   # Production Line Fill Weights. 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 σ = .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 α = .05. a. State the hypothesis test for this quality control application. b. If a sample mean of x ¯ = 16.32 ounces were found, what is the p -value? What action would you recommend? c. If a sample mean of 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? ### Essentials Of Statistics For Busin...

9th Edition
David R. Anderson + 4 others
Publisher: South-Western College Pub
ISBN: 9780357045435

#### Solutions

Chapter
Section ### Essentials Of Statistics For Busin...

9th Edition
David R. Anderson + 4 others
Publisher: South-Western College Pub
ISBN: 9780357045435
Chapter 9, Problem 64SE
Textbook Problem
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## Production Line Fill Weights. 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 σ = .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 α = .05. a. State the hypothesis test for this quality control application. b. If a sample mean of x ¯ = 16.32 ounces were found, what is the p-value? What action would you recommend? c. If a sample mean of 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?

a.

To determine

Identify the null and alternative hypotheses for the given situation.

The null and alternative hypotheses are given below:

Null hypothesis:H0:μ=16.

Alternative hypothesis:Ha:μ16 (claim).

### Explanation of Solution

The given information is that the mean filling weight is 16 ounces.

Here, the claim is that overfilling or underfilling exists. This can be written as μ16. The complement of the claim is μ=16. In the given experiment, the alternative hypothesis indicates the claim.

The null and alternative hypotheses are given below:

Null hypothesis:

H0:μ=16.

Alternative hypothesis:

Ha:μ16 (claim).

b.

To determine

Find the p-value when x¯=16.32 and explain the recommended action.

The p-value is 0.0286.

### Explanation of Solution

Calculation:

The given information is that n=30, x¯=16.32, and σ=0.8.

Test statistic and p-value:

The formula for finding the test statistic is as follows:

z=x¯μ0σn

Here, x¯ represents the sample mean, μ0 represents the hypothesized value of the population mean, σ represents the population standard deviation, and n represents the sample size.

Substitute 16.32 for x¯, 16 for μ0, 0.8 for σ, and 30 for n in z formula.

z=16.32160.830=0.320.85.4772=0.320.1461=2.19

Thus, the value of the test statistic is 2.19.

In this case, z is greater than 0. Therefore, the p-value is two times the upper tail area.

That is,

p-value=2(Upper tail area)=2(1Lower tail area)

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value 2.1 in the first column.
• Locate the value 0.09 in the first row.
• The intersecting value that corresponds to 2.19 is 0.9857.

p-value=2(10.9857)=2(0.0143)=0.0286

Thus, the p-value is 0.0286.

Rejection rule:

If p-valueα, reject the null hypothesis.

If p-value>α, do not reject the null hypothesis.

Conclusion:

Here, the p-value is less than the level of significance.

That is, p-value(=0.0286)<α(=0.05).

By the rejection rule, the null hypothesis is rejected.

Hence, there is evidence that overfilling or underfilling exists. That is, the production line operates properly. Thus, the actions shut down and readjusting the production line would be recommended.

c.

To determine

Find the p-value when x¯=15.82 and explain the recommended action.

The p-value is 0.218.

### Explanation of Solution

Calculation:

The given information is that n=30, x¯=15.82, and σ=0.8.

Test statistic and p-value:

Substitute 15.82 for x¯, 16 for μ0, 0.8 for Ïƒ, and 30 for n in z formula.

z=15.82160.830=0.180.85.4772=0.180.1461=1.23

Thus, the value of the test statistic is –1.23.

In this case, z is less than 0. Therefore, the p-value is two times the lower tail area.

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value –1.2 in the first column.
• Locate the value 0.03 in the first row.
• The intersecting value that corresponds to–1.23 is 0.1093.

Thus, the p-value is 0.2186 (=2×0.1093).

Conclusion:

Here, the p-value is greater than the level of significance.

That is, p-value(=0.2186)>α(=0.05).

By the rejection rule, the null hypothesis is not rejected.

Hence, there is no evidence that overfilling or underfilling exists. That is, the production line does not operate properly. Thus, the action to continue the production process would be recommended.

d.

To determine

Find the rejection rule using the critical value approach.

Find the conclusion when x¯=16.32 and x¯=15.82 and then compare the result with Parts (b) and (c).

The rejection rule using the critical value is as follows:

If z1.96 or z1.96, reject the null hypothesis.

The conclusion for x¯=16.32; there is evidence that overfilling or underfilling exists.

The conclusion for x¯=15.82; there is no evidence that overfilling or underfilling exists.

### Explanation of Solution

Calculation:

The rejection rule for a two-tailed test using the critical value is as follows:

If zzα2 or zzα2, reject the null hypothesis.

Critical value:

If zzα2 or zzα2, reject the null hypothesis.

Critical value:

α=0.05α2=0.052=0.025

The cumulative area to the left is calculated as follows:

Area to the left=1Area to the right=10.025=0.975

Use Table 1: Cumulative probabilities for the standard normal distribution to find zα2.

• Locate the area of 0.975 in the body of Table 1.
• Move left until the first column and note the value as 1.9.
• Move upward until the top row is reached and note the value as 0.06.

Thus, the critical values of zα2 and zα2 are 1.96 and –1.96, respectively.

Rejection rule:

If z1.96 or z1.96, reject the null hypothesis.

Conclusion for x¯=16.32:

Here, the test statistic is greater than the positive critical value.

That is, z(=2.19)>zα(=1.96).

By the rejection rule, the null hypothesis is rejected.

Hence, there is evidence that overfilling or underfilling exists. That is, the production line operates properly. Thus, the actions shut down and readjusting the production line would be recommended.

Conclusion for x¯=15.82:

Here, the test statistic is greater than the negative critical value and less than the positive critical value.

That is, zα(=1.96)>z(=1.23)>zα(=1.96).

By the rejection rule, the null hypothesis is not rejected.

Hence, there is no evidence that overfilling or underfilling exists. That is, the production line does not operate properly. Thus, the action to continue the production process would be recommended.

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