Statistics for Business and Economics (13th Edition)
13th Edition
ISBN: 9780134506593
Author: James T. McClave, P. George Benson, Terry Sincich
Publisher: PEARSON
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Chapter 8.5, Problem 8.64LM
Enough money has been budgeted to collect independent random
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Chapter 8 Solutions
Statistics for Business and Economics (13th Edition)
Ch. 8.2 - The purpose of this exercise is to compare the...Ch. 8.2 - Prob. 8.2LMCh. 8.2 - In order to compare the means of two populations,...Ch. 8.2 - Prob. 8.4LMCh. 8.2 - Prob. 8.5LMCh. 8.2 - Prob. 8.6LMCh. 8.2 - Independent random samples from normal populations...Ch. 8.2 - Two independent random samples nave been...Ch. 8.2 - Independent random samples of n1 = 233 and n2 =...Ch. 8.2 - Independent random samples from approximately...
Ch. 8.2 - Independent random samples selected from two...Ch. 8.2 - Lobster trap placement. Refer to the Bulletin of...Ch. 8.2 - Last name and acquisition timing. The speed with...Ch. 8.2 - Shared leadership in airplane crews. Human Factors...Ch. 8.2 - Performance-based logistics. Refer to the Journal...Ch. 8.2 - Drug content assessment. Refer to Exercise 4.123...Ch. 8.2 - Buy-side vs. sell-side analysts earnings...Ch. 8.2 - Homework assistance for accounting students. How...Ch. 8.2 - Comparing taste-test rating protocols....Ch. 8.2 - Producer willingness to supply biomass. The...Ch. 8.2 - Does rudeness really matter in the workplace?...Ch. 8.2 - Prob. 8.22ACICh. 8.2 - Is honey a cough remedy? Refer to the Archives of...Ch. 8.2 - Gender diversity of board of directors. The gender...Ch. 8.2 - Prob. 8.25ACACh. 8.3 - A paired difference experiment produced the...Ch. 8.3 - A paired difference experiment yielded nd pairs of...Ch. 8.3 - The data for a random sample of six paired...Ch. 8.3 - The data for a random sample of 10 paired...Ch. 8.3 - A paired difference experiment yielded the...Ch. 8.3 - Summer weight-loss camp. Camp Jump Start is an...Ch. 8.3 - Performance ratings of government agencies. The...Ch. 8.3 - Twinned drill holes. A traditional method of...Ch. 8.3 - Prob. 8.34ACBCh. 8.3 - I am not selling anything surveys. To improve...Ch. 8.3 - Consumers attitudes toward advertising. The two...Ch. 8.3 - Taking power naps during work breaks. Lack of...Ch. 8.3 - Acidity of mouthwash. Acid has been found to be a...Ch. 8.3 - Solar energy generation along highways. The...Ch. 8.3 - Impact of red light cameras on car crashes. To...Ch. 8.3 - Prob. 8.41ACICh. 8.3 - Alcoholic fermentation in wines. Determining...Ch. 8.4 - Consider making an inference about p1 p2, where...Ch. 8.4 - For each of the following values of , find the...Ch. 8.4 - Prob. 8.45LMCh. 8.4 - Prob. 8.46LMCh. 8.4 - Prob. 8.47LMCh. 8.4 - Prob. 8.48LMCh. 8.4 - The winners curse in auction bidding. In auction...Ch. 8.4 - Prob. 8.50ACBCh. 8.4 - Prob. 8.51ACBCh. 8.4 - Web survey response rates. Response rates to Web...Ch. 8.4 - Hospital administration of malaria patients. One...Ch. 8.4 - Traffic sign maintenance. The Federal Highway...Ch. 8.4 - Salmonella in produce. Salmonella infection is the...Ch. 8.4 - Angioplastys benefits challenged. More than 1...Ch. 8.4 - Entrepreneurial careers of MBA alumni. Are African...Ch. 8.4 - Predicting software defects. Refer to the PROMISE...Ch. 8.4 - Prob. 8.59ACICh. 8.4 - Religious symbolism in TV commercials. Gonzaga...Ch. 8.5 - Assuming that n1 = n2, find the sample sizes...Ch. 8.5 - Find the appropriate values of n1 and n2 (assume...Ch. 8.5 - Suppose you want to estimate the difference...Ch. 8.5 - Enough money has been budgeted to collect...Ch. 8.5 - Last name and acquisition timing. Refer to the...Ch. 8.5 - Homework assistance for accounting students. Refer...Ch. 8.5 - Vulnerability of relying party Web sites. Refer to...Ch. 8.5 - Conducting a political poll. A pollster wants to...Ch. 8.5 - Shared leadership in airplane crews. Refer to the...Ch. 8.5 - Solar energy generation along highways. Refer to...Ch. 8.5 - Angioplastys benefits challenged. Refer to the...Ch. 8.5 - Traffic sign maintenance. Refer to the Journal of...Ch. 8.6 - Use Tables V, VI, VII, and VIII in Appendix D to...Ch. 8.6 - Given v1 and v2, find the following probabilities:...Ch. 8.6 - For each of the following cases, identify the...Ch. 8.6 - For each of the following cases, identify the...Ch. 8.6 - Specify the appropriate rejection region for...Ch. 8.6 - Independent random samples were selected from each...Ch. 8.6 - Independent random samples were selected from each...Ch. 8.6 - Lobster trap placement. Refer to the Bulletin of...Ch. 8.6 - Prob. 8.81ACBCh. 8.6 - Mental health of workers and the unemployed. A...Ch. 8.6 - Drug content assessment Refer to Exercise 8.16 (p....Ch. 8.6 - Last name and acquisition timing. Refer to the...Ch. 8.6 - Prob. 8.85ACICh. 8.6 - Oil content of fried sweet potato chips. Refer to...Ch. 8.6 - Shopping vehicle and judgment. Refer to the...Ch. 8.6 - Is honey a cough remedy? Refer to the Archives of...Ch. 8 - List the assumptions necessary for each of the...Ch. 8 - Two independent random samples were selected from...Ch. 8 - Independent random samples were selected from two...Ch. 8 - Independent random samples were selected from two...Ch. 8 - Two independent random samples are taken from two...Ch. 8 - A random sample of five pairs of observations were...Ch. 8 - Prob. 8.95ACBCh. 8 - Prob. 8.96ACBCh. 8 - Prob. 8.97ACBCh. 8 - Hull failures of oil tankers. Refer to the Marine...Ch. 8 - Effectiveness of teaching software. The U.S....Ch. 8 - Durability of shock absorbers. A manufacturer of...Ch. 8 - NHTSA new car crash tests. Refer to the National...Ch. 8 - Diamonds sold at retail. Refer to the data for 308...Ch. 8 - Childrens recall of TV ads. A study examined...Ch. 8 - Is steak your favorite barbeque food? July is...Ch. 8 - Planning habits survey. American Demographics...Ch. 8 - Turnover rates in the United States and Japan....Ch. 8 - Smartphone usage differs by gender. The role of...Ch. 8 - Life expectancies of working women and housewives....Ch. 8 - Comparing purchasers and nonpurchasers of...Ch. 8 - Prob. 8.110ACICh. 8 - Racial profiling by the LAPD. Racial profiling is...Ch. 8 - State SAT scores. Refer to Exercise 2.27 (p. 60)...Ch. 8 - Rat damage to sugarcane fields. Poisons are used...Ch. 8 - Environmental impact study. Some power plants are...Ch. 8 - Instrument precision. When new instruments are...Ch. 8 - Cooling method for gas turbines. Refer to the...Ch. 8 - Average housing space per person. Even though...Ch. 8 - Positive spillover effects from self-managed work...Ch. 8 - Impact of gender on advertising. How does gender...Ch. 8 - Salaries of postgraduates. Refer to the Economics...Ch. 8 - Gambling in public high schools. With the rapid...Ch. 8 - CareerBank.com annual salary survey....Ch. 8 - Facility layout study. Facility layout and...
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- URGENTTarrow_forwardA sample of size n = 20 is drawn from a normal population. Find the critical value tα/2 needed to construct a 99% confidence interval.arrow_forwardA simple random sample of size n=18 is drawn from a population that is normally distributed. The sample mean is found to be x=62 and the sample standard deviation is found to be s=19 Construct a 90% confidence interval about the population mean.arrow_forward
- A random sample of 52 measurements was selected from a population with σ=3.6. The sample mean was 139.3. Calculate a 95% confidence interval for μarrow_forwardWhat is the value of tα/2 for a sample of size n=39 at the 90% level of confidence?arrow_forwardConstruct a 99 % confidence interval for the difference between the mean lifetimes of two kinds of light bulbs, given that a random sample of 40 light bulbs of the first kind lasted on the average 41 hours of continuous use and 50 light bulbs of the second kind lasted on the average 40. hours of continuous. use. The population standard deviations are known to be o1 = 26 and o2 = 22. Solution For a = 0.06, we find from Table III that z0.03 = 94% confidence interval for µi – 1.88. Therefore, the uz is 262 (418 - 402) – 1.88 · V 40. 222 < H1 - 42 50 262 222 + 40 < (418 – 402) +1.88 · 50arrow_forward
- Bob selects independent random samples from wo populations and obtains the values p1 0.700 and p2 = 0.500. He constructs the 95% confidence interval for p1 – P2 and gets: 0.200 + 1.96(0.048) = 0.200 ± 0.094. Note that 0.048 is called the estimated standard error of p1 – p2 (the ESE of the estimate). Tom wants to estimate the mean of the success ates: Pi + P2 2 (a) Calculate Tom's point estimate. (b) Given that the estimated standard er- ror of (pi + P2)/2 is 0.024, calculate the 95% confidence interval estimate of (pi + P2)/2. Hint: The answer has our usual form: Pt. est. ±1.96 × ESE of the estimate. Carl selects one random sample from a popu- ation and calculates three confidence intervals for p. His intervals are below. A C p+0.080 p±0.040 p± 0.072 Match each confidence interval to its level, with evels chosen from: 80%, 90%, 95%, 98%, and 99%. Note: Clearly, two of these levels will not be used. You do not need to explain your reasoningarrow_forwardA random sample of size 35, taken from a normal population with a variance of 5, has a mean of 80. A second random sample of size 36, taken from a different normal population with a variance of 3, has a mean of 75. Find a 95% confidence interval for μ1 − μ2.arrow_forwardA random sample of size n = 40 is drawn from a population with the 9) (unknow population mean µ. Suppose the sample mean and the sample variance from this sample are given, respectively, as = 23 and s? = 10.24. Find a 98 percent confidence interval for the population mean µ.arrow_forward
- To study the water PH level, two independent lakes were selected for sampling. In a duration of one year, 12 samples (one in each month) are selected from each lake. The PH mean and standard deviation of lake 1 sample were found to be 6.5 and 0.2, while the PH mean and standard deviation of lake 2 sample were found to be 7.2 and 0.1. (a) Find a 90% confidence interval for µ1 – µ2 (i.e., the difference between the population means for the two lakes) assuming that the populations are approximately normally distributed with equal variances. (b) You decide to change the above significance level from 0.1 to 0.05. Everything else is kept similar to above. Explain how your calculated confidence interval will change, and why? (i.e., will it become wider or narrower). (No need to calculate the new confidence interval) (- :arrow_forwardIndependent random samples from two normally distributed populations give the following results:nx = 10 x̄ = 480 sx = 30ny = 12 ȳ = 520 sy = 25a. If we assume that the unknown population variances are equal, find the 90% confidence interval for the difference of population means.b. If we do not assume that the unknown population variances are equal, find the 90% confidence interval for the difference between population means.arrow_forwardshow complete solutionarrow_forward
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