Question

A simple random sample of size nequals24 is drawn from a population that is normally distributed. The sample mean is found to be x overbar equals 58 and the sample standard deviation is found to be sequals10. Construct a 90​% confidence interval about the population mean. The 90​% confidence interval is ​( nothing​, nothing​). ​(Round to two decimal places as​ needed.)

Answer 1

Given data

Sample Size (n)=24

Sample mean

Sample standard deviation (S)=10

confidence level =90%

hence

now confidence interval from the formula is

where

ME= margine of error

where

criticle value of test statistic

here the sample size is < 30 so it willl be a t test   so for t criticle value

Degree of freedom (v)=n-1=24-1=23

From Standard   t disribution table the t criticle value with respect to degree of freedom v=23 and is

now

now

lower level

Upper level

So the confidence interval is