SOLUTION: A supplier of USB flash claims that more than 11% of the USBs are defective. In a random sample of 167 USBs, it is found that 37 are defective, but the supplier claims that this is
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Question 1200911: A supplier of USB flash claims that more than 11% of the USBs are defective. In a random sample of 167 USBs, it is found that 37 are defective, but the supplier claims that this is only a sample fluctuation. At the 0.05 level of significance, test the supplier's claim that more than 11% are defective.
a. Is this (a: one sample mean test, b: two sample mean test, c: one proprtion test, d: ch-square test)
b. Since the test statistic (round to 2 decimals): Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! population p = .11
population q = 1 - .11 = .89
sample size is 167.
37 are defective.
sample p = 37/167 = .2215568862
standard error of the test is equal to sqrt(population p * population q / 167) = sqrt(.11 * .89 / 167) = .0242121363.
z-score = (.2215568862 - .11) / .0242121363 = 4.607478034
area to the right of that = .000002040040862.
that's significantly less than the critical p-value of .05.
the probability that the resuls is a fluctuation in smple means is pretty close to 0.
the conclusion i that the preponderance of the evidence indicates that more than 11% of the USBs are defective, rejecting the claim that it's no more than 11%.
this is a one sample proportion test.
the critical p-value is .05
the test p-value is 0, rounded to two decimal places.
differences between the calculator numbers and the ones i calculated have an insignificant impact to the result and are more than likely due to rounding differences.
