Mathematics, 09.12.2020 21:10 JustinLaucirica

A supplier of digital memory cards claims that less than 1% of the cards are defective. In a random sample of600 memory cards, it is found that 3% are defective, but the supplier claims that this is only a sample fluctuation. At the 0.01 level of significance, test the supplier's claim that less than 1% are defective. Use the p-value method.

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Answer from: AdoNice

The decision rule is

Fail to reject the null hypothesis

The conclusion is

The is no sufficient evidence to support the claim that the less than 1% of the cards are defective

Step-by-step explanation:

From the question we are told that

The sample size is n = 600

The population proportion of defective samples is p = 0.01

The sample proportion of defective samples is $\^ p = 0.03$

The level of significance is $\alpha = 0.01$

The null hypothesis is H_o : p = 0.01

The alternative hypothesis is H_a: p < 0.01

Generally the test statistics is mathematically represented as

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

=> $t = \frac{0.03 - 0.01 }{ \sqrt{ \frac{0.01( 1 - 0.01 )}{600} } }$

=> t = 0.00406

From the z table the area under the normal curve to the left corresponding to 0.00406 is

p-value = P(Z < 0.00406 ) = 0.50162

comparing the p-valve and the level of significance we see that the

$p-value \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

The is no sufficient evidence to support the claim that the less than 1% of the cards are defective

Answer from: Quest

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Answer from: Quest

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