Salary information regarding two independent random samples of male and female employees of a large company is shown below.Female90$41$42Sample sizeSample meanPopulation standard deviationMale111$73$48Let us develop a 99% confidence interval estimate of the difference between the salary of the two populations. Calculate the margin of errorfor the confidence interval. (Answer to two decimal places)

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Answered: Population standard deviation Let us…

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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Question Help ▼ 7.2.7-1 Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t/2, (b) find the critical value za/2, or (c) 40- state that neither the normal distribution nor the t distribution applies. 30- The confidence level is 90%, o = 3775 thousand dollars, and the histogram of 64 player salaries (in thousands of dollars) of football players on a team is as shown. 20- 10- 0- 4000 8000 1200 16000 2000 Salary (thousands of dollars) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. OA. ta/2= (Round to two decimal places as needed.) O B. Za/2= (Round to two decimal places as needed.) OC. Neither the normal distribution nor the t distribution applies.
Overweight Men For a random sample of 55 overweight men, the mean of the number of pounds that they were overweight was 31. The standard deviation of the population is 3.9 pounds. Use a graphing calculator and round the final answers to one decimal place. Part 1 of 4 The best point estimate of the mean is 31 pounds. Part: 1 / 4 Part 2 of 4 Find the 90% confidence interval of the mean of these pounds. <μ< X
Gasoline Use A random sample of 36 drivers used on average 741 gallons of gasoline per year. The standard deviation of the population is 33 gallons. Part: 0 / 2 Part 1 of 2 (a) Find the 95% confidence interval of the mean for all drivers. Round intermediate answers to at least three decimal places. Round your final answers to the nearest whole number. <µ< X 5
(5 points) Rock band The Rolling Stones have played scores of concerts in the last twenty years. For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.12 million dollars. Part a) Assuming a population standard deviation gross earnings of 0.51 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions). Confidence interval: ( Part b) Which of the following is the correct interpretation for your answer in part (a)? O A. We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the interval O B. If we repeat the study many times, 99% of the calculated confidence intervals will contain the mean gross earning of all Rolling Stones concerts. O C. There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the interval O D. None of the above
Constructing a Confidence Interval for a Population Mean A biologist measures the lengths of a random sample 30 mature brown trout in a large lake and finds that the sample a mean weight of 41 pounds. Assume the population standard deviation is 3.7 pounds. Based on this, construct a 90% confidence interval for the mean weight of all mature brown trout in the lake. Round your anwers to two decimal places.
Scores on an English test from 30 students in Class One has a mean of 33 and a standard deviation of 7.2. Scores on an English test from 40 students in Class Two has a mean of 39 and a standard deviation of 5.5. Assuming the two classes are independent, construct a 90%confidence interval for difference between the mean scores for Class One and Class Two. Round your answer to 3 decimal places. Base on the confidence interval, what conclusion can you draw? Group of answer choices (-8.530,-3.470) Group 2 is bigger (-8.530,-3.470) Group 1 is bigger (-8.639,-3.361) No conclusion (-8.639,-3.361) Group 2 is bigger (-6.769,-5.231) Group 2 is bigger
upper and lower limit
Just need help with part 3, thank you!
Please do not give solution in image format thanku A dietitian wants to know the average time spent on breakfast in a primary school. The dietitian randomly samples 16 students and finds that the average is 15.8 minutes with a standard deviation of 2.31 minutes. Assume that the distribution of the time spent on the breakfast is normally distributed. The dietitian finds a 90% confidence interval for this sample is (14.788, 16.812). Select one or more: a. The margin of error is 1.012. b.The margin of error is 2.024 c. The margin of error is 0.950. d. We believe that the true mean time spent on breakfast in this primary school is between 14.788 and 16.812 minutes. e. If we take many other samples from this population, 90% of them will have a sample mean that is between 14.788 and 16.812. f. There is a 90% chance that the true mean is between 14.788 and 16.812 minutes.
Find the left and right chi square values, then give the confidence interval. Sample Variance s 15.99 %3D Confidence Level c = 0.90 and Sample Size n = 13 and x %3D %3D At the 90 % level of confidence the population variance is between and (Round all answers to 3 decimal places)
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A sample is used to construct a confidence interval for an unknown population Which of the following is the least likely to result in a decrease in the margin of error? Increasing the sample size Increasing the confidence level Decreasing the confidence level A change in the standard deviation of the

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Population standard deviation Let us develop a 99% confidence interval estimate of the difference between the salary of the two populations. Calculate the margin of error for the confidence interval. (Answer to two decimal places)