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kodGreya [7K]
3 years ago
8

Now suppose that the manufacturer decides to accept the shipment if there is at most one defective part in the sample. How large

does K have to be to ensure that the probability that the manufacturer accepts an unacceptable shipment is less than 0.1?
Mathematics
1 answer:
Mashcka [7]3 years ago
4 0

Answer:

hello your question is incomplete below is the missing parts of the question

answer : K = 80

Step-by-step explanation:

P [ x ≤ 1 ]  < 0.1

determine how large k have to be

note : n = k

sample proportion = 0.01

E = 0.05 - 0.01 = 0.04

applying this formula below

E = Z_{c} \sqrt{\frac{pq}{n} }  

0.04 = 1.645 \sqrt{\frac{0.05*0.95}{n} }    make n subject of the equation

n = 80.3353

hence  the value of K ≈ 80

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A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student
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The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

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CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

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CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}

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