How many samples are needed? (a)Given that, the required Confidence Level is 99%
ID: 3134705 • Letter: H
Question
How many samples are needed? (a)Given that, the required Confidence Level is 99%, Margin of Error is 3% and the assumed population proportion is 30%.(b)How many samples are needed? Given that, the required Confidence Level is 90%, Margin of Error is 2% and the assumed population proportion is 40%. (c)How many samples are needed? Given that, the required Confidence Level is 95%, Margin of Error is 3% and the population proportion is NOT available. (d)How many samples are needed? Given that, the required Confidence Level is 99%, Margin of Error is 2% and the population proportion is NOT available.Explanation / Answer
A)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.005
Using a table/technology,
z(alpha/2) = 2.575829304
Also,
E = 0.03
p = 0.3
Thus,
n = 1548.14254
Rounding up,
n = 1549 [ANSWER]
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b)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.05
Using a table/technology,
z(alpha/2) = 1.644853627
Also,
E = 0.02
p = 0.4
Thus,
n = 1623.326072
Rounding up,
n = 1624 [ANSWER]
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c)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.025
As there is no previous estimate for p, we set p = 0.5.
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
E = 0.03
p = 0.5
Thus,
n = 1067.071895
Rounding up,
n = 1068 [ANSWER]
*********************
d)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.005
As there is no previous estimate for p, we set p = 0.5.
Using a table/technology,
z(alpha/2) = 2.575829304
Also,
E = 0.03
p = 0.5
Thus,
n = 1843.026834
Rounding up,
n = 1844 [ANSWER]
***********************
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