Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (±.05, ±.025, ±.005) to get the Z scores.
% = mean ± Z SEm
SEm = SD/√n
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 275 boys with the disorder was reported as 2.31 with a standard deviation of 1.11. (Round your answers to four decimal places.) Find 90%, 95%, and 99% confidence intervals.
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