Expand Your Knowledge: Confidence Intervals for μd Using techniques from Section 7.2, we can find a confidence interval for μd. Consider a random sample of n matched data pairs A, B. Let d = B – A be a random variable representing the difference between the values in a matched data pair. Compute the sample mean d of the differences and the sample standard deviation sd. If d has a
where
c = confidence level (0 < c < 1)
tc = critical value for confidence level c and d.f. = n – 1
- (a) Using the data of Problem 9, find a 95% confidence interval for the mean difference between percentage increase in company revenue and percentage increase in CEO salary.
- (b) Use the confidence interval method of hypothesis testing outlined in Problem 25 of Section 8.2 to test the hypothesis that population mean percentage increase in company revenue is different from that of CEO salary. Use a 5% level of significance.
9. Business: CEO Raises Are America’s top chief executive officers (CEOs) really worth all that money? One way to answer this question is to look at row B, the annual company percentage increase in revenue, versus row A, the CEO’s annual percentage salary increase in that same company (Source: Forbes, Vol. 159, No. 10). A random sample of companies such as John Deere & Co., General Electric, and Dow Chemical yielded the following data:
Do these data indicate that the population mean percentage increase in corporate revenue (row B) is different from the population mean percentage increase in CEO salary? Use a 5% level of significance.
25. Expand Your Knowledge: Confidence Intervals and Two-Tailed Hypothesis Tests Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let c be the level of confidence used to construct a confidence interval from sample data. Let a be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean.
For a two-tailed hypothesis test with level of significance α and null hypothesis H0: µ = k, we reject H0 Whenever k falls outside the c = 1 – α confidence interval for µ based on the sample data When k falls within the c = 1 – α confidence interval, we do not reject H0.
(A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as p, μ1 – μ2, and p1 – p2, which we will study in Sections 8.3 and 8.5.) Whenever the value of k given in the null hypothesis falls outside the c = 1 – α confidence interval for the parameter, we reject H0. For example, consider a two-tailed hypothesis test with α = 0.01 and
H0: µ = 20 H1: µ ≠ 20
A random sample of size 36 has a sample mean
- (a) What is the value of c = 1 – α? Using the methods of Chapter 7, construct a 1 – α confidence interval for μ from the sample data. What is the value of m given in the null hypothesis (i.e., what is k)? Is this value in the confidence interval? Do we reject or fail to reject H0 based on this information?
- (b) Using methods of this chapter, find the P-value for the hypothesis test. Do we reject or fail to reject H0? Compare your result to that of part (a).
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Chapter 8 Solutions
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- Question 1 The data shown in Table 1 are and R values for 24 samples of size n = 5 taken from a process producing bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals recorded (i.e., 34.5 should be 0.50345). Table 1: Bearing Diameter Data Sample Number I R Sample Number I R 1 34.5 3 13 35.4 8 2 34.2 4 14 34.0 6 3 31.6 4 15 37.1 5 4 31.5 4 16 34.9 7 5 35.0 5 17 33.5 4 6 34.1 6 18 31.7 3 7 32.6 4 19 34.0 8 8 33.8 3 20 35.1 9 34.8 7 21 33.7 2 10 33.6 8 22 32.8 1 11 31.9 3 23 33.5 3 12 38.6 9 24 34.2 2 (a) Set up and R charts on this process. Does the process seem to be in statistical control? If necessary, revise the trial control limits. [15 pts] (b) If specifications on this diameter are 0.5030±0.0010, find the percentage of nonconforming bearings pro- duced by this process. Assume that diameter is normally distributed. [10 pts] 1arrow_forward4. (5 pts) Conduct a chi-square contingency test (test of independence) to assess whether there is an association between the behavior of the elderly person (did not stop to talk, did stop to talk) and their likelihood of falling. Below, please state your null and alternative hypotheses, calculate your expected values and write them in the table, compute the test statistic, test the null by comparing your test statistic to the critical value in Table A (p. 713-714) of your textbook and/or estimating the P-value, and provide your conclusions in written form. Make sure to show your work. Did not stop walking to talk Stopped walking to talk Suffered a fall 12 11 Totals 23 Did not suffer a fall | 2 Totals 35 37 14 46 60 Tarrow_forwardQuestion 2 Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty samples of five parts each are collected, and the compressive strengths (in psi) are shown in Table 2. Table 2: Strength Data for Question 2 Sample Number x1 x2 23 x4 x5 R 1 83.0 2 88.6 78.3 78.8 3 85.7 75.8 84.3 81.2 78.7 75.7 77.0 71.0 84.2 81.0 79.1 7.3 80.2 17.6 75.2 80.4 10.4 4 80.8 74.4 82.5 74.1 75.7 77.5 8.4 5 83.4 78.4 82.6 78.2 78.9 80.3 5.2 File Preview 6 75.3 79.9 87.3 89.7 81.8 82.8 14.5 7 74.5 78.0 80.8 73.4 79.7 77.3 7.4 8 79.2 84.4 81.5 86.0 74.5 81.1 11.4 9 80.5 86.2 76.2 64.1 80.2 81.4 9.9 10 75.7 75.2 71.1 82.1 74.3 75.7 10.9 11 80.0 81.5 78.4 73.8 78.1 78.4 7.7 12 80.6 81.8 79.3 73.8 81.7 79.4 8.0 13 82.7 81.3 79.1 82.0 79.5 80.9 3.6 14 79.2 74.9 78.6 77.7 75.3 77.1 4.3 15 85.5 82.1 82.8 73.4 71.7 79.1 13.8 16 78.8 79.6 80.2 79.1 80.8 79.7 2.0 17 82.1 78.2 18 84.5 76.9 75.5 83.5 81.2 19 79.0 77.8 20 84.5 73.1 78.2 82.1 79.2 81.1 7.6 81.2 84.4 81.6 80.8…arrow_forward
- Name: Lab Time: Quiz 7 & 8 (Take Home) - due Wednesday, Feb. 26 Contingency Analysis (Ch. 9) In lab 5, part 3, you will create a mosaic plot and conducted a chi-square contingency test to evaluate whether elderly patients who did not stop walking to talk (vs. those who did stop) were more likely to suffer a fall in the next six months. I have tabulated the data below. Answer the questions below. Please show your calculations on this or a separate sheet. Did not stop walking to talk Stopped walking to talk Totals Suffered a fall Did not suffer a fall Totals 12 11 23 2 35 37 14 14 46 60 Quiz 7: 1. (2 pts) Compute the odds of falling for each group. Compute the odds ratio for those who did not stop walking vs. those who did stop walking. Interpret your result verbally.arrow_forwardSolve please and thank you!arrow_forward7. In a 2011 article, M. Radelet and G. Pierce reported a logistic prediction equation for the death penalty verdicts in North Carolina. Let Y denote whether a subject convicted of murder received the death penalty (1=yes), for the defendant's race h (h1, black; h = 2, white), victim's race i (i = 1, black; i = 2, white), and number of additional factors j (j = 0, 1, 2). For the model logit[P(Y = 1)] = a + ß₁₂ + By + B²², they reported = -5.26, D â BD = 0, BD = 0.17, BY = 0, BY = 0.91, B = 0, B = 2.02, B = 3.98. (a) Estimate the probability of receiving the death penalty for the group most likely to receive it. [4 pts] (b) If, instead, parameters used constraints 3D = BY = 35 = 0, report the esti- mates. [3 pts] h (c) If, instead, parameters used constraints Σ₁ = Σ₁ BY = Σ; B = 0, report the estimates. [3 pts] Hint the probabilities, odds and odds ratios do not change with constraints.arrow_forward
- Solve please and thank you!arrow_forwardSolve please and thank you!arrow_forwardQuestion 1:We want to evaluate the impact on the monetary economy for a company of two types of strategy (competitive strategy, cooperative strategy) adopted by buyers.Competitive strategy: strategy characterized by firm behavior aimed at obtaining concessions from the buyer.Cooperative strategy: a strategy based on a problem-solving negotiating attitude, with a high level of trust and cooperation.A random sample of 17 buyers took part in a negotiation experiment in which 9 buyers adopted the competitive strategy, and the other 8 the cooperative strategy. The savings obtained for each group of buyers are presented in the pdf that i sent: For this problem, we assume that the samples are random and come from two normal populations of unknown but equal variances.According to the theory, the average saving of buyers adopting a competitive strategy will be lower than that of buyers adopting a cooperative strategy.a) Specify the population identifications and the hypotheses H0 and H1…arrow_forward
- You assume that the annual incomes for certain workers are normal with a mean of $28,500 and a standard deviation of $2,400. What’s the chance that a randomly selected employee makes more than $30,000?What’s the chance that 36 randomly selected employees make more than $30,000, on average?arrow_forwardWhat’s the chance that a fair coin comes up heads more than 60 times when you toss it 100 times?arrow_forwardSuppose that you have a normal population of quiz scores with mean 40 and standard deviation 10. Select a random sample of 40. What’s the chance that the mean of the quiz scores won’t exceed 45?Select one individual from the population. What’s the chance that his/her quiz score won’t exceed 45?arrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
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