Conduct a full hypothesis test and determine if the given claim is supported or not supported at the 0.05 significance level. A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 85 items, the defect rate is 5.8% but the manager claims that this is only a sample fluctuation and production is not really out of control. Test whether the manufacturer's claim that production is out of control is supported or not supported.

Given that the manufacturer claims the production process is not out of control, we want to test if the observed defect rate of 5.8% is only due to sample fluctuation or if the process truly has a defect rate exceeding 3%.

Step 1: State the null and alternative hypotheses. The null hypothesis (H₀) should represent the status quo, which in this case is that the production is not out of control (defect rate ≤ 3%). The alternative hypothesis (H₁) is the hypothesis that the data provide evidence in favor of. In this test, it is that the defect rate is greater than 3% (indicating the production might be out of control). So, we have: H₀: p ≤ 0.03 (where p is the true defect rate) H₁: p > 0.03

Step 2: Set the level of significance (α). The significance level is given as 0.05. This is the probability of rejecting the null hypothesis when it is actually true.

Step 3: Calculate the test statistic. For this problem, we will use a z-test since we have a sample proportion and we assume the sample size is large enough. The test statistic for a proportion is calculated as:

\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]

where \(\hat{p}\) is the sample proportion, \(p_0\) is the null hypothesis proportion, and \(n\) is the sample size.

From the problem, we have: \(\hat{p}\) = 5.8/100 = 0.058 \(p_0\) = 3/100 = 0.03 \(n\) = 85

Calculate the test statistic: \[ z = \frac{0.058 - 0.03}{\sqrt{\frac{0.03(1-0.03)}{85}}} \] Solve for z: \[ z ≈ \frac{0.028}{\sqrt{\frac{0.03(0.97)}{85}}} \] \[ z ≈ \frac{0.028}{\sqrt{0.00033}} \] \[ z ≈ \frac{0.028}{0.0182} \] \[ z ≈ 1.5385 \]

Step 4: Determine the p-value. Next, we find the p-value associated with our calculated z-test statistic, which is the probability of obtaining test results at least as extreme as the results observed during the test, under the assumption that the null hypothesis is correct.

With a z-value of 1.5385, you can use a z-table or statistical software to find that the p-value is greater than 0.05 (because the z-value for 0.05 significance level in a one-tailed test is roughly 1.645, which is higher than our z-value).

Step 5: Make a decision to reject or fail to reject the null hypothesis. Since the p-value is greater than the significance level (α = 0.05), we fail to reject the null hypothesis.

Conclusion: There is not enough statistical evidence at the 0.05 significance level to support the claim that the production is out of control (that is, the defect rate is greater than 3%).