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:

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.

Final answer:

This is a hypothesis testing problem where we test the claim that less than 4% of widgets are defective. We set the null and alternative hypotheses, confirm conditions for a binomial distribution, compute the test statistic, find the p-value and then make a conclusion based on the comparison of p-value with the given significance level.

Explanation:

In this scenario, we are interested in testing the hypothesis about the proportion of defective widgets. We define our null hypothesis (H0) and the alternative hypothesis (Ha) as follows:

H0: p = 0.04  (The proportion of defective widgets is 4%)

Ha: p < 0.04  (The proportion of defective widgets is less than 4%)

The conditions for a binomial distribution are met here, as each widget is either defective or not, and each widget is tested independently. Also, the quantities np and nq (where n is the sample size and q is the probability of failure) are greater than five, so we can approximate by the normal distribution.

We calculate the test statistic using the formula: z = (p' - p) / sqrt [ (p * q) / n ]

Where, p' is the sample proportion, which is 6/250, p is the hypothesized proportion which is 0.04, q is 1 - p and n is the sample size (250). This gives us a z value. Then, we find the p-value from the standard normal distribution using this z value. If p-value < α (0.01), we reject the null hypothesis. Otherwise, we do not reject it.

At the end, you will conclude. If we reject the null, we say, 'At the 1 percent significance level, there is sufficient evidence to conclude that less than 4% of the company's widgets are defective'. If we don't reject the null, 'At the 1 percent significance level, there is insufficient evidence to conclude that less than 4% of the company's widgets are defective.'

Learn more about Hypothesis testing here:

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