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.

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,

Compute the test statistic value as follows:  

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:  

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.

22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

happy to

pls mark as brainliest

step-by-step explanation:

the answer is 128