91Ó°ÊÓ

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

Step by step solution

01

State the Hypotheses

In this scenario, we establish two hypotheses. The null hypothesis \( H_0 \) is that the mean filling weight is 16 ounces: \( H_0: \mu = 16 \). The alternative hypothesis \( H_a \) is that the mean filling weight is not 16 ounces: \( H_a: \mu eq 16 \). This is a two-tailed test because deviations in either direction are concerning.

02

Calculate Test Statistic for Part (b)

Given \( \bar{x} = 16.32 \), \( \mu = 16 \), \( \sigma = 0.8 \), and \( n = 30 \), we calculate the test statistic using the formula \( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \).\[ z = \frac{16.32 - 16}{0.8/\sqrt{30}} = \frac{0.32}{0.146} \approx 2.192 \]

03

Determine p-value for Part (b)

Using a standard normal distribution table, find the p-value corresponding to \( z = 2.192 \). The p-value is approximately 0.0282 for one tail, and since it's a two-tailed test, multiply this by 2 to get 0.0564.

04

Decision for Part (b)

Compare the p-value (0.0564) with \( \alpha = 0.05 \). Since 0.0564 > 0.05, we fail to reject the null hypothesis. Therefore, do not shut down the production line.

05

Calculate Test Statistic for Part (c)

For \( \bar{x} = 15.82 \), use the same calculation with the mean difference.\[ z = \frac{15.82 - 16}{0.8/\sqrt{30}} = \frac{-0.18}{0.146} \approx -1.233 \]

06

Determine p-value for Part (c)

For \( z = -1.233 \), the p-value for one tail is approximately 0.1093. As it's a two-tailed test, the p-value is 0.2186.

07

Decision for Part (c)

Compare the p-value (0.2186) with \( \alpha = 0.05 \). Since 0.2186 > 0.05, we fail to reject the null hypothesis. Therefore, do not shut down the production line.

08

Critical Value Approach

For a significance level of 0.05 in a two-tailed test, the critical z-values are approximately ±1.96. The rejection region for \( H_0 \) is when \( z < -1.96 \) or \( z > 1.96 \).

09

Reevaluate with Critical Values for Part (b)

For \( z = 2.192 \), since 2.192 > 1.96, we reject \( H_0 \) based on the critical value approach, suggesting to shut down the line.

10

Reevaluate with Critical Values for Part (c)

For \( z = -1.233 \), since -1.233 is not less than -1.96, we fail to reject \( H_0 \). Therefore, do not shut down the production line.

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Key Concepts

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

Quality Control

Quality control is a key aspect in manufacturing that ensures products meet a defined standard. In the context of the provided exercise, the focus is on the filling weight of containers on a production line. Ensuring each container is filled to precisely 16 ounces is crucial. Overfilling or underfilling can lead to wastage or dissatisfaction, impacting both costs and customer satisfaction.

To monitor this, a quality control inspector checks a sample of the filled containers every hour. By doing this, they can determine if the process needs adjusting. The sampling involves measuring the weight of each item in a small group (here, 30 items). Quality control helps detect any deviations early, ensuring the production line operates efficiently with minimal errors.

Implementing regular checks as described involves hypothesis testing. It helps decide whether to keep the production running smoothly or to halt it for adjustments when inconsistencies are detected.

Significance Level

The significance level, denoted by \( \alpha \), is a threshold used in hypothesis testing to decide whether to reject the null hypothesis. In our example, the level of significance is 0.05, meaning there is a 5% chance of rejecting the null hypothesis incorrectly. This is a standard threshold in many scientific and industrial processes.

Choosing a significance level of 0.05 strikes a balance between being too lenient and too strict. If it's too lenient, you might miss significant variations (Type II errors). If too stringent, you might overreact to normal variations (Type I errors).

In the case of the production line, setting the \( \alpha = 0.05 \) implies you are willing to take a 5% risk of falsely shutting down the line, when, in fact, there is no substantial deviation from the mean filling weight.

Two-tailed Test

A two-tailed test is used when deviations in either direction—higher or lower—are important. For the production line example, the concern is both overfilling and underfilling the containers, which makes the two-tailed approach suitable.

The null hypothesis in this situation is that the mean weight of the container filling is 16 ounces. The alternative hypothesis, which signifies a problem, is that the mean weight is not 16 ounces, \( H_a: \mu eq 16 \).

By using a two-tailed test, you'd assess how extreme the sample mean is on both ends of the spectrum. This kind of analysis helps in determining if changes in the mean are significant enough to justify shutting down the production line for adjustments. In essence, it provides a more thorough check to ensure quality control without bias towards overfilling or underfilling alone.