Answer:
Please the read the answer below
Explanation:
In order to find the 95% confidence interval for the difference of the two populations, you use the following formula (which is available when the population size is greater than 30):
CI =
(1)
where:
p1: proportion of one population = 52/9853 = 0.0052
p2: proportion of the other population = 41/11541 = 0.0035
α: tail area = 1 - 0.95 = 0.05
Z_α/2: Z factor of normal distribution = Z_0.025 = 1.96
n1: sample of the first population = 52
n2: sample of the second population = 41
You replace the values of all parameters in the equation (1) :

By the result obtained in the solution, you can conclude that the sample is not enough, because the margin error is greater that the difference of proportion of each sample population.