Mathematics, 16.12.2020 16:40 floressavanna15

requires that the percentage of the plastic shelves that are defective be no more than 10%. If more than 10% are defective, the Prime Factor Company will reject the lot. To test this, the National Applicance Company randomly selects 573 of the shelves and has them inspected. It turns out that 70 are defective. Use this information and a 5% level of significance to determine if National Applicance should reject the batch.

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Answer from: shanekamayo1120

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The National Appliance Company makes refrigerators. One of its suppliers, the Exeter company, produces plastic shelves for the refrigerators and ships them in large lots. The National Appliance Company requires that the percentage of the plastic shelves that are defective be no more than 10%. If more than 10% are defective, the Prime Factor Company will reject the lot. To test this, the National Appliance Company randomly selects 573 of the shelves and has them inspected. It turns out that 70 are defective. Use this information and a 5% level of significance to determine if National Appliance should reject the batch.

The decision rule is

Reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that proportion of defective shelf is less than or equal to 0.10 hence the National Appliance Company should reject that batch of shelfs

Step-by-step explanation:

From the question we are told that

The population of defective plastics is p = 0.10

The sample size is n = 573

The number of shelf that are defective is k = 70

The level of significance is $\alpha = 0.05$

Generally the sample proportion is mathematically represented as

$\^ p = \frac{k}{n}$

=> $\^ p = 0.1222$

The null hypothesis is $H_o : p \le 0.10$

The alternative hypothesis is H_a : p 0.10

Generally the test statistics is mathematically represented as

$z = \frac{\^ p - p }{ \sqrt{\frac{p (1 - p)}{ n } } }$

=> $z = \frac{ 0.1222 - 0.10 }{ \sqrt{\frac{0.10 (1 - 0.10 )}{ 573 } } }$

=> z = 1.77

From the z table the area under the normal curve to the right corresponding to 1.77 is

p-value = P(Z 1.77 ) = 0.038364

From the question we are told that $p-value < \alpha$ hence

The decision rule is

Reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that proportion of defective shelf is less than or equal to 0.10 hence the company should reject theat batch of shelfs

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It’s hard to understand what you put.. can you show a picture ?

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idek

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