posted by Anonymous on .
American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens reveled
they consumed the following amounts, reported in cups, of coffee yesterday.
3.1 3.3 3.5 2.6 2.6 4.3 4.4 3.8 3.1 4.1 3.1 3.2
At the .05 significance level, does this sample data suggest there is a difference between the national
average and the sample information from senior citizens?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√(n-1)
Since only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.