SOLUTION: ----------- A simple random sample of 300 is selected from a large shipment and testing reveals that 4% of selected samples are defective. The supplier claims that no more than 2%

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Question 1183159: -----------
A simple random sample of 300 is selected from a large shipment and testing reveals that 4% of selected samples are defective. The supplier claims that no more than 2% of the shipment are defective.Testing at 1% of significance. What conclusion can you draw from this statement

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
H%5B0%5D%3A+p+%3C=+0.02
H%5B1%5D%3A++p%3E+0.02
We are given the info n = 300. alpha+=+0.01, p%5B1%5D+=+0.04, and we will implement a one-tailed test with the rejection region on the far right side of the normal distribution.
==> , using 5 d.p.

Using the calculator on https://stattrek.com/online-calculator/normal.aspx, we find out that P%28Z+%3C+2.47436%29+=+0.993,
hence the p-value is 1 - 0.993 = 0.007 < 0.01,
the significance level alpha. Therefore we reject H%5B0%5D,
and conclude that the proportion of defective items is greater than 2%.