Question 1160699: A production run is not acceptable for shipment to customers if a sample of 100 items
contains 5% or more defective items. If a production run has a population proportion
defective of P 0.100 , what is the probability that P will be at least 0.05?
(a) In an effort to estimate the mean amount spent per customer for dinner at a major
restaurant, data were collected for a sample of 49 customers over a three-week
period.
(i) Assume a population standard deviation of $10, 000. What is the standard
error of the mean?
(ii) With a .95 probability, what statement can be made about the sampling
error?
(iii) If the sample mean is $90, 400, what is the 95% confidence interval for
population mean?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Have a high enough np for normal approximation,
the mean for the population is np=10 defective
the variance is np(1-p)=9
sd is sqrt(V)=3
probability >=5 (in other words 0.05 probability for that group) is z > (4.5-10)/3=-1.83 for probability 0.9664
exact would be probability for 0-4 given n=100 and p=0.1, and taking the complement of that answer, which is 0.0237, so 1-0.0237=0.9763.
SEM is s/sqrt(n)=10000/7=$1428.57
CI 95%: half-interval is t0.975, df=48)=2.01*s/sqrt(n)=$2872.28
the CI is mean +/- the half-interval
=($87,528, $93,272)
The sampling error is approximately $2872.
We are 95% confident that the population mean is in the above interval.
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