/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 The fan blades on commercial jet... [FREE SOLUTION] | 91Ó°ÊÓ

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The fan blades on commercial jet engines must be replaced when wear on these parts indicates too much variability to pass inspection. If a single fan blade broke during operation, it could severely endanger a flight. A large engine contains thousands of fan blades, and safety regulations require that variability measurements on the population of all blades not exceed \(\sigma^{2}=0.18 \mathrm{mm}^{2} .\) An engine inspector took a random sample of 61 fan blades from an engine. She measured each blade and found a sample variance of \(0.27 \mathrm{mm}^{2} .\) Using a 0.01 level of significance, is the inspector justified in claiming that all the engine fan blades must be replaced? Find a \(90 \%\) confidence interval for the population standard deviation.

Short Answer

Expert verified
Yes, the inspector is justified in replacing the fan blades. The test supports variance exceeding 0.18 mm², with a standard deviation confidence interval of (0.20, 0.25) mm.

Step by step solution

01

Understand the Hypotheses

We need to determine if the inspector can claim that the variance exceeds the safety threshold. For this, we perform a hypothesis test for a single population variance. The null hypothesis (\(H_0\)) is that the population variance (\(\sigma^2\)) is \(0.18\ mm^2\), and the alternative hypothesis (\(H_a\)) is that the population variance is greater than \(0.18\ mm^2\). Symbolically, these are:\[ H_0: \sigma^2 = 0.18 \ mm^2 \]\[ H_a: \sigma^2 > 0.18 \ mm^2 \]
02

Calculate the Test Statistic

The test statistic for a chi-square test on variance is given by:\[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \]where \(n\) is the sample size, \(s^2\) is the sample variance, and \(\sigma^2\) is the hypothesized population variance. Substituting the given values:\[ \chi^2 = \frac{(61-1)\times 0.27}{0.18} = \frac{60 \times 0.27}{0.18} = 90\]
03

Determine the Critical Value

For a significance level (\(\alpha\)) of 0.01 and a one-tailed test with \(n-1 = 60\) degrees of freedom, we find the critical value from the chi-square distribution table. The critical chi-square value \(\chi^2_{0.01, 60}\) is approximately 82.53.
04

Compare Test Statistic with Critical Value

We compare the test statistic (90) with the critical value (82.53). Since 90 > 82.53, we reject the null hypothesis \(H_0\). This suggests that the variance indeed exceeds \(0.18\ mm^2\), justifying the inspector's claim to replace all fan blades.
05

Calculate a 90% Confidence Interval for Population Standard Deviation

To find a 90% confidence interval for the population standard deviation \(\sigma\), we use the formula for the chi-square distribution:\[\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{upper}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{lower}}} \right)\]where \(\chi^2_{upper}\) and \(\chi^2_{lower}\) are the chi-square critical values corresponding to the two tails of the distribution. Using 60 degrees of freedom:\(\chi^2_{0.05, 60} \approx 78.23\) and \(\chi^2_{0.95, 60} \approx 43.19\).Thus, the confidence interval is:\[\left( \sqrt{\frac{60 \times 0.27}{78.23}}, \sqrt{\frac{60 \times 0.27}{43.19}} \right) \approx (0.20, 0.25)\ mm\]
06

Conclusion and Decision

The test shows statistical evidence that the variance of the fan blades exceeds the regulation limit. Additionally, the 90% confidence interval for the standard deviation is between 0.20 and 0.25 mm, both higher than the variance-equivalent standard deviation of \(\sqrt{0.18} \approx 0.424\ mm\), supporting the decision to replace all fan blades.

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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 compare and evaluate whether an observed variance significantly differs from a specified value in the population.
In our problem, this test helps to determine whether the variability in the fan blades exceeds the safe limit set by regulations.
To perform this test, you formulate two hypotheses:
  • The null hypothesis (\( H_0 \)): assumes there is no effect or difference, in other words, that the variance is equal to 0.18 mm².
  • The alternative hypothesis (\( H_a \)): which states that the variance is greater than 0.18 mm².

To calculate the chi-square statistic, you use the formula: \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \] where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2 \) is the hypothesized population variance.
This calculated value is then compared against a critical value from the chi-square distribution table to decide whether to reject the null hypothesis.
In this case, since the computed chi-square statistic exceeded the critical value (90 > 82.53), we reject the null hypothesis, supporting the inspector's claim that the variance is indeed beyond the allowable limit.
Population Variance
Population variance is a measure of the variability within a whole group or population. It indicates how much the individual data points differ from the mean of the data set.
A low variance suggests that the data points tend to be very close to the mean, whereas a high variance implies that the data points are spread out over a wider range of values.
In the context of this exercise, the population variance is critical for ensuring safety standards for jet engine fan blades.
The regular variability or variance allowed is \( \sigma^2 = 0.18 \, mm^2 \), which represents a safe threshold for operation.
In our case, a sample was taken to make inferences about the entire engine. The sample revealed a variance of \( 0.27 \, mm^2 \), which prompted the use of hypothesis testing to determine if this level of variability was truly different from the targeted population variance of 0.18 mm².
Using the sample variance and size, the test helps infer whether the variability meets safety criteria and whether actions, like replacement, should be taken.
Level of Significance
The level of significance (\( \alpha \)) in hypothesis testing is a threshold set by the researcher to determine when to reject the null hypothesis.
It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true.
Using a 0.01 level of significance means that there is only a 1% chance of incorrectly concluding that the population variance exceeds the safety limit when it does not.
This stricter level of significance is vital in safety-critical applications like jet engine inspections, as making incorrect conclusions could have serious consequences.
Therefore, by setting a low \( \alpha \), the inspector ensures greater confidence in the decision to replace the fan blades if the null hypothesis is rejected.
In simpler terms: at a 1% significance level, the evidence must be very strong before we decide that the population variance exceeds the safe threshold of 0.18 mm².
Confidence Interval
A confidence interval provides a range of values within which we can say, with a certain level of confidence, that the true population parameter lies.
In this context, we are interested in the confidence interval for the population standard deviation, which is more informative than just knowing if the variance exceeds a certain threshold.
Calculating a 90% confidence interval for the population standard deviation involves using the chi-square distribution: \[\left( \sqrt{\frac{(n-1)s^2}{\chi^2_{\text{upper}}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{\text{lower}}}} \right)\] Here, \(\chi^2_{\text{upper}}\) and \(\chi^2_{\text{lower}}\) are the critical values that mark the upper and lower bounds of the interval, respectively.
The computed confidence interval in our problem was approximately (0.20, 0.25) mm, indicating that we are 90% confident the true standard deviation lies within this range.
This range being higher than \( \sqrt{0.18} \approx 0.424 \text{ mm} \), further supports the decision to replace all fan blades to ensure safety.

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