)33 participants. mean .23 SD .05 what is the margin of error, assuming a 95% confidence level? (Assume a normal distribution.)
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A 95% CI corresponds to tails of 2.5% on each side of the mean in the normal distribution. We need to find the Z-score which corresponds to 100-2.5%=97.5%=0.975. Z=1.96=(X-0.23)/0.05, X=0.23+0.098=0.328. This is the high side of the CI, while 0.23-0.098=0.132 is the low side of the CI. CI=[0.132,0.328].

However, the title of the question has mean=0.25, so:

Z=1.96=(X-0.25)/0.05, X=0.25+0.098=0.348. This is the high side of the CI, while 0.25-0.098=0.152 is the low side of the CI. CI=[0.152,0.348].

The population size of 33 is high enough to apply normal distribution. The CI tells us that we can be 95% sure that the results in the population will lie between 0.152 and 0.348 (for mean=0.25) or between 0.132 and 0.328 (for mean=0.23).

If 33 is the size of a sample, rather than the population, and the SD is the population SD, then we need to adjust this SD to fit the sample, making s=0.05/√33=0.0087 approx.

Z=1.96=(X-0.23)/0.0087, X=0.23+0.17=0.4 (high) and 0.23-0.17=0.06 (low), or if mean=0.25, 0.42 (high) and 0.08 (low).

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