Question 5You are interested in finding a 95% confidence interval for the average number of days of classthat college students miss each year. The data below show the number of missed days for 15randomly selected college students. Round answers to 3 decimal places where possible.7▼ <2 12 11 6 107 5 3 10Submit Question5 0 72 5a. To compute the confidence interval use a ? distribution.b. With 95% confidence the population mean number of days of class that college students missis betweenanddays.c. If many groups of 15 randomly selected non-residential college students are surveyed, then adifferent confidence interval would be produced from each group. Aboutpercent ofthese confidence intervals will contain the true population mean number of missed class daysand aboutpercent will not contain the true population mean number of missed classdays.

The confidence level is 1-α=0.95. The sample size is n=15. Given data: X 7 2 12 11 6…

Answered: are interested in finding a 95%…

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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0 9 10 10 10 4 12 7 11 5 0 8 3 a. To compute the confidence interval use a ? t z distribution. b. With 90% confidence the population meannumber of days of class that college students miss is between and days. c. If many groups of 13 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.
The data below represent a random sample of 9 scores on a statistics quiz. (The maximum possible score on the quiz is 10.) Assume that the scores are normally distributed with a standard deviation of 2.5. Estimate the population mean with 95% confidence. 6,9,6,5,4,7,3,10,6 Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. Confidence Interval =
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Out of 500 people sampled, 365 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids.Give your answers as decimals, to three places______ < p < _____
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You are interested in finding a 99% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 15 randomly selected college students. Round answers to 3 decimal places where possible. 97192 38 90 10 10 5 10 8 5 a. To compute the confidence interval use a V distribution. b. With 99% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 15 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about will not contain the true population mean number of missed class days. percent
Use the Student's t distribution to find t. for a 0.90 confidence level when the sample is 3. (Round your answer to three decimal places.) 2.353 792ABO2284 hp & #3 $ 3 4 5 y e 00
A service data on the body mass index of 654 young women the mean body mass index in the sample was X bar equals 26.8 service data on the body mass index of 654 young women the mean body mass index. We treated these data as an SRS from a normal distribution population with a standard deviation of 7.5 A. given 3 confidence intervals for the mean BMI in the population using 90 95 and 99% confidence. B. What are the margins of air for 90, 95 and 99% confidence? how does increasing the confidence level change the margin of error of a confidence interval when the sample size and population standard deviation remain the same

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