posted by James .
In a study of children with a particular disorder, parents were asked to rate their child on a variety of items related to how well their child performs different tasks. One item was "Has difficulty organizing work," rated on a five-point scale of 0 to 4 with 0 corresponding to "not at all" and 4 corresponding to "very much." The mean rating for 278 boys with the disorder was reported as 2.28 with a standard deviation of 1.14. (Round your answers to four decimal places.)
Compute the 90% confidence interval.
( 2.167 , 2.393) This is right
Compute the 95% confidence interval.
( 2.1459, 2.4141 ) This is wrong
Compute the 99% confidence interval.
( 2.1035, 2.4565 ) This is wrong
It depends if you are looking at a distribution of scores or a distribution of means.
Since the scores cannot vary more than 2 SD in either direction, I assume that you are looking for means.
90% = mean ± 2.04 SEm
SEm = SD/√n
95% = mean ± 1.96 SEm
99% = mean ± 2.575 SEm
I'll let you do the calculations.