pleae explain confiedence intercals and how to calculate part f only   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 σ2 = 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: σ2 = 0.18; H1: σ2 < 0.18H0: σ2 > 0.18; H1: σ2 = 0.18    H0: σ2 = 0.18; H1: σ2 ≠ 0.18H0: σ2 = 0.18; H1: σ2 > 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 > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, 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. (f) Find a 90% confidence interval for the population standard deviation. (Round your answers to two decimal places.) lower limit  mm upper limit      mm Interpret the results in the context of the application. We are 90% confident that σ lies below this interval.We are 90% confident that σ lies outside this interval.    We are 90% confident that σ lies within this interval.We are 90% confident that σ lies above this interval.

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pleae explain confiedence intercals and how to calculate part f only

 

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 σ2 = 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: σ2 = 0.18; H1: σ2 < 0.18H0: σ2 > 0.18; H1: σ2 = 0.18    H0: σ2 = 0.18; H1: σ2 ≠ 0.18H0: σ2 = 0.18; H1: σ2 > 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 > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, 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.

(f) Find a 90% confidence interval for the population standard deviation. (Round your answers to two decimal places.)
lower limit  mm
upper limit      mm

Interpret the results in the context of the application.
We are 90% confident that σ lies below this interval.We are 90% confident that σ lies outside this interval.    We are 90% confident that σ lies within this interval.We are 90% confident that σ lies above this interval.
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