SOLVE FOR BOTH K's PLS
1 answer:
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Answer:
a) The minimum sample size is 601.
b) The minimum sample size 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 is:
For this problem, we have that:
We dont know the true proportion, so we use , which is when we are are going to need the largest sample size.
95% confidence level
So , z is the value of Z that has a pvalue of
, so
.
a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)
This is n for which M = 0.04. So
Rounding up
The minimum sample size is 601.
b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.
Now we want n for which M = 0.02. So
The minimum sample size is 2401.