MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- please answer a, b, carrow_forwardhe Computer Anxiety Rating Scale (CARS) measures an individual’s level of computer anxiety, on a scale from 20 (no anxiety) to 100 (highest level of anxiety). Researchers at Miami University administered CARS to 172 business students. One of the objectives of the study was to determine whether there are differences in the amount of computer anxiety experienced by students with different majors. They found the following: Major n Mean Marketing 19 44.37 Management 11 43.18 Other 14 42.21 Finance 45 41.80 Accountancy 36 37.56 MIS 47 32.21 (A) Complte the following ANOVA table: Source of Variation SS df MS F-ratio F-crit Between groups 3172 5 634.4 4.9567 2.2685 Within groups 21246 166 127.988 Total 24418 171 (A) At the 0.05 level of significance, is there evidence of a difference in the mean computer anxiety…arrow_forwardA consumer group wanted to determine if there was a difference in customer perceptions about prices for a specific type of toy depending on where the toy was purchased. In the local area there are three main retailers: W-Mart, Tag, and URToy. For each retailer, the consumer group randomly selected 5 customers, and asked them to rate how expensive they thought the toy was on a 1-to-10 scale (1= not expensive, to 10 = very expensive). The toy was priced the same at all retail stores. Compute the percentage of variance explained by the group differences for these data. Q: Percentage and variance explained = ?arrow_forward
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- Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1) and the annual credit card charges for dining out (population 2) Using the difference data, the sample mean difference was d = $850, and the sample standard deviation was så = $1,123. a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. Ho : μα ♦ Select your answer - Ha d + - Select your answer - b. Use a = = 0.05 level of significance. Can you conclude that the population means differ? - Select your answer - ◆ What is the p-value? (to 6 decimals) c.…arrow_forwardBank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment. Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1) and the annual credit card charges for dining out (population 2). Using the difference data, the sample mean difference was d = $850, and the sample standard deviation was Sd = $1,123. a. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out. Hoμd Haμd - Select your answer - - Select your answer - b. Use a 0.05 level of significance. Can you conclude that the population means differ? - Select your answer - ✓ What is the p-value? (to 6 decimals) c. Which…arrow_forwardTwenty-nine college students, identified as having a positive attitude about Mitt Romney as compared to Barack Obama in the 2012 presidential election, were asked to rate how trustworthy the face of Mitt Romney appeared, as represented in their mental image of Mitt Romney’s face. Ratings were on a scale of 0 to 7, with 0 being “not at all trustworthy” and 7 being “extremely trustworthy.” Here are the 29 ratings: 2.6 3.2 3.7 3.3 3.4 3.6 3.7 3.8 3.9 4.1 4.2 4.9 5.7 4.2 3.9 3.2 4.5 5.0 5.0 4.6 4.6 3.9 3.9 5.3 2.8 2.6 3.0 3.3 3.7 a 95% confidence interval for the mean rating. Is there significant evidence at the 5% level that the mean rating is greater than 3.5 (a neutral rating)?arrow_forward
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