Question 1088741: A crossover trial is planned to evaluate the impact of an educational intervention program to reduce alcohol consumption in patients determined to be at risk for alcohol problems. The plan is to measure alcohol consumption (the number of drinks on a typical drinking day) before the intervention and then again after participants complete the educational intervention program. How many participants would be requires to ensure that a 95% confidence interval for the mean difference in the number of drinks is within 2 drinks of the true mean difference? Assume that the standard deviation of the difference in the mean number of drinks is 6.7 drinks.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 95% CI is mean +/- 1.96 * s/ sqrt(n), assuming n is large enough. If not, will need to use t.
the half interval is width 2.
therefore, +/-1.96 *6.7/(sqrt(n)=2
13.132/sqrt(n)=2
square both sides
172.45=4n
n=43.11 or 44.
This requires a t, since the t-value is 2.017
will still use 44 as it was rounded up.
2.017*6.7/sqrt (44)=2.037.
try n=45
t=2.015, interval is 2.012
try n=47
t=2.013, interval is 1.97
try n=46
2.014; interval is 1.99
46.
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