Question

In a sample of 500 parts manufactured by a company, the number of defective parts was found to be 42. The company, however, claimed that 6% of their product is defective. Test the claim at 5% level of significance.

Answer 1

This rate is incorrect. Using the equation 500/42 to calculate the percentage of defective parts in the sample we can tell that approximately 12% of the product is defective.

Answer 2

Their claim is false. More than 6% of their products are defective.

Given:

The number of parts in the sample is n= 500.

The number of parts found to be defective, x = 42.

To Find:

Testing the claim that 6% of their product is defective  at 5% level of significance.

Solution:

n = 500, x = 42

Sample proportion p = $\frac{x}{n}$ = $\frac{42}{500}$ = 0.084

Population proportion P = 6% or  P = 0.06 and Q = 0.94

We need to test the null hypothesis H$_{o}$: P = 0.06 against

the alternative hypothesis H$_{1}$: P > 0.06   ( right tailed test)

The test statistic we use is ,

   z = $\frac{p-P}{\sqrt{\frac{PQ}{n} }}$

⇒ z = $\frac{0.084-0.06}{\sqrt{\frac{(0.06)(0.94)}{500} }}$

⇒ z = $\frac{0.024}{\sqrt{0.0001128} }$

⇒ z = 2.25988

Calculated value of z = 2.25988

At 5% significance level tabulated value of $z_{\alpha }$= 1.164

|Calculated value| $\leq$ Tabulated value ⇒ Accept H$_{o}$

Here |2.25988 | > 1.164

So we reject the null hypothesis.

Hence, their claim is false. More than 6% of their products are defective.

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