Do your calculations in R In a report you encounter a 95 con

In a report you encounter a 95% confidence interval (1.6,7.8) for the parameter n of an N(mu,sigma^2) distribution. The interval is based on 16 observations. constructed according to the studentized mean procedure. What is the moan of the (unknown) dataset? You prefer to have a 99% confidence interval for mu. Construct it.

Solution

a) z-score for 95% conf. interval =  1.959964..........
upper limit of the conf. interval= 7.8 = mean+ ( 1.959964* sigma / 4 )
and lower limit of conf. interval = 1.6 = mean - (  1.959964* sigma / 4 ) ....

so, adding the above 2 expressions we have, mean = ( 7.8 + 1.6) / 2 = 4.7....

and margin of error = ( 7.8 - 4.7 ) = 3.1 = (  1.959964* sigma / 4 )........so, sigma = 6.326647....

b) 99% conf. interval of mean.....

upper limit = 4.7 + ( z * 6.326647 / 4 ) and lower limit = 4.7 - ( z * 6.326647 / 4 ).....
and z for 99% conf. interval =  2.575829....

so, confidence interval = [ 0.6259098 , 8.77409 ]