Answer: The northern California study with a margin of error of 3.2%.
Step-by-step explanation:
The formula for margin of error for proportion is given by :-
[tex]z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}[/tex]
Given : Significance level : [tex]\alpha:1-0.98=0.02[/tex]
Critical value : [tex]z_{\alpha/2}=2.33[/tex]
For northern California , the sample size was takes : n= 1000
The proportion of patients experienced flu like symptoms during the month of December = 0.74
Then , the margin of error will be :-
[tex](2.33)\sqrt{\dfrac{0.74(1-0.74)}{1000}}\approx0.032=3.2\%[/tex]
For southern California , the sample size was takes : n= 500
The proportion of patients experienced flu like symptoms during the month of December = 0.34
Then , the margin of error will be :-
[tex](2.33)\sqrt{\dfrac{0.34(1-0.34)}{500}}\approx0.049=4.9\%[/tex]
Clearly, the northern California study with a margin of error of 3.2% has the smallest margin of error for a 98% confidence interval.
