Help on this please.
1 answer:
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Answer:
A sample of 1032 is needed.
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 is:
A sample of 300 components showed that 20 were defective.
This means that
99% confidence level
So , z is the value of Z that has a pvalue of
, so
.
How large a sample is needed to estimate the true proportion of defective components to within 2.5 percentage points with 99% confidence?
A sample of n is needed.
n is found when M = 0.025. So
Rounding up
A sample of 1032 is needed.