Many firms use on-the-job training to teach their employees computer programming. Suppose you work in the personnel department of a firm that just finished training a group of its employees to program, and you have been requested to review the performance of one of the trainees on the final test that was given to all trainees. The mean and standard deviation of the test scores are 77 and 2 respectively, and the distribution of scores is bell-shaped and symmetric. Suppose the trainee in question received a score of 74. Compute the trainee's z-score. O A. z = 0.94 O B. z=-1.50 O C. z=1.50 O D. z= -0.94 Click to select your answer. Copyright 2021 Pearson Education Inc. All rights reserved. Terms of Use Privacy Policy Permissions Contact U-

Many firms use on-the-job training to teach their employees computer programming. Suppose you work in the personnel department of a firm that just finished training a group of its employees to program, and you have been requested to review the performance of one of the trainees on the final test that was given to all trainees. The mean and standard deviation of the test scores are 77 and 2 respectively, and the distribution of scores is bell-shaped and symmetric. Suppose the trainee in question received a score of 74. Compute the trainee's z-score. O A. z = 0.94 O B. z=-1.50 O C. z=1.50 O D. z= -0.94 Click to select your answer. Copyright 2021 Pearson Education Inc. All rights reserved. Terms of Use Privacy Policy Permissions Contact U-

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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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You hypothesize that people in stats classes are happier than people in all other classes. You compare happiness scores in three of your classes: Statistics, Developmental Psychology, and Social Psychology. In Statistics there are 15 students and the mean happiness rating is 23.1 with a standard deviation of 9.78. In Developmental Psychology there are 15 students and the mean happiness rating is 24.4 with a standard deviation of 2.92. In Social Psychology there are 15 students with a mean rating of 17.7 and standard deviation of 8.46. The sum of squares between groups is equal to 384. The sum of squares within groups is equal to 2461. Use the One-Way ANOVA calculation table to help calculate. What is the degrees of freedom within samples?
Lucas and Alex began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. Lucas took a test in Math and earned a 73.4, and Alex took a test in English and earned a 70.5. Use the fact that all the students' test grades in the Math class had a mean of 74.1 and a standard deviation of 11.3, and all the students' test grades in English had a mean of 60.9 and a standard deviation of 8.1 to answer the following questions.a) Calculate the z-score for Lucas's test grade.z=z=b) Calculate the z-score for Alex's test grade.z=z=
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the degrees of freedom within samples?
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7.  For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square between samples?
1.     In 1955, Life Magazine reported that a 25-year-old mother of 3 children, worked an average of 80 hours each week. Recently, some researchers have been studying whether the women’s movement has resulted in an increase in the work week hours for women (combining employment and at-home work). A study was done to find out if the mean work week has, in fact, increased. A random sample of 81 women were surveyed and the mean was 83 hours with a standard deviation of 10. Using a 95% confidence level, does the sample data support the researcher’s claim that the average number of weekly hours that women work, is greater than 80? a.     What statistical test will you use to test this data ? b.     State the null and alternative hypothesis. i.     H0: ii.     Ha: c.     What are the confidence intervals? d.     What is the value of the test statistic? e.     What is the p-value? f.      Circle/highlight the correct decision about the null hypothesis. Reject H0…
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 150 students in evening classes and finds that they have a mean test score of 88.8. He knows the population standard deviation for the evening classes to be 8.4 points. A random sample of 250 students from morning classes results in a mean test score of 89.9. He knows the population standard deviation for the morning classes to be 5.4 points. Test his claim with a 99% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 3 of 3 :   Draw a conclusion and interpret the decision.
April and Heather began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. April took a test in Science and earned a 74.4, and Heather took a test in English and earned a 69.7. Use the fact that all the students' test grades in the Science class had a mean of 71.3 and a standard deviation of 8.9, and all the students' test grades in English had a mean of 60.5 and a standard deviation of 8.5 to answer the following questions. a) Calculate the z-score for April's test grade. b) Calculate the z-score for Heather's test grade. c) Which person did relatively better? April Heather O They did equally well. Submit Question logi $22 *3 F4 F8 7. 8. %4
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You are comparing data from two similar experiments that have a mean of 53. Experiment A has a standard deviation of 3.4 and experiment B has a standard deviation of 1.5. What does this tell you about the data in the experiments?
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
The state Government gives grants to low-income undergraduate students. According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, the average Pell grant award for 2007-2008 was $2,600. We wonder if the mean amount is different this year for Pell grant recipients at San Jose State University. Suppose that we randomly select 50 Pell grant recipients from San Jose State University. For these 50 students, the mean Pell grant award is $2,450 with a standard deviation of $600. Let µ = the mean amount of Pell grant awards received by San Jose State University Pell grant recipients this year. We test the hypotheses: H0: μ = 2,600 , Ha: μ ≠ 2,600. The sample size is greater than 30, so a t-model is a good fit for the sampling distribution. What is the t-test statistic? What is the P-value?
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the mean square within samples?