QualityAssociatesQuality Associates, Inc., a consulting firm, advises its clientsabout sampling and statistical procedures that can be used tocontrol their manufacturing processes. In one particularapplication, a client gave Quality Associates a sample of 800observations taken during a time in which that client's processwas operating satisfactorily. The sample standard deviation forthese data was .21; hence, with so much data, the populationstandard deviation was assumed to be .21. Quality Associatesthen suggested that random samples of size 30 be takenperiodically to monitor the process on an ongoing basis. Byanalyzing the new samples, the client could quickly learnwhether the process was operating satisfactorily. When theprocess was not operating satisfactorily, corrective action couldbe taken to eliminate the problem. The design specificationindicated the mean for the process should be 12. The hypothesistest suggested by Quality Associates follows.Ho: μ = 12Ha: μ # 12Corrective action will be take any time H0 is rejected.The following samples were collected at hourly intervalsduring the first day of operation of the new statistical processcontrol procedure. Sample 1 Sample 2 Sample 3 Sample 411.55 11.62 11.9112.0211.62 11.69 11.3612.0211.52 11.59 11.7512.0511.75 11.82 11.9512.1811.9011.97 12.1412.1111.64 11.71 11.7212.0711.80 11.87 11.61 12.0512.03 12.1011.8511.6411.94 12.01 12.16 12.3911.9212.13 12.20 12.1211.99 11.91 11.6512.1112.0912.1611.61 11.9011.93 12.00 12.21 12.2212.21 12.28 11.5611.8812.3211.93 12.00 12.0112.39 11.95 12.0312.3511.85 11.92 12.06 12.09011.76 11.83 11.76 11.7712.16 12.23 11.82 12.2011.77 11.84 12.12 11.7912.00 12.07 11.60 12.3012.04 12.11 11.95 12.2711.98 12.05 11.96 12.2912.30 12.37 12.22 12.4712.18 12.25 11.75 12.0311.97 12.04 11.96 12.1712.17 12.24 11.95 11.9411.85 11.92 11.89 11.9712.30 12.37 11.99 12.2312.15 12.22 11.93 12.25Managerial Report:1) Compute the standard deviation for each of the four samples.Does the assumption of .21 for the population standarddeviation appear reasonable?2) Compute limits for the sample mean around u-12 such that,as long as a new sample mean is within those limits, the processwill be considered to be operating satisfactorily. If sample meanexceeds the upper limit or if sample mean is below the lowerlimit, corrective action will be taken. These limits are referredto as upper and lower control limits for quality controlpurposes.3) Discuss the implications of changing the level of significanceto a larger value. What mistakes or error could increase if thelevel of significance in increased?

Given the data asSample 1Sample 2Sample 3Sample…

Answered: Quality Associates, Inc., a consulting…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question 5 A common claim is that garlic lowers cholesterol levels. In a test of the effectiveness of garlic, 45 subjects were treated with doses of raw garlic, and their cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/mL) have a mean of 4.94 mg/mL and a standard deviation of 18.51 mg/mL. Construct a 95% confidence interval for the average net change in LDL cholesterol after the garlic treatment. i. Procedure: Select an answer Select an answer One proportion Z procedure One mean T procedure One mean Z procedure One variance x² procedure ii. Assumptions: (select everything that applies) Simple random sample Sample size is greater than 30 Population standard deviation is unknown Population standard deviation is known The number of positive and negative responses ar Normal population iii. Unknown parameter: Select an answer Select an answer u, population mean o², population variance p, population proportion
Question 03: Arbitron Media Research Inc. conducted a study of the iPod listening habits of men and women. One facet of the study involved the mean listening time. It was discovered that the mean listening time for a sample of 10 men was 35 minutes per day. The standard deviation was 10 minutes per day. The mean listening time for a sample of 12 women was also 35 minutes, but the standard deviation of the sample was 12 minutes. At the .10 significance level, can we conclude that there is a difference in the variation in the listening times for men and women?
Part C. Compute the sample mean and sample deviation.
Jessica Salas, President of Salas Products, is reviewing the warranty policy for her company's new model of automobile batteries. Life tests performed on a sample of 100 batteries indicated: (1) an average life of 75 months, (2) a standard deviation of 5 months, and (3) a bell-shaped battery life distribution. What percentage of the batteries will fail within the first 65 months of use? 5% 1% 0.5% 7.5% 2.5%
QUESTION 1 A hardware store determined the current balances for a random sample of 100 customers with non-zero balances. The sample mean of the amount owed was $100.00 and the standard deviation was $25. Use the assumption that the sample standard deviation is equal to the population standard deviation. You are asked to find a confidence interval for the average amount owed by all customers with a non-zero balance. Which Minitab procedure would be most appropriate to use for this task? O Power and Sample Size/1-Sample Z O Power and Sample Size/Sample Size for Estimation/1-Sample Z O Basic Statistics/1-Sample Z O Basic Statistics/1-Sample t
If the actual deviation rate in the sample is greater than the expected deviation rate: The achieved upper deviation rate is less than the tolerable rate. The achieved upper deviation rate is greater than the tolerable rate. The sample result supports reliance on the control. The sample size was too large.
Question 3: The waiting times (in minutes) of a random sample of 22 people at a bank have a sample standard deviation of 3.6 minutes. Assume the sample is from a normally distributed population. Use a 98% level of confidence to construct the confidence intervals for (a) the population variance and (b) the population standard deviation. Interpret the results.
An instructor has decided to introduce a greater component of independent study into an intermediate microeconomics course as a way of motivating students to work independently and think more carefully about the course material. A colleague cautions that a possible consequence may be increased variability in student performance. However, the instructor responds that she would expect less variability. From her records she found that in the past, student scores on the final exam for this course followed a normal distribution with standard deviation 18.2 points. For a class of 25 students using the new approach, the standard deviation of scores on the final exam was 15.3 points. Assuming that these 25 students can be viewed as a random sample of all those who might be subjected to the new approach, test the null hypothesis that the population standard deviation is at least 18.2 points against the alternative that it is lower.
QUESTION 11 Provide an appropriate response. A local bank needs information concerning the savings account balances of its customers. A random sample of 15 accounts was checked. The mean balance was $686.75 with a standard deviation of $256.20. Find a 98% confidence interval for the true mean. Assume that the account balances are normally distributed. Round to the nearest cent. ($487.31, $563.80) ($513.17, $860.33) O ($326.21, $437.90) O($238.23, $326.41)
Estimating income tax based on dividends is also not straightforward, due in part to Australia’s franking credit scheme. Mike is looking to simplify total annual tax projections for a share portfolio based simply on the share portfolio's current value. To do this, he wishes to determine if there is any statistically significant linear relationship between the two. Mike has randomly sampled 30 individuals with share portfolios over AUD$100,000 and mapped the assessed value of the portfolio (in AUD $1000) to the total taxes paid on that amount (in AUD$1). Mike's data is in the table below. Assessed Value Taxes 426.1 5014 368.8 3698 309.8 3691 373.9 5189 445.7 4455 505.1 5189 342.3 3416 343.6 3931 432.3 4393 329.2 3267 291.2 2871 399.3 4013 392.7 4728 482.2 4885 480.6 4888 390.8 3917 371.6 3738 311.9 3098 343.7 3633 304.7 3538 274.6 3427 457.1 4633 231.5 4009 450.5 4565 510.1 5946 308.4 3045 348.8 3504 410.4 3375…
Question 3 A. The management of White Industries is considering a new method of assembling its golf cart. The present method requires a mean time of 42.3 minutes to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster? B. Do married and unmarried women spend the same amount of time per week using Facebook? A random sample of 45 married women who use Facebook spent an average of 3.0 hours per week on this social media website. A random sample of 39 unmarried women who regularly use Facebook spent an average of 3.4 hours per week. Assume that the weekly Facebook time for married women has a population standard deviation of 1.2 hours, and the population standard deviation for unmarried Facebook users is 1.1 hours per week. Using the .05…
1. Active management of labor (AML) is a group of interventions designed to help reduce the length of labor and the rate of cesarean deliveries. Physicians from the Department of Obstetrics and Gynecology at the University of New Mexico Health Sciences Center were interested in determining whether AML would also translate into reduced cost of delivery. According to their study, 500 AML deliveries had a sample mean cost of $2,480 dollars with a sample standard deviation of $766. The average cost of all AML deliveries is to be estimated with 90% confidence interval. The ? − or ? −value that should be used to construct this interval is 1.96 2.33 1.645 1.28 The margin of error is equal to 67.14 79.82 43.85 56.35 The confidence interval is (2423.65, 2536.35) (2421.86, 2547.14) (2409.18, 2559.82) (2445.15, 2523.85)

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