Recall that a bank manager has developed a new system to reduce the time customers spend waiting to be served by tellers duringpeak business hours. The mean waiting time during peak business hours under the current system is roughly 9 to 10 minutes. Thebank manager hopes that the new system will have a mean waiting time that is less than six minutes. The mean of the sample of 93bank customer waiting times is 2-5.44. If we let udenote the mean of all possible bank customer waiting times using the new systemand assume that o equals 2.44:(a) Calculate 95 percent and 99 percent confidence intervals for u. (Round your answers to 3 decimal places.)95 percent confidence intervals for p is99 percent confidence intervals for u in(b) Using the 95 percent confidence interval, can the bank manager be 95 percent confident that is less than six minutes? Explain.95 percent intervalis16(c) Using the 99 percent confidence interval, can the bank manager be 99 percent confident that is less than six minutes? Explain99 percent interval extends6

The question is about confidence interval.Given :Sample no. of customers ( n ) = 93Sample mean…

Answered: Recall that a bank manager has…

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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An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 250 Fundash runners and 285 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 250 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.7 minutes. The 285 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.6 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, u, of Fundash is not equal to the mean completion time, µa, of Coolsprint? Perform a two-tailed test.…
It is known that average house prices today is $371,040 higher compared to 10 years ago and median in the current time also $382,000 higher than house prices 10 years ago. Provide an interpretation of the change of mean and median based on the information above. Compare and contrast these values and comment on the distribution of the increases in house prices.
A social psychologist recently developed a childhood-memories test that is intended to measure how people recall pleasant and unpleasant experiences from childhood. Scores on the test range from 1 (very unpleasant) to 100 (very pleasant). The psychologist standardized the test with a large group of college students (ages 18-25), and the mean score was p=60. The test was then given to a sample of n=25 individuals who were all between the ages of 40 and 45. The average score for this sample was 64.3 with SS (sum of squares)=600. Note: S?=SS÷df (variance is equal to sum of squares divided by the degrees of freedom). What is your standard deviation? Answer:
Suppose the distribution of typing speeds in words per minute for experienced typists can be approximated by a normal curve with a mean of 60 wpm and a standard deviation of 15 wpm. To be considered for a position an applicant needs to have a typing speed of at least 70 wpm. If 62 experienced typists apply for the job, how many of these would be considered based on their typing speed?
The second slide is the first part of the question and the first slide is the second part of the question.
An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 240 Fundash runners and 200 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 240 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.2 minutes. The 200 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.4 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, μ1 , of Fundash is not equal to the mean completion time, μ2 , of Coolsprint?…
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A researcher wanted to find the best viewing distance for watching a TV screen. He recruited the following number of participants and asked them to evaluate their satisfaction of the viewing distance. Then, he conducted an ANOVA to see if there is any significant mean differences. Here are the number of participants by each group: Group 1 (4 feet) - 5 people Group 2 (7 feet) - 5 people Group 3 (10 feet) - 6 people Group 4 (13 feet) - 5 people Group 5 (16 feet) - 4 people Given the numbers as presented above, what is the degrees of freedom for within groups for this analysis? a. 4 b. 20 O c. 21 d. 25
13. Dave likes to grow cucumbers, and each year he experiments with ways to improve his crop. This summer, he wants to determine whether the new fertilizer he has seen advertised will increase the mean number of cucumbers produced per plant. He uses the old fertilizer on 15 cucumber plants and the new fertilizer on the other 13 plants. Over the course of the growing season, he calculates a mean of 17.4 cucumbers per plant for the old fertilizer, with a standard deviation of 2.0 cucumbers. He calculates a mean of 21.6 cucumbers per plant for the new fertilizer, with a standard deviation of 2.7 cucumbers. You may assume that both samples came from populations that are approximately normal. Assume that the population variances are not equal. a. Determine the best point estimate for the true mean difference in cucumber production. Correct notation counts. b. Construct a 95% confidence interval for the true mean difference in cucumber production. For credit, your work must include a…
The post-aesthesia care area (recovery room) at Centre Hospital was recently enlarged. The hope was that with the enlargement the mean number of patients per day would be more than 25. A random sample of 15 days revealed the following number of patients X
There are many ways to measure the reading ability of children. One frequently used test is the DRP or Degree of Reading Power. The national average score in DRP test is 32. We want to determine if there is sufficient evidence at the 5% level to suggest that the mean score of all third graders in YOUR district is higher than the national mean. The DRP scores for a simple random sample of 41 third graders in that suburban school in YOUR district yielded a mean of 35.2. Assume that the population of DRP scores of third graders in that suburban school district are normally distributed with a population std.dev of 11.
The mean number of hours that Raphael worked over the past four weeks is 36.75. If he works 39 hours this week, what will the mean number of hours be that he will have worked over the five-week period? What is the five-week mean?

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