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January 30, 2015

January 30, 2015

Posted by **Anonymous** on Saturday, June 5, 2010 at 8:46pm.

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?

- math -
**PsyDAG**, Sunday, June 6, 2010 at 11:03amZ = (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.

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