91Ó°ÊÓ

Lazurus Steel Corporation produces iron rods that are supposed to be 36 inches long. The machine that makes these rods does not produce each rod exactly 36 inches long. The lengths of the rods are normally distributed, and they vary slightly. It is known that when the machine is working properly, the mean length of the rods is 36 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to \(.035\) inch. The quality control department at the company takes a sample of 20 such rods every week, calculates the mean length of these rods, and tests the null hypothesis, \(\mu=36\) inches, against the alternative hypothesis, \(\mu \neq 36\) inches. If the null hypothesis is rejected, the machine is stopped and adjusted. A recent sample of 20 rods produced a mean length of \(36.015\) inches. a. Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control inspector decide to stop the machine and adjust it if he chooses the maximum probability of a Type I error to be \(.02 ?\) What if the maximum probability of a Type I error is .10? b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.02 .\) Does the machine need to be adjusted? What if \(\alpha=.10\) ?

Step by step solution

01

Calculate the Test Statistic

First, the test statistic needs to be calculated. This is done using the formula: \(Z = \frac{(\overline{X} - \mu)}{(\sigma / \sqrt{n})}\). Where, \(\overline{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size. Plugging in the given values, \(Z = \frac{(36.015 - 36)}{(.035 / \sqrt{20})}\).

02

Calculate the P-value

The p-value represents the probability of observing a statistic as extreme as the test statistic. Because we are running a two-tailed test, we need to find the probability that a Z-score is less than our negative test statistic or greater than our positive test statistic. Using the standard normal distribution table or a calculator that is programmed to solve such problems, obtain the p-value.

03

Decide based on P-value and Type I Error Tolerance

If the determined p-value is less than our chosen level of significance (.02 or .10), we reject the null hypothesis. This implies that the machine requires adjustments. If the p-value is not less than the level of significance, we do not reject the null hypothesis, concluding the machine doesn't need adjustments.

04

Apply Critical-value Approach

The critical-value approach involves comparing the absolute value of the test statistic with the critical value associated with the selected level of significance. If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. The z critical values for two-tailed tests with \(\alpha = .02\) and \(\alpha = .10\) are commonly obtained from statistical tables or calculators.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

p-value calculation

In hypothesis testing, the p-value is a crucial metric that shows the probability of obtaining results at least as extreme as the observed results, given that the null hypothesis is true. For the exercise, we need to calculate the p-value for a test statistic resulting from a two-tailed test of iron rod lengths. To find the p-value, we first calculate the test statistic using the formula:

• \[ Z = \frac{(\overline{X} - \mu)}{(\sigma / \sqrt{n})} \]

Here, \( \overline{X} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the standard deviation, and \( n \) is the sample size. In this case, the calculation yields a Z-score, which can then be used with a standard normal distribution table to find the p-value. The p-value tells us how extreme the observed or more extreme results are under the null hypothesis, which in this case is that the mean rod length is exactly 36 inches.

Type I error

A Type I error occurs when we wrongly reject a true null hypothesis. This is a critical consideration in hypothesis testing and quality control. In the context of the Lazurus Steel Corporation exercise, if the null hypothesis that the average iron rod length is 36 inches is rejected unjustly, it leads to unnecessary machine adjustments. Our control over making a Type I error is set by the significance level \( \alpha \), chosen as 0.02 or 0.10.

• A significance level of 0.02 means there is a 2% risk of making a Type I error.
• Similarly, 0.10 allows for a larger, 10% risk.

Balancing the significance level is essential since a too-low \( \alpha \) might make us miss significant deviations, while a too-high \( \alpha \) increases the rate of false alarms.

two-tailed test

A two-tailed test is used in hypothesis testing when deviations in both directions from the null hypothesis are considered significant. For the exercise, Lazurus Steel's quality control involves checking if the mean rod length deviates from exactly 36 inches in either direction.

• This means, we are interested in both instances where the rod length could be lesser or greater than 36 inches.
• The test checks not only for the mean being smaller or larger but checks for any significant variations in both directions.

The critical values or p-values are thus found in both tails of the standard normal distribution curve to determine the extent of deviation from the assumed mean. If the p-value found is smaller than the set \( \alpha \), the null hypothesis is rejected, indicating a significant deviation in either direction from the expected outcome.

standard normal distribution

The standard normal distribution is a foundational concept in statistics, especially in hypothesis testing. It is a normal distribution with a mean of 0 and a standard deviation of 1. When assessing the quality of Lazurus Steel's iron rods, the normal distribution of the rod lengths is considered.

• Once we compute the Z-score from the sample data, we refer it to the standard normal distribution.
• The Z-score tells us how many standard deviations an element is from the mean.

In this exercise, using the standard normal distribution allows us to find the probability associated with the test statistic, guiding conclusions about whether to adjust the manufacturing machine.

quality control

Quality control in manufacturing processes, like those used by Lazurus Steel Corporation, ensures products meet specified standards. Ensuring rods are close to the desired 36 inches requires regular hypothesis tests. This process involves:

• Sampling rods regularly to check for significant deviation from the target mean length.
• Stopping and adjusting the machinery when deviations are statistically significant.

Through continuous quality control, the company can maintain the precision in rod lengths, minimizing defective products and maintaining customer satisfaction. The hypothesis testing plays a pivotal role here, ensuring the machinery operates within acceptable variability limits.