Question 1156490: A survey of 53 randomly selected "iPhone" owners showed that the purchase price has a mean of $412 with a sample standard deviation of $180.
a. Compute the standard error of the sample mean. (Round the final answer to the nearest whole number.)
Standard error
b. Compute the 80% confidence interval for the mean. (Round the final answers to 2 decimal places.)
The confidence interval is between
and
.
c. How large a sample is needed to estimate the population mean within $14 at a 80% degree of confidence? (Round the final answer to the nearest whole number.)
Sample size
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! the t value (0.90) is +/-1.298
SE is s/sqrt(n)=180/sqrt(53)=24.72 or 25
80%CI is t*SE=depends on how it is rounded. If I round only at the end, then it is 32.09
that is on either side of mean 412
(379.91, 444.09) $ units
$14=1.28*180/sqrt(n). Because I don't know the sample size, I don't know t, but I can use z to approximate.
square both sides
196=53084/n
n=270.84 or 271. At this point, use the t value for that which make the interval a little wider, so then use the next higher and the next higher.
n=273 gives ($398, $412)
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