SOLUTION: The sample mean (n = 175) for student housing is $616.91 and the sample standard deviation is $128.65. Construct a 95% confidence interval and evaluate for $650.

Question 1201709: The sample mean (n = 175) for student housing is $616.91 and the sample standard deviation is $128.65. Construct a 95% confidence interval and evaluate for $650.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
sample mean is 616.91
sample standard deviation is 128.65
sample size is 175
standard error = standard deviation / square root of sample size = 128.65/sqrt(175) = 9.725025891.
t-score is used, rather than z-score.
since the sample size is so large, the difference between using the t-score or the z-score will be small.
t-score 95% confidence interval with 174 degrees of freedom is t = plus or minus 1.9736914.
use the t-score formula to find the equivalent raw scores.
t-score formula is t = (x - m) / s
in this problem, t is the critical t-score, x is the critical raw score, m is the sample mean, s is the standard error.
on the low side, you get -1.9736914 = (x - 616.91) / 9.72502591.
solve for x to get:
x = -1.9736914 * 9.72502591 + 616.91 = 597.7158.
on the high side, you get 1.9736914 = (x - 616.91) / 9.72502591.
solve for x to get:
x = 1.9736914 * 9.72502591 + 616.91 = 636.1042.
your 95% confidence interval is from 597.7158 to 636.1042.
evaluating a raw score of 650 is done in the folllowig manner.
use the t-score formula to get:
t = (x - m) / s becomes t = (650 - 616.91) / 9.72502591 = 3.402561636.
area to the right of t-score, with 174 degrees of freedom = .0004141885659.
since that is considerably less than the 95% confidence interval alpha on the high end of .025, the results are considered significant.
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