Answer:
A. The northern California study with a margin of error of 3.2%.
Step-by-step explanation:
We know that,
\(\text{M.E}=Z_{critical}\cdot \sqrt{\dfrac{p(1-p)}{n}}\)
Where,
M.E = margin of error,
\(Z_{critical}\) = z score of the confidence interval,
for 98% confidence interval \(Z_{critical}=2.33\)
p = proportion,
n = sample size.
One study in northern California involved 1,000 patients; 74% of them experienced flu like symptoms during the month of December.
Putting the values,
\(\text{M.E}=2.33\cdot \sqrt{\dfrac{0.74(1-0.74)}{1000}}=0.032=3.2\%\)
The other study, in southern California, involved 500 patients; 34% of them experienced flu like symptoms during the same month.
Putting the values,
\(\text{M.E}=2.33\cdot \sqrt{\dfrac{0.34(1-0.34)}{500}}=0.049=4.9\%\)
The smallest margin of error is 3.2% of the northern California study.
