What is the difference between commuting patterns for studentsand professors. 11 students and 14 professors took part in a studyto find mean commuting distances. The mean number of milestraveled by students was 5.6 and the standard deviation was 2.8.The mean number of miles traveled by professors was 14.3 and thestandard deviation was 9.1. Construct a 95% confidence intervalfor the difference between the means. What assumption have wemade?

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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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Kira recently switched her primary doctor to one specializing in caring for elderly patients. On her new doctor's website, it says that the mean systolic blood pressure among elderly females is 120 millimeters of mercury (mmHg). Kira believes the value is actually higher. She bases her belief on a recently reported study of 11 randomly selected, elderly females. The sample mean systolic blood pressure was 124mmHg, and the sample standard deviation was 24mmHg . Assume that the systolic blood pressures of elderly females are approximately normally distributed. Based on the study, at the 0.10 level of significance, can it be concluded that μ , the population mean systolic blood pressure among elderly females, is greater than 120 mmHg ? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the…
Do shoppers at the mall spend the same amount of money on average the day after Thanksgiving compared to the day after Christmas? The 56 randomly surveyed shoppers on the day after Thanksgiving spent an average of $145. Their standard deviation was $46. The 43 randomly surveyed shoppers on the day after Christmas spent an average of $123. Their standard deviation was $43. What can be concluded at the a = 0.05 level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: Select an answer v Select an answer Select an answer v H: Select an answer v Select an answer v Select an answer v b. The test statistic ? v = (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is ? v a e. Based on this, we should Select an answer v the null hypothesis. f. Thus, the final conclusion is that ... O The results are statistically significant at a = 0.05, so there is…
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?
One year Terry had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 2.98. Also, Karla had the lowest ERA of any female pitcher at the school with an ERA of 3.37. For the males, the mean ERA was 4.456 and the standard deviation was 0.557. For the females, the mean ERA was 4.166 and the standard deviation was 0.635. Find their respective z-scores. Which player had the better year relative to their peers, Terry or Karla? (Note: In general, the lower the ERA, the better the pitcher.) Terry had an ERA with a z-score of Karla had an ERA with a z-score of (Round to two decimal places as needed.) Which player had a better year in comparison with their peers? OA. Karla had a better year because of a lower z-score. B. Terry had a better year because of a lower z-score. OC. Terry had a better year because of a higher z-score. OD. Karla had a better year because of a higher z-score.
Two classes were given identical quizzes. Class A had a mean score of 7.5 and a standard deviation of 0.8 Class B had a mean score of 8.7 and a standard deviation of 0.4 Which class scored better on average? | Select an answer v Which class had more consistent scores? Select an answer ♥
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?
A random sample of 30 words from Jane Austen's Pride and Prejudice had a mean length of 4.08 letters with a standard deviation of 2.40. A random sample of 30 words from Henry James's What Maisie Knew had a mean length of 3.85 letters with a standard deviation of 2.26. Which of the following is a correct expression for the 99% confidence interval for the difference in mean word length for these two novels? 2.40, 2.26 2.40, 2.26 (a) (4.08– 3.85)±2.756| (b) (4.08 – 3.85) ±2.045 30 30 30 30 2.402 2.26? |2.40? 2.26? (c) (4.08 – 3.85)± 2.045| 30 (d) (4.08 – 3.85)±2.045 30 30 30 |2.40² + 30 2.262 (e) (4.08 – 3.85)±2.756| 30
Data of the weights of 81 pet dogs was collected. The sample mean was 30. The standard deviation is 8. Construct a 99% confidence interval to estimate the mean weight of the dogs.
It seems these days that college graduates who are employed full-time work more than 40-hour weeks. Data are available that can help us decide if this is true. A survey was recently sent to a group of adults selected at random. There were 15 respondents who were college graduates employed full-time. The mean number of hours worked per week by these 15 respondents was 43 hours, with a standard deviation of 8 hours. Assume that the population of hours worked per week by college graduates employed full-time is normally distributed with mean μ. Can we conclude that μ is greater than 40 hours? Use the 0.05 level of significance. State the null hypothesis H0 and the alternative hypothesis H1 Find the t-test df & Solve the t-test value Find the critical value Can we conclude, at the 0.05 level of significance, that the mean number of hours worked per week by college graduates is greater than 40 hours? Yes or No?
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?
You would like to estimate the variance and standard deviation for credit scores based on FICO. These scores are thought to be normally distributed. You randomly select 36 credit scores to get a standard deviation of 56.5. Using a 95% confidence interval, estimate the variance and standard deviation of credit scores.
Say you have 50 samples, each with 15 puppies. All have been rated in regards to cuteness. You are asked by the kennel association to calculate the Standard Error of the Mean for them. They just couldn’t do it and they need it for accreditation purposes. However, they were able to tell you that the mean for one of the samples was 8.4, and it’s standard deviation was 1.3, while the range was 5. So, please calculate the Standard Error of Mean, which is:

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