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In a sample of 5 courses with no support course the average final of 200 students was 70% with a standard deviation of 10%. Calculate a 95% confidence interval for the mean of students who did not take a support course. (Type numbers in Excel without percent sign) (68.61, 71.39)

In a sample of 5 courses with no support course the average final of 200 students was 70% with a standard deviation of 10%. Calculate a 95% confidence interval for the mean of students who did not take a support course. (Type numbers in Excel without percent sign) (68.61, 71.39)

MATLAB: An Introduction with Applications
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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In a sample of 5 courses with no support course the average final of 200 students was 70% with a standard deviation of 10%. Calculate a 95% confidence interval for the mean of students who did not take a support course. (Type numbers in Excel without percent sign) (68.61, 71.39) (68.61, 71.38) O (68.60, 71.39) O (69.96, 70.04) O 5 is not larger than 40 so we cannot compute a confidence interval.
Transcribed Image Text:In a sample of 5 courses with no support course the average final of 200 students was 70% with a standard deviation of 10%. Calculate a 95% confidence interval for the mean of students who did not take a support course. (Type numbers in Excel without percent sign) (68.61, 71.39) (68.61, 71.38) O (68.60, 71.39) O (69.96, 70.04) O 5 is not larger than 40 so we cannot compute a confidence interval.
If a sample of 35 employees had been selected, could we find a 95% confidence interval for the average salary of the entire company? Yes because when I plug 35 into the calculator with the previous average of $55,000 and standard deviation of $5,000 I get a confidence interval of ($53,343.53, $56,656.47) O No since 35 is less than 40.
Transcribed Image Text:If a sample of 35 employees had been selected, could we find a 95% confidence interval for the average salary of the entire company? Yes because when I plug 35 into the calculator with the previous average of $55,000 and standard deviation of $5,000 I get a confidence interval of ($53,343.53, $56,656.47) O No since 35 is less than 40.
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