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Chapter 10: Problem 10

A machine fills bottles with 64 fluid ounces of liquid. The quality-control manager determines that the fill levels are normally distributed with a mean of 64 ounces and a standard deviation of 0.42 ounce. He has an engineer recalibrate the machine in an attempt to lower the standard deviation. After the recalibration, the quality-control manager randomly selects 19 bottles from the line and determines that the standard deviation is 0.38 ounce. Is there less variability in the filling machine? Use the \(\alpha=0.01\) level of significance.

Short Answer

Expert verified
No, there is insufficient evidence to conclude that the variability in the filling machine has decreased at the 伪 = 0.01 level of significance.

Step by step solution

01

State the Hypotheses

Identify the null hypothesis (H_0) and the alternative hypothesis (H_a). The null hypothesis (H_0): 蟽 = 0.42 ounce (the standard deviation has not decreased). The alternative hypothesis (H_a): 蟽 < 0.42 ounce (the standard deviation has decreased).
02

Choose the Significance Level

Determine the level of significance. The problem gives 伪 = 0.01.
03

Identify the Test Statistic

Since we are dealing with standard deviations and sample size is relatively small, we use the Chi-square (蠂^2) statistic. The test statistic formula for standard deviation is: 蠂^2 = (n-1)s^2 / 蟽^2, where n is the sample size, s is the sample standard deviation, and 蟽 is the population standard deviation.
04

Calculate the Test Statistic

Substitute the given values into the formula. Here are the values: n = 19, s = 0.38, and 蟽 = 0.42. 蠂^2 = (19-1)(0.38)^2 / (0.42)^2 = 18 * 0.1444 / 0.1764 鈮� 14.73
05

Determine the Critical Value

Find the critical 蠂^2 value for df = n - 1 degrees of freedom at 伪 = 0.01 significance level using Chi-square distribution tables. Degrees of freedom = 19 - 1 = 18. Critical value at 伪 = 0.01 for 蠂^2 with 18 df is approximately 蠂^2_{critical} = 10.28.
06

Compare the Test Statistic to the Critical Value

Compare the calculated 蠂^2 value to the critical 蠂^2 value. If the test statistic is less than the critical value, we reject the null hypothesis. Here, 蠂^2 = 14.73 is greater than 蠂^2_{critical} = 10.28.
07

Conclusion

Based on the comparison, we do not reject the null hypothesis. There is insufficient evidence at the 伪 =0.01 level of significance to conclude that the standard deviation has decreased.

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Key Concepts

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

Chi-square test
The Chi-square test is a statistical method used to determine if there is a significant difference between the expected and observed data. In this exercise, we're evaluating whether the standard deviation of the bottle-fill levels has decreased after recalibration. The Chi-square test is appropriate here because it specifically tests variance and standard deviation. This test uses the formula: \( \ chi^2 = (n-1)s^2 / \sigma^2 \), where \( n \) is the sample size, \( s \) is the sample standard deviation, and \( \sigma \) is the hypothesized population standard deviation. By comparing the test statistic against a critical value from the Chi-square distribution table, we can decide whether to reject or accept the null hypothesis.
Standard deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the individual data points deviate from the mean of the data set. In the exercise, the initial standard deviation is \(0.42\) ounces, meaning on average, the fill levels of the bottles differ by \(0.42\) ounces from the mean of \(64\) ounces. After recalibration, the sample standard deviation is \(0.38\) ounces. The goal is to determine if this reduction is statistically significant, indicating less variability in the bottle-filling process.
Significance level
The significance level, denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It defines the threshold for determining whether a test statistic is extreme enough to reject the null hypothesis. In this exercise, \( \alpha \) is set to \(0.01\). This means we are 99% confident in our results, and there is a 1% chance of committing a Type I error 鈥� falsely claiming that the recalibration has decreased the standard deviation when it has not.
Null hypothesis
The null hypothesis \( (H_0)\) represents the default or initial assumption that there is no effect or no difference. In the exercise, \( H_0 \) states that the standard deviation has not changed from \(0.42\) ounces after the machine's recalibration. This hypothesis is tested against the alternative hypothesis to determine if there is sufficient evidence to support a change.
Alternative hypothesis
The alternative hypothesis \( (H_a) \) is what we aim to support with our test. It proposes that there is a significant effect or difference. In this exercise, \( H_a \) states that the standard deviation has decreased from \(0.42\) ounces. To support \( H_a \), the test statistic must be less than the critical Chi-square value; a lower test statistic suggests reduced variability, implying that the recalibration was effective.

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Most popular questions from this chapter

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