According to the Alzheimer’s Association, Alzheimer’s disease affects 1 in 10 pe
ID: 3128135 • Letter: A
Question
According to the Alzheimer’s Association, Alzheimer’s disease affects 1 in 10 people over the age of 65. A study was conducted on health issues facing people over the age of 65. What would be the shape, mean, and standard error of the sampling distribution of the proportion that suffer from Alzheimer’s disease if the sample size is 200 compared to if the sample size were 800?
A. If the sample size were 200 what is the probability, using a Z-table, that the sample proprtion would be at least 12.5%?
B. Does this outcome seem likely? Explain
(Please show work)
Explanation / Answer
According to the Alzheimer’s Association, Alzheimer’s disease affects 1 in 10 people over the age of 65. A study was conducted on health issues facing people over the age of 65. What would be the shape, mean, and standard error of the sampling distribution of the proportion that suffer from Alzheimer’s disease if the sample size is 200 compared to if the sample size were 800?
Both distributions will be bell shaped with mean at p = 0.10.
However, they will have different standard errors.
For n = 200, sigma(p^) = 0.021213203
For n = 800, sigma(p^) = 0.010606602
Hence, the distrbution when n = 800 is slimmer, that is, values are closer to p = 0.10.
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a)
If the sample size were 200 what is the probability, using a Z-table, that the sample proprtion would be at least 12.5%?
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 0.125
u = mean = p = 0.1
s = standard deviation = sqrt(p(1-p)/n) = 0.021213203
Thus,
z = (x - u) / s = 1.18
Thus, using a table/technology, the right tailed area of this is
P(z > 1.18 ) = 0.1190 [ANSWER]
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B. Does this outcome seem likely? Explain
(Please show work)
As 0.1190 > 0.05, then yes, this outcome is LIKELY. [ANSWER]
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