In a random sample of 8 people, the mean commute time to work was 34.5 minutes and the standard deviation was 7.2 minutes. A 90% confidence interval using the t-distribution was calculated to be (29.7,39.3). After researching commute times to work, it was found that the population standard deviation is 8.8 minutes. Find the margin oferror and construct a 90% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results.CThe margin of error of μ is.(Round to two decimal places as needed.)A 90% confidence interval using the standard normal distribution is (0.(Round to one decimal place as needed.)Compare the results. Choose the correct answer below.O A. The confidence interval found using the standard normal distribution has smaller lower and upper confidence interval limits.OB. The confidence interval found using the standard normal distribution is the same as the confidence interval found using the student's t-distribution.OC. The confidence interval found using the standard normal distribution is narrower than the confidence interval found using the student's t-distribution.O D. The confidence interval found using the standard normal distribution is wider than the confidence interval found using the student's t-distribution.

Given n=8 Standard deviations=8.8

Answered: af error of u is o decimal places as…

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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Use technology to construct the confidence intervals for the population variance o² and the population standard deviation o. Assume the sample is taken from a normally distributed population. ntents Iccess c = 0.98, s = 38, n = 15 The confidence interval for the population variance is (, ). (Round to two decimal places as needed.) a Librai ptions tor
Construct a confidence interval for p, - p, at the given level of confidence. X1 = 398, n1 = 543, x2 = 408, n2 = 578, 99% confidence The researchers are % confident the difference between the two population proportions, p, - p2, is between and (Use ascending order. Type an integer or decimal rounded to three decimal places as needed.)
n1=16 x‾1=124 s1=35 n2=12 x‾2=86 s2=31 Use this data to find the 98% confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed. Copy Data Step 2 of 3 : Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places. Step 3 of 3 : Please also find the high and low endpoints and round to four decimal points.
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
Construct a confidence interval for p, - P, at the given level of confidence. X1 = 34, n1 = 273, x2 = 32, n2 = 313, 90% confidence The researchers are % confident the difference between the two population proportions, p, - P2, is between and (Use ascending order. Type an integer or decimal rounded to three decimal places as needed.)
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.
Construct a confidence interval of the population proportion at the given level of confidence. x equalsx=540540, n equalsn=11001100, 9595% confidence Click here to view the standard normal distribution table (page 1). LOADING... Click here to view the standard normal distribution table (page 2). LOADING... The lower bound of the confidence interval is nothing . (Round to three decimal places as needed.) The upper bound of the confidence interval is nothing . (Round to three decimal places as needed.) Enter your answer in each of the answer boxes.
Use the given confidence interval to find the margin of error and the sample mean. (15.2,21.8) The sample mean is (Type an integer or a decimal.) The margin of error is | (Type an integer or a decimal.)

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