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 ?2 = 0.18 mm2. An engine inspector took a random sample of 71 fan blades from an engine. She measured each blade and found a sample variance of 0.28 mm2. Using a 0.01 level of significance, is the inspector justified in claiming that all the engine fan blades must be replaced? (a) What is the level of significance? State the null and alternate hypotheses. H0: ?2 = 0.18; H1: ?2 < 0.18H0: ?2 = 0.18; H1: ?2 ≠ 0.18 H0: ?2 = 0.18; H1: ?2 > 0.18H0: ?2 > 0.18; H1: ?2 = 0.18 (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) What are the degrees of freedom? What assumptions are you making about the original distribution? We assume a normal population distribution. We assume a exponential population distribution. We assume a uniform population distribution. We assume a binomial population distribution. (c) Find or estimate the P-value of the sample test statistic. P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > ?, we fail to reject the null hypothesis. Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis. Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 1% level of significance, there is insufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is not justified in claiming the blades must be replaced.At the 1% level of significance, there is sufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is justified in claiming the blades must be replaced. At the 1% level of significance, there is sufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is not justified in claiming the blades must be replaced.At the 1% level of significance, there is insufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is justified in claiming the blades must be replaced.

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 ?2 = 0.18 mm2. An engine inspector took a random sample of 71 fan blades from an engine. She measured each blade and found a sample variance of 0.28 mm2. Using a 0.01 level of significance, is the inspector justified in claiming that all the engine fan blades must be replaced? (a) What is the level of significance? State the null and alternate hypotheses. H0: ?2 = 0.18; H1: ?2 < 0.18H0: ?2 = 0.18; H1: ?2 ≠ 0.18 H0: ?2 = 0.18; H1: ?2 > 0.18H0: ?2 > 0.18; H1: ?2 = 0.18 (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) What are the degrees of freedom? What assumptions are you making about the original distribution? We assume a normal population distribution. We assume a exponential population distribution. We assume a uniform population distribution. We assume a binomial population distribution. (c) Find or estimate the P-value of the sample test statistic. P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Since the P-value > ?, we fail to reject the null hypothesis. Since the P-value > ?, we reject the null hypothesis. Since the P-value ≤ ?, we reject the null hypothesis. Since the P-value ≤ ?, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 1% level of significance, there is insufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is not justified in claiming the blades must be replaced.At the 1% level of significance, there is sufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is justified in claiming the blades must be replaced. At the 1% level of significance, there is sufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is not justified in claiming the blades must be replaced.At the 1% level of significance, there is insufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is justified in claiming the blades must be replaced.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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(a) What is the level of significance?

State the null and alternate hypotheses.

H₀: ?² = 0.18; H₁: ?² < 0.18H₀: ?² = 0.18; H₁: ?² ≠ 0.18 H₀: ?² = 0.18; H₁: ?² > 0.18H₀: ?² > 0.18; H₁: ?² = 0.18

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)

What are the degrees of freedom?

What assumptions are you making about the original distribution?

We assume a normal population distribution.

We assume a exponential population distribution.

We assume a uniform population distribution.

We assume a binomial population distribution.

P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > ?, we fail to reject the null hypothesis.

Since the P-value > ?, we reject the null hypothesis.

Since the P-value ≤ ?, we reject the null hypothesis.

Since the P-value ≤ ?, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 1% level of significance, there is insufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is not justified in claiming the blades must be replaced.At the 1% level of significance, there is sufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is justified in claiming the blades must be replaced. At the 1% level of significance, there is sufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is not justified in claiming the blades must be replaced.At the 1% level of significance, there is insufficient evidence to conclude that the variance of measurements on the fan blades is higher than the specified amount. The inspector is justified in claiming the blades must be replaced.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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