Question

In manufacturing processes, it is of interest to know with confidence the proportion of defective parts. Suppose that we want to be reasonably certain that less than 4% of a company's widgets are defective. To test this, we obtain a random sample of 250 widgets from a large batch. Each of the 250 widgets is tested for defects, and 6 are determined to be defective, based upon the manufacturer's standards. Using α = 0.01, is this evidence that less than 4% of the company's widgets are defective? State the hypotheses, list and check the conditions, calculate the test statistic, find the p-value, and make a conclusion in a complete sentence related to the scenario.

Answer 1

Answer:

Less than 4% of a company's widgets are defective.

Step-by-step explanation:

In this case we want to be reasonably certain that less than 4% of a company's widgets are defective.

The significance level of the test is, α = 0.01.

The hypothesis can be defined as follows:  

H₀: At least 4% of a company's widgets are defective, i.e. p ≥ 0.04.  

Hₐ: Less than 4% of a company's widgets are defective, i.e. p < 0.04.  

The information provided is:  

n = 250

x = 6

The sample proportion is, $\hat p=\frac{x}{n}=\frac{6}{250}=0.024$

Compute the test statistic value as follows:  

 $z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}\\\\=\frac{0.024-0.04}{\sqrt{\frac{0.04(1-0.04)}{250}}}\\\\=-1.29$

The test statistic value is -1.29.  

The decision rule is:  

The null hypothesis will be rejected if the p-value of the test is less than the significance level.  

Compute the p-value as follows:  

 $p-value=P(Z<-1.29)=0.0985$

So,

p-value = 0.0985 > α = 0.01.  

The null hypothesis will not be rejected at 1% significance level.  

Thus, there is not enough evidence to support the claim.

Conclusion:

Less than 4% of a company's widgets are defective.