Twenty observations on etch uniformity on silicon wafers are taken during a qualification experiment for a plasma etcher. The data are as follows: Etch Uniformity 5.34 6.65 4.76 5.98 7.25 6.00 7.55 5.54 5.62 6.21 5.97 7.35 5.44 4.39 4.98 5.25 6.35 4.61 6.00 5.32 (a) Construct a 95 percent confidence interval estimate of s2. (b) Test the hypothesis that 2= 1.0. Use a = 0.05. What are your conclusions?

Twenty observations on etch uniformity on silicon wafers are taken during a qualification experiment for a plasma etcher. The data are as follows: Etch Uniformity 5.34 6.65 4.76 5.98 7.25 6.00 7.55 5.54 5.62 6.21 5.97 7.35 5.44 4.39 4.98 5.25 6.35 4.61 6.00 5.32 (a) Construct a 95 percent confidence interval estimate of s2. (b) Test the hypothesis that 2= 1.0. Use a = 0.05. What are your conclusions?

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter8: Sequences, Series, And Probability

Section8.7: Probability

Problem 58E: What is meant by the sample space of an experiment?

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What is meant by the sample space of an experiment?
Determinations of saliva pH levels were made in two independent random samples of seventh-grade schoolchildren. Sample A children were caries-free while sample B children had a high incidence of caries. The results were as follows: A 7.14 7.11 7.61 7.98 7.21 7.16 7.89 7.24 7.86 B 7.36 7.04 7.19 7.41 7.10 7.15 7.36 7.57 7.19 Construct and interpret a 90% confidence interval for the difference between the population means.
A U.S. Food Survey showed that Americans routinely eat beef in their diet. Suppose that in a study of 49 consumers in Illinois and 64 consumers in Texas the following results were obtained from two samples regarding average yearly beef consumption: Illinois Texas = 49 = 64 = 54.1lb = 60.4lb S1 = 7.0 S2 = 8.0 Develop a 95% confidence Interval Estimate for the difference between the two population means.
During a laboratory analysis of human blood composition, tests are repeated and are known to be normally distributed. Fifteen tests on a given sample of blood yielded the following values1.003 1.01 1.03 1.045 1.0061.024 1.019 1.009 1.014 1.0211.031 1.08 1.058 1.012 1.001 Calculate the 98% confidence interval for true composition of the blood in repeated tests of the sample.b) Data on investments in the high-tech industry by venture capitalists are compiled by VentureOne Corporation. A random sample of 16 venture capital investments in the fiber optics business sector yielded the following data, in millions of dollars. 5.60 6.27 5.96 10.51 2.04 5.48 5.74 5.588.63 5.95 6.67 4.21 7.71 9.21 4.98 8.64 Determine a 99% confidence interval for the mean amount, μ, of all venture-capital investments in the fiber optics business sector. Assume that the population standard deviation is $1.98 million.
The following are interest rates for a 30-year mortgage from a sample of lenders in Atlanta, Georgia, on August 9, 2010. Assume that the population is normally distributed. 4.327 4.531 4.461 4.365 4.547 4.407 4.398 4.639 4.199 4.804 4.460 4.842 Construct a 95% confidence interval for the population standard deviation
In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.a. Suppose that a random sample of 400 television ads in the United Kingdom reveals that 142 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor.b. Suppose a random sample of 500 television ads in the United States reveals that 122 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor.c. Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor? Explain.
2. A food truck serves lunch every Tuesday and Thursday on the plaza in front of Van Metre Hall on the GMU Arlington campus. The number of customers varies every day, so it wants to order supplies based on the average number of customers served per day. The truck recorded how many customers were served on 40 randomly selected days, and obtained a sample mean and sample standard deviation of 145 and 17, respectively. Construct a 90% confidence interval for the mean number of customers served per day.
During a laboratory analysis of human blood composition, tests are repeated and are known to be normally distributed. Fifteen tests on a given sample of blood yielded the following values 1.003 1.01 1.03 1.045 1.006 1.024 1.019 1.009 1.014 1.021 1.031 1.08 1.058 1.012 1.001 Calculate the 98% confidence interval for true composition of the blood in repeated tests of the sample.
Payments In a May 2007 Experian/Gallup PersonalCredit Index poll of 1008 U.S. adults aged 18 and over,8% of respondents said they were very uncomfortablewith their ability to make their monthly payments ontheir current debt during the next three months. A moredetailed poll surveyed 1288 adults, reporting similaroverall results and also noting differences among fourage groups: 18–29, 30–49, 50–64, and 65+.a) Do you expect the 95% confidence interval for the true proportion of all 18- to 29-year-olds who are wor-ried to be wider or narrower than the 95% confidence interval for the true proportion of all U.S. consumers?Explain.b) Do you expect this second poll’s overall margin oferror to be larger or smaller than the Experian/Galluppoll’s? Explain.
The drug Viagra became available in the U.S. in May 1988, in the wake of an advertising campaign that was unprecedented in scope and intensity. A Gallup poll found that by the end of the first week of May, 643 out of 1,005 adults were aware that Viagra was an impotency medication. a) Is the sample size large enough to compute a confidence interval for the proportion of adults who were aware that Viagra was an impotency medication after the first week of May? Yes, because more than 10 adults were not aware of Viagra Yes, because the sample size is greater than 30 Yes, because more than 10 adults were aware of Viagra Yes, because both np̂ and n(1-p̂) are greater than 10 b)Using the information from questions 1-3, suppose that p̂ is 0.40 and the standard error of p̂ is 0.10. Compute a 90% confidence interval for the proportion of adults that were aware of Viagra. (0.2355, 0.5645) (0.3, 0.5) (0.2355, 0.4) (0.3, 0.4)
A clinic offers a weight-loss program. A review of its records found the following amounts of weight loss, in pounds, for a random sample of 10 clients at the conclusion of the program:18.2 25.9 6.3 11.8 15.4 20.3 16.8 18.5 12.3 17.2Find a 90% confidence interval for the population variance of weight loss for clients of this weight-loss program.
The following data were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol. New Drug (n=75) Placebo (n=75) Total Sample (n=150) Mean (SD) Total Serum Cholesterol 182.0 (24.5) 206.3 (21.8) 194.15 (23.2) % Patients with Total Cholesterol < 200 78.0% 65.0% 71.5% Generate the 95% confidence interval for the difference in mean total cholesterol levels between treatments Generate a 95% confidence interval for the difference in proportions of all patients with total cholesterol < 200. How many patients would be required to detect the difference in proportions observed in the current study with 80% power. A new investigation will be enrolling subjects to demonstrate the efficacy of the drug in individuals with high cholesterol at an early age. A two-sided test is planned at α = 0.05