Homework SourseCalculate the approximate size required for each of the foll
Calculate the approximate size required for each of the following at 95 percent confidence:
Standard deviation = 10, (ME) = 0.5, two tail
Standard deviation = 10, (ME) = 0.5, one tail
Standard deviation = 10, (ME) = 1.5, two tail
Standard deviation = 10, (ME) = 1.5, one tail
Standard deviation = 35, (ME) = 5, two tail
Standard deviation = 35, (ME) = 5, one tail
Standard deviation = $3,200, (ME) = 400, two tail
Standard deviation = $3,200, (ME) = 400, one tail
Solution
a)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
s = sample standard deviation = 10
E = margin of error = 0.5
Thus,
n = 1536.583528
Rounding up,
n = 1537 [ANSWER]
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b)
Note that
n = z(alpha)^2 s^2 / E^2
where
alpha = (1 - confidence level) = 0.05
Using a table/technology,
z(alpha) = 1.644853627
Also,
s = sample standard deviation = 10
E = margin of error = 0.5
Thus,
n = 1082.217382
Rounding up,
n = 1083 [ANSWER]
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c)
Note that
n = z(alpha/2)^2 s^2 / E^2
where
alpha/2 = (1 - confidence level)/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
s = sample standard deviation = 10
E = margin of error = 1.5
Thus,
n = 170.7315031
Rounding up,
n = 171 [ANSWER]
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D)
Note that
n = z(alpha)^2 s^2 / E^2
where
alpha = (1 - confidence level) = 0.05
Using a table/technology,
z(alpha) = 1.644853627
Also,
s = sample standard deviation = 10
E = margin of error = 1.5
Thus,
n = 120.2463757
Rounding up,
n = 121 [answer]
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