This is a sample proportion problem which can solved easily using z table and finding margin of error.
Given data,
Sample size, n = 3492
People wo have started paying bills online, x = 1453
So, the sample proportion, = = = 0.416
The confidence interval(C.I.) for population proportion is given by-
C.I. = where z is the z score .
For 99% confidence interval, z = 2.58
Now, = = 0.0215 = 0.022
So, the C.I. is= ( 0.416-0.022, 0.416+0.022) = (0.394, 0.438)
The correct option is B
With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online last year is between the endpoints of the confidence interval.
Test: Exam#2 Submit Test sti This Question: 1 pt 10 of 19 (13 complete) This Test:...
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