in a survey of 1008 adults a poll asked \" Are you worried or not about having e
ID: 3154004 • Letter: I
Question
in a survey of 1008 adults a poll asked " Are you worried or not about having enough money for retirement?" Of 1008 surveyed, 573 states that they were worried about having enough money for retirement. A 99% confidence interval for the proportion of adults who are worried about having enough money for retirement is?( use ascending order. Round up four decimal places as needed). in a survey of 1008 adults a poll asked " Are you worried or not about having enough money for retirement?" Of 1008 surveyed, 573 states that they were worried about having enough money for retirement. A 99% confidence interval for the proportion of adults who are worried about having enough money for retirement is?( use ascending order. Round up four decimal places as needed). in a survey of 1008 adults a poll asked " Are you worried or not about having enough money for retirement?" Of 1008 surveyed, 573 states that they were worried about having enough money for retirement. A 99% confidence interval for the proportion of adults who are worried about having enough money for retirement is?( use ascending order. Round up four decimal places as needed).Explanation / Answer
Note that
p^ = point estimate of the population proportion = x / n = 0.568452381
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.015600235
Now, for the critical z,
alpha/2 = 0.005
Thus, z(alpha/2) = 2.575829304
Thus,
Margin of error = z(alpha/2)*sp = 0.040183543
lower bound = p^ - z(alpha/2) * sp = 0.528268838
upper bound = p^ + z(alpha/2) * sp = 0.608635923
Thus, the confidence interval is
( 0.528268838 , 0.608635923 ) [ANSWER]
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