An estimate of the percentage of defective pins in a large batch of pins supplie
ID: 3023453 • Letter: A
Question
An estimate of the percentage of defective pins in a large batch of pins supplied by a vendor is desired to be estimated within 1% with a 90% confidence level. The actual percentage of defective pins is guessed to be 4%.
a) What is the minimum sample size (n) required to satisfy this requirement?
b) If the actual percentage of defective pins may be anywhere between 3% and 6%, tabulate the minimum sample size required for actual percentages from 3% to 6%.
c) If the cost of sampling and testing n pins is (25+6n) dollars, tabulate the cost for the same range of percentages as in part (b).
Explanation / Answer
a)
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 = 0.01
p = 0.04
Thus,
n = 1038.928686
Rounding up,
n = 1039 [ANSWER]
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b)
For p^ = 0.03:
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 = 0.01
p = 0.03
Thus,
n = 787.3131451
Rounding up,
n = 788
For p^ = 0.05:
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 = 0.01
p = 0.05
Thus,
n = 1285.133141
Rounding up,
n = 1286
For p^ = 0.06:
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 = 0.01
p = 0.06
Thus,
n = 1525.926508
Rounding up,
n = 1526
Hence, the table:
[ANSWER]
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c)
[ANSWER]
p^ n 0.03 788 0.04 1039 0.05 1286 0.06 1526