A supervisor records the repair cost for 17 randomly selected VCRs. A sample mean of $68.83 and standard deviation of $12.17 are
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Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error of the interval is given by:
In this problem, we have that:
99.5% confidence level
So , z is the value of Z that has a pvalue of
, so
.
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07