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Chapter 10: Problem 11

The piston diameter of a certain hand pump is 0.5 inch. The quality-control manager determines that the diameters are normally distributed, with a mean of 0.5 inch and a standard deviation of 0.004 inch. The machine that controls the piston diameter is recalibrated in an attempt to lower the standard deviation. After recalibration, the quality-control manager randomly selects 25 pistons from the production line and determines that the standard deviation is 0.0025 inch. Was the recalibration effective? Use the \(\alpha=0.01\) level of significance.

Short Answer

Expert verified
The recalibration was effective.

Step by step solution

01

- State the Hypotheses

Formulate the null hypothesis (\textbf{H}_0) and the alternative hypothesis (\textbf{H}_a).\(\textbf{H}_0\): The population standard deviation is the same before and after recalibration, \(\sigma = 0.004\). \(\textbf{H}_a\): The population standard deviation has decreased after recalibration, \(\sigma < 0.004\).
02

- Identify the Test Statistic

The sample standard deviation after recalibration is S = 0.0025 inch from a sample size of n = 25 pistons. Use the chi-squared test for standard deviation: \[\chi^2 = \frac{(n-1)S^2}{\sigma^2}\].
03

- Calculate the Test Statistic

Substitute the given values into the test statistic formula: \[\chi^2 = \frac{(25-1)\times (0.0025)^2}{(0.004)^2} = \frac{24\times 0.00000625}{0.000016} = 9.375\].
04

- Determine the Critical Value

At \(\alpha = 0.01\), the degrees of freedom (df) is \(n-1 = 24\). Find the critical value from the chi-squared distribution table for a one-tailed test. \(\chi^2_{0.01, 24} = 13.848\).
05

- Make a Decision

Compare the test statistic to the critical value. If \(\chi^2\) is less than the critical value, reject the null hypothesis. In this case, \(9.375 < 13.848\), so reject \(\textbf{H}_0\).
06

- Conclusion

Since the null hypothesis is rejected, there is sufficient evidence to conclude that the recalibration was effective in lowering the standard deviation of the piston diameter.

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Key Concepts

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

Chi-Squared Test
The Chi-Squared Test is a statistical method used to test the variance or standard deviation of a sample against a known value or another sample's variance. In the given exercise, it helps determine if the recalibration of the machine was effective in reducing the piston diameter's standard deviation. Chi-squared tests are especially handy when the data is normally distributed.

The formula for the Chi-squared test statistic is \[ \chi^2 = \frac{(n-1)S^2}{\sigma^2} \].

In our case:
  • \( n \) is the sample size, which is 25.
  • \( S \) is the sample standard deviation after recalibration, which is 0.0025 inches.
  • \( \sigma \) is the population standard deviation before recalibration, which is 0.004 inches.
Using these values, the test statistic is calculated to be 9.375. To reach a conclusion, we also need a critical value from the Chi-squared distribution table based on the chosen significance level and degrees of freedom.
Significance Level
The significance level, often denoted by \( \alpha \), is the threshold used to decide whether a result is statistically significant. In simpler terms, it's the probability of rejecting the null hypothesis when it is actually true. For this problem, the significance level is set at 0.01, meaning there is a 1% risk of concluding that the recalibration was effective when it actually wasn't.

To find the critical value at this significance level, we use the degrees of freedom, which is defined as \( n-1 \). In our scenario, the degrees of freedom are 24 (because \( 25-1 = 24 \)).

Looking up in the Chi-squared distribution table for \( \alpha = 0.01 \) and 24 degrees of freedom, we find the critical value to be 13.848. Therefore, our test statistic (9.375) is compared against this critical value to make our decision.
Standard Deviation
Standard deviation measures the amount of variation or dispersion in a set of values. A low standard deviation means the values are close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In the context of the exercise, the standard deviation of the piston diameter before recalibration was 0.004 inches. The goal of recalibration was to reduce this standard deviation, and a sample of 25 pistons was tested afterwards, showing a new standard deviation of 0.0025 inches.

Using the Chi-squared test, we were able to determine if this reduction was statistically significant. Since our test statistic (9.375) was less than the critical value (13.848), we rejected the null hypothesis and concluded that the recalibration was indeed effective at reducing the standard deviation. This implies that the piston diameters are now more consistent and closer to the target diameter of 0.5 inches.

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Most popular questions from this chapter

A simple random sample of size \(n=200\) individuals with a valid driver's license is asked if they drive an American-made automobile. Of the 200 individuals surveyed, 115 responded that they drive an American-made automobile. Determine if a majority of those with a valid driver's license drive an American made automobile at the \(\alpha=0.05\) level of significance.

To test \(H_{0}: p=0.25\) versus \(H_{1}: p \neq 0.25,\) a simple random sample of \(n=350\) individuals is obtained and \(x=74\) successes are observed. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, compute the probability of making a Type II error if the true population proportion is 0.23. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.28 .

If we do not reject the null hypothesis when the statement in the alternative hypothesis is true, we have made a Type ____ error.

What's the Problem? The head of institutional research at a university believes that the mean age of full-time students is declining. In \(1995,\) the mean age of a full-time student was known to be 27.4 years. After looking at the enrollment records of all 4934 full-time students in the current semester, he found that the mean age was 27.1 years, with a standard deviation of 7.3 years. He conducted a hypothesis of \(H_{0}: \mu=27.4\) years versus \(H_{1}: \mu<27.4\) years and obtained a \(P\) -value of \(0.0019 .\) He concluded that the mean age of full-time students did decline. Is there anything wrong with his research?

Student loan debt has reached record levels in the United States. In a random sample of 100 individuals who have student loan debt, it was found the mean debt was 23,979 dollar with a standard deviation of 31,400 dollar . Data based on results from the Federal Reserve Bank of New York. (a) What do you believe is the shape of the distribution of student loan debt? Explain. (b) Use this information to estimate the mean student loan debt among all with such debt at the \(95 \%\) level of confidence. Interpret this result. (c) What could be done to increase the precision of the estimate?
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