01:21:0In an effort to improve scores on college-entrance exams, the principal at a large high school chooses 35 studentsat random who have already taken the exam. The students are enrolled in a six-week prep course to improve theirscores. After the prep course, the students take the college-entrance exam again. The mean difference (first -second) in exam scores is-1.75 points with a standard deviation of 7.12 points. Assuming the conditions forinference are met, what is the test statistic for testing the hypotheses Ho: H Di=0; H HDi <0?-1.75 -07.12V 350+1.757.1235-1.75-07.12V351.75-07.12V35SubmitNextSave and ExitMark this and return

GivenHypotheses isH0:μDiff=0Ha:μDiff<0μ=0Mean(x¯)=-1.75standard deviation(s)=7.12sample…

Answered: In an effort to improve scores on…

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 small company is looking to see if they have reached their goal of $30.20, in saving per family. To see if goal was met their goal they randomly surveyed 50 of their 1000 customers, found that they saved average of $32.12 with a standard deviation of $4.83. Have they attained their saving goal.
Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
According to Kaiser Family Foundation survey in 2011 and 2010, the average premium for employer-sponsored health insurance for family coverage was $15,073 in 2011 and $13,770 in 2010 (USA TODAY, September 29, 2011). Suppose that these averages were based on random samples of 250 and 200 employees who had such employer-sponsored health insurance plans for 2011 and 2010, respectively. Further assume that the population standard deviations for 2011 and 2010 were $2160 and $1990, respectively.
To estimate the benefits of an SAT prep course, a random sample of 10 students enrolled in the course is selected. For each of these students, their entrance score on the exam taken at the beginning of the course is recorded. Their exit score on the exam they take at the end of the course is recorded as well. The table displays the scores. The mean of the differences is 193 points, and the standard deviation of the differences is 62.73 points. The conditions for inference are met. A 98% confidence interval for the mean difference (after – before) in score is 137.04 points to 248.96 points. What is the correct interpretation of this interval? 98% of students in this sample increased their score, on average, by 137.04 to 248.96 points. 98% of students who take this course will increase their score, on average, by 137.04 to 248.96 points. The school can be 98% confident that students who take this course increase their SAT score, on average, by 137.04 to 248.96 points. The school can…
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A bottled water distributor wants to determine whether the mean amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company is actually 1 gallon. You know from the water bottling company specifications that the standard deviation of the amount of water is 0.02 gallon. You select a random sample of 50 bottles, and the mean amount of water per 1-gallon bottle is 0.992 gallon. What is the test statistic?
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Historically, the mean age of all Memphis residents has been 40 years. You believe this has changed. You take a random sample of 130 Memphis residents and find that their mean age is 36 years with a standard deviation of 17 years. Can you conclude that the mean age of all Memphis residents has changed from 40 years. For the hypothesis testing scenario above, compute the test statistic. For the hypothesis testing scenario above, you conclude that the mean age of all Memphis residents has changed from 40 years at the 0.05 significance level. Compute the 95% confidence interval to estimate the mean age of all Memphis residents.

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