Posted by sweet on Monday, August 16, 2010 at 12:58pm.
Use the processes from your previous post.
However for d, you are using a distribution of means rather than Individual scores.
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 that Z score.
That table can be used for your other problems too, finding Z scores to convert from percentile to score.
Am I close? a- 99%, b- 1.29, c- yes, unusual because only 1% of the healthy adult population would be higher,d- 24.5 of the 50 are to be at or below 98.98, e- 4.52, an unusual result and the adult is probably not healthy because the 101 is outside the niormal range by a large percentage. f- 98.97
Am I close? a- 99%, b- 1.29, c- yes, unusual because only 1% of the healthy adult population would be higher,d- 24.5 of the 50 are to be at or below 98.98, e- 4.52, an unusual result and the adult is probably not healthy because the 101 is outside the niormal range by a large percentage. f- 98.97
Am I close? a- 99%, b- 1.29, c- yes, unusual because only 1% of the healthy adult population would be higher,d- 24.5 of the 50 are to be at or below 98.98, e- 4.52, an unusual result and the adult is probably not healthy because the 101 is outside the niormal range by a large percentage. f- 98.97
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