Question 1001739: Finding confidence intervals.
Find the margin of error and the confidence interval for the population mean. Assume that the population has a normal distribution.
Math SAT scores for women: 95% confidence, n = 15, xbar = 496, s = 108
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! This is a t-distribution, so 95% CI for t(df=14) is
496 +/- 2.145(108/sqrt(15); mean +/- t-value *SE, where SE= s/sqrt(n)
the SE here is 27.89. Without rounding until the end, the CI=496+/-59.81
(436.19, 555.81).
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