Question 1182597
the sample mean is equal to 3.5 hours.
the sample standard deviation is equal to .8 hours.
the sample size is 20.


t-score would be used with 19 degrees of freedom is what i think.
at 95% confidence level, the two tail critical t-score with 19 degrees of freedom would be equal to plus or minus 2.093.


19 degrees of freedom is because the sample size is 20.


the standard error would be equal to .8 / sqrt(20) = .1789.
that's the standard deviation of the sample divided by the square root of the sample size.


the t-score formula becomes plus or minus 2.093 =  ​(x - 3.5) / .1789


solve for x to get:


x = .1789 * 2.093 + 3.5 on the high side and x = .1789 * -2.093 + 3.5 on the low side.


x will be equal to 3.8744377 on the high side.
x will be equal to 3.1255623 on the low side.


your confidence interval is from 3.1256 on the low side to 3.8744 on the high side when the solution is rounded to 4 decimal places.


the graph of the critical t-score would look like this:


<img src = "http://theo.x10hosting.com/2021/072801.jpg" >


you need to find the standard error and use the t-score formula to find the raw scores from the t-scores.


i did that above.


standard error = standard deviation of sample divided by square root of sample size.


t-score formula is t = (x = m) / s
t is the t-score
x is the raw score
m is the mean
s is the standard error