160.1/2 ÷ 70.1/3I Mark brainlies
1 answer:
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Answer:
The sample size needed is 2401.
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:
95% confidence level
So , z is the value of Z that has a pvalue of
, so
.
Estimate the sample size needed if no estimate of p is available so that the confidence interval will have a margin of error of 0.02.
We have to find n, for which M = 0.02.
There is no estimate of p available, so we use
The sample size needed is 2401.