Find the difference of 7 2/3 and 6 3/7
2 answers:
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Answer:
(a) The p-value of the test statistic should be 0.33.
(b) No, there is not sufficient evidence to reject the entire shipment.
Step-by-step explanation:
We are given that for each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts.
A quality control manager randomly selects 50 of the parts from the shipment and finds that 6 parts are defective.
Let p = <u><em>proportion of defective parts among the current shipment.</em></u>
So, Null Hypothesis, : p
10% {means that the defective parts in the shipment is at most 10%}
Alternate Hypothesis, : p > 10% {means that the defective parts in the shipment is greater 10%}
The test statistics that would be used here <u>One-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of parts that are defective among the current shipment =
= 12%
n = sample of parts from the shipment = 50
So, <u><em>test statistics</em></u> =
= 0.44
The value of z test statistics is 0.44.
(a) <u>Now, P-value of the test statistics is given by the following formula;</u>
P-value = P(Z > 0.44) = 1 - P(Z 0.44)
= 1 - 0.67003 = 0.3299 ≈ 0.33
(b) Since, the P-value of test statistics is more than the level of significance as 0.33 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.
Therefore, we conclude that the defective parts in the shipment is at most 10%.