A random sample of 25 students were asked to self-identify as "statistics-lovers" or"statistics-adorers." 16 students self-identified as "lovers" and 9 students as"adorers." Below table summarizes the mean of a statistics test and standarddeviation that was given after the self-identification. Test the following hypothesis atalpha=5% (0.05).HO: Mean Statistics Lovers = Mean Statistics AdorersHA: Mean Statistics Lovers = Mean Statistics AdorersYou can assume equal variance.Statistics StatisticsAdorersLovers16mean9092Variance7 What do you conclude using the correct statistical language?) You fail to reject the null.) You reject the null.You accept the alternative.You accept the null.

Denote μ1, μ2 as the population mean number of students who “love” statistics and who “adore”…

Answered: A random sample of 25 students were…

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 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?
Assume that an intelligence test scores are distributed normallt with a mean of 100 and a standard deviation of 15. If you scored 124 on this intelligence test, what is your z-score on this distribution.
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.
Suppose 36% of movie goers say comedy is their favorite genre. If a random sample of 120 movie goers is taken, what are the mean and standard deviation of the number of movie goers that prefer comedy?
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.
It has been found that with the standard dosage of a drug, the change in people’s uric acid levelis normally distributed with a mean of 240 and a standard deviation of 60. What proportionof people receiving the standard dosage of this drug experience changes in uric acid levelbetween 210 and 270?
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?
Research was being done on who makes more money right after college, a person with a sociology degree or a psychology degree. To test this, the researcher randomly selected 45 recent soclology graduates and found that the mean income they are recelving Is $53,900 with a standard devlation of $2,050. The researcher also randomly selected 40 psychology majors and found that the mean income they are receiving is $55,200 with a standard deviation of $3,850.Test the claim that the mean Income for psychology majors is more than the mean Income for soclology majors at the a-0.05 level of significance. Calculator Function [ Select] P-Value [ Select] Conclusion
A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test.What is the 90% confidence interval for the mean score of all students taking the test?
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 88.1. He knows the population standard deviation for the evening classes to be 4.2 points. A random sample of 150 students from morning classes results in a mean test score of 89.1. He knows the population standard deviation for the morning classes to be 7.7 points. Test his claim with a 95 % 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.
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 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1__μ2

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