Respuesta :
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 = proportion of defective parts among the current shipment.
So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 10% {means that the defective parts in the shipment is at most 10%}
Alternate Hypothesis, [tex]H_A[/tex] : p > 10% {means that the defective parts in the shipment is greater 10%}
The test statistics that would be used here One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of parts that are defective among the current shipment = [tex]\frac{6}{50}[/tex] = 12%
n = sample of parts from the shipment = 50
So, test statistics = [tex]\frac{0.12-0.10}{\sqrt{\frac{0.12(1-0.12)}{50} } }[/tex]
= 0.44
The value of z test statistics is 0.44.
(a) Now, P-value of the test statistics is given by the following formula;
P-value = P(Z > 0.44) = 1 - P(Z [tex]\leq[/tex] 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 we fail to reject our null hypothesis.
Therefore, we conclude that the defective parts in the shipment is at most 10%.
