25 mpg Manufacturer’s claim is supported; average mileage is indeed what is claimed  3  3   4  4  7 𝐻𝑎: μ < 25 mpg Manufacturer’s claim rejected; average mileage less than stated For σ = 3 and a .02 significance, what sample size would you recommend if you want an 80% chance of detecting that μ is less than 25 mpg when it is actually 24 mpg? ____________ 12. Consider the following hypothesis test. 𝐻𝑜: μ ≤ 2,400 𝐻𝑎: μ > 2,400 A simple random sample of n = 50 items will be selected and the population standard deviation is  = 200. Use a = .05. Compute the probability of making a Type II error if the actual population mean is: (a) m = 2,425. ____________ (b) m = 2,450. ____________ (c) m = 2,475. ____________ (d) What is the “power of the test” when the actual mean is 2,475? That is, what is the probability of correctly rejecting the null hypothesis when it is false and (moreover) when m = 2,475. ____________ 13. Recall from our discussion in class that food and beverage companies seeking to hire “tasters” who have above average taste-discrimination ability can apply the hypothesis-testing framework. Suppose a brewer intends to use the triangle taste test method to identify the best applicants for the position of “taster.” (It does so because the company will want their tasters to have sensitive palates, and it will need someway of determining beforehand whether a candidate for a taster’s job can detect subtle differences in taste.) Remember how this works: in a single trial of this test, the applicant is presented with 3 samples of beer---two of which are alike---and is asked to identify the odd sample. Except for the taste difference, the samples are as alike as possible (same color, same temperature, same cup, and so on). To check the applicant’s ability, he/she is presented with a series of triangle tests. The order of the presentation is randomized within each trial. Clearly, in the absence of any ability at all to distinguish tastes, the probability the applicant will correctly identify the odd sample in a single trial is one-third (i.e., 1/3). The question---and the question the brewing company wants to answer---is whether the applicant can do better than this. More specifically, if an applicant with no ability to distinguish tastes were presented with 15 trials, we would expect that he/she would correctly identify the odd sample 5 times, simply by luck 8 alone. On the other hand, if an applicant with a sensitive palate is presented with 15 trials, we wouldn’t be surprised to see him/her correctly identify the odd sample more frequently than this, and perhaps much more frequently. Suppose our null hypothesis is that a certain applicant has no ability at all to discern differences in beer samples, and that to her one beer tastes pretty much the same as another. Obviously, the alternative hypothesis states that the applicant does have taste discrimination ability. Our job is to present this person with a series of triangle taste tests with the purpose of collecting data (the number of correct identifications made in a series of trials) which help the brewing company classify the job applicant into one of the two groups qualified or unqualified. If the company administers n=15 identical triangle taste tests to this job applicant and if we say that ‘x’ is the number of correct identifications made (in n=15 trials), then the Rejection Region is the “set of values which ‘x’ could assume that will lead us to reject the null hypothesis, and prefer the alternative hypothesis.” We could choose any Rejection Region we like, but suppose the company decides it should be: 7, 8, 9, 10, 11, 12, 13, 14, or 15. That is, if after an applicant is presented with n=15 triangular taste tests (or 15 trials), she correctly identifies the odd sample at least 7 times, we reject the null hypothesis (that the applicant has no taste sensitivity) and prefer the alternative hypothesis (that the applicant has taste discrimination ability), and we make her an offer of employment as a taster. (a) With a Rejection Region of {7, 8, 9, 10, 11, 12, 13, 14, or 15}, what is the probability of a Type I error? _______________ (b) With a Rejection Region of {7, 8, 9, 10, 11, 12, 13, 14, or 15}, what is the probability of a Type II error, if the job applicant has a probability of identifying the odd sample with p = 0.7? _______________ (c) With a Rejection Region of {9, 10, 11, 12, 13, 14, or 15} what is the probability of a Type I error? _______________ (d) With a Rejection Region of {9, 10, 11, 12, 13, 14, or 15}, what is the probability of a Type II error, if the job applicant has a probability of identifying the odd sample with p = 0.7? __________ Binomial probabilities n = 15 and p = 1/3 Binomial probabilities n = 15 and p = .70 p(x=0|n=15, p= 1/3) = 0.0023 p(x=0|n=15, p= .70) = 0.0000 p(x=1|n=15, p = 1/3) = 0.0171 p(x=1|n=15, p = .70)= 0.0000 p(x=2|n=15, p = 1/3) = 0.0599 p(x=2|n=15, p = .70)= 0.0000 p(x=3|n=15, p = 1/3) = 0.1299 p(x=3|n=15, p = .70)= 0.0000 p(x=4|n=15, p = 1/3) = 0.1948 p(x=4|n=15, p = .70)= 0.0006 p(x=5|n=15, p = 1/3) = 0.2143 p(x=5|n=15, p = .70)= 0.0030 p(x=6|n=15, p = 1/3) = 0.1786 p(x=6|n=15, p = .70)= 0.0116 p(x=7|n=15, p = 1/3) = 0.1148 p(x=7|n=15, p = .70)= 0.0348 9 p(x=8|n=15, p = 1/3) = 0.0574 p(x=8|n=15, p = .70)= 0.0811 p(x=9|n=15, p = 1/3) = 0.0223 p(x=9|n=15, p = .70)= 0.1472 p(x=10|n=15, p = 1/3)= 0.0067 p(x=10|n=15, p = .70)= 0.2061 p(x=11|n=15, p = 1/3)= 0.0015 p(x=11|n=15, p = .70)= 0.2186 p(x=12|n=15, p = 1/3)= 0.0003 p(x=12|n=15, p = .70)= 0.1700 p(x=13|n=15, p = 1/3)= 0.0000 p(x=13|n=15, p = .70)= 0.0916 p(x=14|n=15, p = 1/3)= 0.0000 p(x=14|n=15, p = .70)= 0.0305 p(x=15|n=15, p = 1/3)= 0.0000 p(x=15|n=15, p = .70)= 0.0047 14. A quality control inspector is always on the lookout for substandard parts and components provided to her manufacturing company by outside suppliers. Because most shipments contain some defective items, each must be subjected to inspection. Naturally, some shipments contain more defectives than others, and it is the job of the inspector to identify the most defective-laden shipments so that they may be returned to the supplier. Suppose the inspector selects a sample of n=72 items from a given shipment for testing. Unbeknownst to the inspector, this particular shipment includes 9% defective components. If the policy is to return any shipment with at least 5% defectives, what is the probability that this particular shipment (with 9% defectives) will be accepted as good anyway?

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