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Chapter 10: Problem 11

A manufacturer of high-strength, lowalloy steel beams requires that the standard deviation of yield strength not exceed 7000 pounds per square inch (psi). The quality-control manager selected a sample of 20 steel beams and measured their yield strength. The standard deviation of the sample was 7500 psi. Assume that yield strengths are normally distributed. Does the evidence suggest that the standard deviation of yield strength exceeds 7000 psi at the \(\alpha=0.01\) level of significance?

Short Answer

Expert verified
Do not reject \textbf{\text{H}}_0. There is insufficient evidence to suggest the standard deviation exceeds 7000 psi at the \(\alpha=0.01 \) level.

Step by step solution

01

- State the Hypotheses

We need to set up the null and alternative hypotheses. The null hypothesis (\textbf{\text{H}}_0) is that the population standard deviation is 7000 psi, i.e., \( \textbf{\text{H}}_0: \sigma = 7000 \). The alternative hypothesis (\textbf{\text{H}}_a) is that the population standard deviation exceeds 7000 psi, i.e., \( \textbf{\text{H}}_a: \sigma > 7000 \).
02

- Determine the Test Statistic

Since we are testing a sample standard deviation against a population standard deviation, we use the chi-square distribution. The test statistic is given by: \[ \chi^2 = \frac{(n-1)s^2}{\theta^2} \] where \( n \) is the sample size, \( s \) is the sample standard deviation, and \( \theta \) is the population standard deviation. Thus, \[ \chi^2 = \frac{(20-1)(7500)^2}{(7000)^2} = \frac{19(56250000)}{49000000} = 21.806 \]
03

- Determine the Degrees of Freedom and the Critical Value

The degrees of freedom for this test is \( n-1=20-1=19 \). We need to find the critical value of \( \chi^2 \) at \(\alpha=0.01 \)for a chi-square distribution with 19 degrees of freedom. Consulting the chi-square table, the critical value is \[ \chi^2_{0.01,19 }= 36.19. \]
04

- Make the Decision

Since our calculated test statistic \( \chi^2 = 21.806 \) is less than the critical value 36.19, we do not reject the null hypothesis.
05

- Conclusion

At the \(\alpha=0.01 \) level of significance, there is not enough evidence to suggest that the standard deviation of yield strength exceeds 7000 psi.

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Key Concepts

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

Chi-Square Distribution
In hypothesis testing, the chi-square distribution is critical for various types of tests, especially tests involving the population variance. It is essential when you want to understand the variability within your data based on a sample. The chi-square distribution is asymmetric and its shape depends on the degrees of freedom (df). The more degrees of freedom, the more the distribution looks like a normal distribution. For our problem, we use the chi-square distribution because we are comparing our sample standard deviation to a known population standard deviation. The formula to compute the test statistic is: $$\chi^2 = \frac{(n-1)s^2}{\theta^2}$$ where \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\theta\) is the population standard deviation. The test statistic follows a chi-square distribution with \(n-1\) degrees of freedom.
Sample Standard Deviation
The sample standard deviation measures the dispersion of sample data points around the sample mean. It is crucial when estimating and making conclusions about the population. The formula for sample standard deviation (s) is: $$s = \sqrt{\frac{\sum (X_i - \overline{X})^2}{n-1}}$$ where \(X_i\) are the sample data points, \(\overline{X}\) is the sample mean, and \(n\) is the sample size. In our exercise, the sample standard deviation is 7500 psi. We use this value in our test statistic formula to compare against the population standard deviation of 7000 psi.
Level of Significance
The level of significance (\(\alpha\)) is the probability of rejecting the null hypothesis when it is actually true. It is a threshold set by the researcher before conducting the hypothesis test. Common levels of significance are \(0.05\), \(0.01\), and \(0.10\). For our exercise, the level of significance is \(0.01\). This means there is a 1% risk of concluding that the population standard deviation is greater than 7000 psi when it actually is not. We use this \(\alpha\) value to determine the critical value from the chi-square distribution table.
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) represents what we are trying to prove. It contradicts the null hypothesis (\(H_0\)). In our scenario, the null hypothesis is that the population standard deviation is 7000 psi (\(\sigma = 7000\)). The alternative hypothesis states that the population standard deviation exceeds 7000 psi (\(\sigma > 7000\)). We seek to gather evidence to support this claim using our sample data. If our test statistic exceeds the critical value from the chi-square distribution, we reject the null hypothesis in favor of the alternative. However, in this problem, the computed chi-square is less than the critical chi-square value, leading us to not reject the null hypothesis.

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Most popular questions from this chapter

According to menstuff.org, \(22 \%\) of married men have "strayed" at least once during their married lives. A survey of 500 married men indicated that 122 have strayed at least once during their married life. Does this survey result contradict the results of menstuff.org? (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, determine the probability of making a Type II error if the true population proportion is \(0.25 .\) What is the power of the test? (c) Redo part (b) if the true proportion is 0.20 .

Simulation The exponential probability distribution can be used to model waiting time in line or the lifetime of electronic components. Its density function is skewed right. Suppose the wait time in a line can be modeled by the exponential distribution with \(\mu=\sigma=5\) minutes. (a) Simulate obtaining 100 simple random samples of size \(n=10\) from the population described. That is, simulate obtaining a simple random sample of 10 individuals waiting in a line where the wait time is expected to be 5 minutes. (b) Test the null hypothesis \(H_{0}: \mu=5\) versus the alternative \(H_{1}: \mu \neq 5\) for each of the 100 simulated simple random samples. (c) If we test this hypothesis at the \(\alpha=0.05\) level of significance, how many of the 100 samples would you expect to result in a Type I error? (d) Count the number of samples that lead to a rejection of the null hypothesis. Is it close to the expected value determined in part (c)? What might account for any discrepancies?

In his book, "The Signal and the Noise," Nate Silver analyzed 733 predictions made by experts regarding political events. Of the 733 predictions, 338 were mostly true. (a) Determine the sample proportion of political predictions that were mostly true. (b) Suppose that we want to know whether the evidence suggests the political predictions were mostly true less thar half the time. State the null and alternative hypotheses. (c) Verify the normal model may be used to determine the \(P\) -value for this hypothesis test. (d) Draw a normal model with the area representing the \(P\) -val shaded for this hypothesis test. (e) Determine the \(P\) -value based on the model from part (d). (f) Interpret the \(P\) -value. (g) Based on the \(P\) -value, what does the sample evidence suggest? That is, what is the conclusion of the hypothesis test? Assume an \(\alpha=0.1\) level of significance.

Throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1 year, 53 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested \(H_{0}: p=0.5\) versus \(H_{1}: p>0.5\) and obtained a \(P\) -value of \(0.2743 .\) Explain what this \(P\) -value means and write a conclusion for the researcher.

A pharmaceutical company manufactures a 200-milligram (mg) pain reliever. Company specifications require that the standard deviation of the amount of the active ingredient must not exceed \(5 \mathrm{mg}\). The quality-control manager selects a random sample of 30 tablets from a certain batch and finds that the sample standard deviation is \(7.3 \mathrm{mg}\). Assume that the amount of the active ingredient is normally distributed. Determine whether the standard deviation of the amount of the active ingredient is greater than \(5 \mathrm{mg}\) at the \(\alpha=0.05\) level of significance.
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