Multiple Choice13.5 + 2.026 × 13.9613.5 + 2.021 × 5.7813.5 + 2.026 × 5.7813.5 + 2.021 × 13.9 A random sample of 40 companies with assets over $10 million was surveyed and asked to indicate their industry and annualcomputer technology expense. The ANOVA comparing the average computer technology expense among three industriesrejected the null hypothesis. The mean square error (MSE) was 195. The following table summarized the results:EducationTax ServicesFood ServicesNumber Sampled101416Mean Expense (1,000,000s)215.520When comparing the mean annual computer technology expense for companies in the tax services and education industries, whichof the following 95% confidence interval can be constructed?

Solution Based on the comparison between the mean computer technology expense for companies in the…

Answered: A random sample of 40 companies with…

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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A professor wants to know which type of instructional mode students had more satisfaction. The professor randomly selected n = 5 students from each of the following types of classes: online only, hybrid, and face-to face only. Student were given a satisfaction survey. Are their differences in student satisfaction among the different instructional modes. The data are below: Online Hybrid Face-to-Face 3 3 4 3 3 1 3 1 3 2 2 2 2 1 3 Calculate Mean Square Within Group (MSWG)
The Management Department recently purchased a smallcopier for faculty use. Although the workload of the office staff has improved somewhat, the secretaries are still mak-ing too many trips to the Dean’s office when the depart-mental copier is out of service. Sylvia, the departmental secretary, has been keeping track of failure rates and ser-vice times. With a mean time between failures of 100 hours and a mean time to repair of 24 hours, how much of thetime is the new copier available for faculty use?
A veterinarian wants to perform a study at a local animal shelter where 200 dogs need training on behavior modification. The purpose of the study would be to determine if using a firm voice during training for 10 minutes improves a dog's learning behavior compared to not speaking during training for 10 minutes. Part A: Describe an appropriate design for the study. Part B: The hypotheses for this study are as follows: Ho: There is no difference in the mean improvement of dogs during behavior modification for either treatment. Ha: The mean improvement of dogs during behavior modification is greater for the firm voice treatment. The shelters will allow trainers to use a firm voice during training if the null hypothesis is rejected. What are the possible Type I and II errors? Describe the consequences of each in the context of this study and discuss which type you think is more serious.n
2. Interpret this data.
Students in an experimental psychology class did research on depression as a sign of stress. A test was administered to a sample of 30 students.The scores are shown below. 43 50 10 92 77 35 64 36 42 72 53 62 35 75 50 D 72 36 28 38 61 48 63 35 41 21 36 50 47 86 14 To find the 10% trimmed mean of a data set, order the data, delete the lowest 10% of the entries and the highest 10% of the entries, and find the mean of the remaining entries. Complete parts (a) through (c). (a) Find the 10% trimmed mean for the data The 10% trimmed mean is- (Round to the nearest tenth as needed.) (b) Compare the four measures of central tendency, including the midrange. Mean = (Round to the nearest tenth as needed.) Median = Mode = (Use a comma to separate answers as needed.) The midrange is found by the following formula. (Maximum data entry) + (Minimum data entry) 2 Midrange = (Round to the nearest tenth as needed.) (c) What is the benefit of using a trimmed mean versus using a mean found using all data…
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(3) The table contains hypothetical data from a study that a bank conducted on the credit card balances of male and female credit card holders (credit cards owned by both men and women were left out of the study). Male Female Mean $3,550 $2,670 Median $2,175 $2,500 Std Dev $1,240 $825 a. What conclusions can we come to by comparing the medians of the males and females? b. What conclusions can we come to by comparing the standard deviations of the males and females? c. Notice that the mean and median of the males are fairly far apart while the mean and median of the females are not. What does this tell us about the credit card balances of the males vs the credit card balances of the females?
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atients suffering from rheumatic diseases or osteoporosis often suffer critical losses in bone mineral density (BMD). Alendronate is one medication prescribed to build or prevent further loss of BMD. Holcomb and Rothenberg (A-3) looked at 96 women taking alendronate to determine if a difference existed in the mean percent change in BMD among five different primary diagnosis classifications. Group 1 patients were diagnosed with rheumatoid arthritis (RA). Group 2 patients were a mixed collection of patients with diseases including lupus, Wegener’s granulomatosis and polyarteritis, and other vasculitic diseases (LUPUS). Group 3 patients had polymyalgia rheumatica or temporal arthritis (PMRTA). Group 4 patients had osteoarthritis (OA) and group 5 patients had osteoporosis (O) with no other rheumatic diseases identified in the medical record. Changes in BMD are shown in the following table. Discuss the Alendronate medication data in terms of variability. To determine if a difference…

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