Question 1118455: A survey of 30 randomly selected "iPhone" owners showed that the purchase price has a mean of $414 with a sample standard deviation of $200.
a. Compute the standard error of the sample mean. (Round the final answer to the nearest whole number.)
e. Compute the 95% confidence interval for the mean. (Round the final answers to 2 decimal places.)
The confidence interval is between $____ and $____
.
f. How large a sample is needed to estimate the population mean within $12 at a 95% degree of confidence? (Round the final answer to the nearest whole number.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! t0.975,df=29 is 2.045
standard error is s/sqrt(n)=200/sqrt(30)=36.51 or 37.
95% CI is mean +/- t0.975,df=29*s/sqrt(n)=2.045*37=mean +/- $75.67
($338.33, $489.67)
Assume a t=2
want half-interval to be $12
12=2*200/sqrt(n)
12*sqrt(n)=400
square both sides
144n=160,000
n=1111.1 or 1112
With this number, t can be approximated with z, and 95% interval is
1.96*200/sqrt(n)=12
392^2=144n
n=1067.1 or 1068
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