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nika2105 [10]
3 years ago
6

For each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts. A large shipmen

t has just arrived. A quality control manager randomly selects 50 of the parts from the shipment and finds that 6 parts are defective. Let μ denote the number of defective parts in the current shipment, and let p denote the proportion of defective parts among the current shipment.
a. The p-value of the test statistic should be:_________
b. Is this sufficient evidence to reject the entire shipment? Use α= 0.05 level of significance to conduct the appropriate hypothesis test.
Mathematics
1 answer:
Stella [2.4K]3 years ago
4 0

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, H_0 : p \leq 10%      {means that the defective parts in the shipment is at most 10%}

Alternate Hypothesis, H_A : 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. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of parts that are defective among the current shipment = \frac{6}{50} = 12%

           n = sample of parts from the shipment = 50

So, <u><em>test statistics</em></u>  =  \frac{0.12-0.10}{\sqrt{\frac{0.12(1-0.12)}{50} } }

                               =  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 \leq 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%.

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