Process time at a workstation is monitored using sample mean and range control charts. Six samples of n= 8 observations have been obtained and the sample means and rangescomputed (in minutes):Sample123456MeanRangeMean3.063.273.163.203.303.21a. Using the factors in Table 10-3, determine the upper and lower limits for sample mean and range control charts. (Round the intermediate calculations to 2 decimal places.Round the final mean answers to 2 decimal places. Round the final range answers to 3 decimal places.)Range0.430.440.440.430.510.40UCLLCL4

the sample size (n)= 8number of samples =6from given data,the overall range the overall range the…

Answered: Process time at a workstation is…

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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Which of the following samples will produce the largest value for a t statistic? Solect one: a. D = 10 with SS 40 b. D= 5 with SS = 20 O c. D= 10 with SS = 20 d. D = 5 with SS = 40
The next five questions refer to the following scenario: • sample mean T = 10 • sample size n = 100 • population standard deviation O = 4
Assuming that the data are normally distributed, a. Determine the z-score such that the area to its right is 18%. z= b. Determine the z-score such that the area to its left is 91%. z=
For a given frequency distribution data table, x= 6, Exf =1200 and Ef = 25 then the sample standard deviation equals: Select one: a. 11.5 b. 2.65 с. 3.54 d. 10.5
For a normal population, if a sample of size 25, sample mean3 2.3 and SD= 9.8. Then The 95% one- sided CI lower bound for the population mean is: a. -1.5416 O b. -0.9144 O C. -1.7454 O d. -1.0536
Consider a sample with data values of 57,45,50,39,50,59,60, and 48. Compute the following values a. Mode b. 40th percentile c. Variance d. Standard deviation e. Coefficient of variation
Determine the point of estimate of the population mean and margin of error from the given information: Lower Bound: 5 Upper Bound: 23 Question 4 options: x¯= 15 E = 10 x¯= 12 E = 9 x¯= 13.5 E = 9 x¯= 14 E = 9
Given Samples A and B below Sample A: 2.0 Sample B: 4.3 2.8 4.5 O a. Calculate the mean and standard deviation for each sample. Sample A: XA = Sample B: XB = ; SB = Round to two decimal places if necessary b. Calculate the coefficient of variation for each sample. Sample A: CVA = ; SA = _% 4.2 4.2 4.3 4.0 Sample B: CVB Round to two decimal places if necessary c. Which sample is more variable? o Sample A Sample B Neither sample is more variable than the other 4.4 4.3 3.7 4.9 _% 3.6 2.6 2.8 2.2 2.3 4.1
Select the single best answer: An investigator is interested in the cigarette use of workers in factory. She asks all the workers who come into the site if they use cigarettes and determines that 20% of the subjects smoke. She then asks every third smoker who comes to the site how many cigarettes they smoke each day; she finds that the mean number of cigarettes smoked is 16. She plots the number of cigarettes smoked by each worker on a frequency distribution and finds that it is normally distributed. She also finds that the number of male smokers is equal to the number of female smokers. She already knows that half of her subjects are men, and half are women. A particular worker, Mr. X, smokes 24 cigarettes a day. What is the corresponding deviation score? A. 40 B. +8 C. 0 D. -8 E. 24
Consider the population with the values {2.1, 3.3, 7.6, 8.5), list all possible samples of size n = 2 WITH REPLACEMENT and find: d. population mean e. population variance f. population standard deviation. Please write your solutions in a legible manner and draw a box around your final answers.
The following data show the values of sample mean X and the range. R for the samples of size 5 each. Calculate the values for central line and control limits for mean-chart and range chart and determine whether the process is in control. Sample No 2 5 7 8 10 Mean (X) I1.2 11.8 10.8 11.6 11.0 96 10.4 9.6 10.6 10.0 Range (R) 4 7 4 8 4 7 (Conversion factors for n= 5 are 4, = 0.577, D; = 0 and D. = 2.115)

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