Question

method of testing hypotheses. Company A uses a new production method to manufacture aircraft altimeters. A simple random sample of new altimeters resulted in errors listed below. Use a 0.05 level of significance to test the claim that the new production method has errors with a standard deviation greater than 32.2​ ft, which was the standard deviation for the old production method. If it appears that the standard deviation is​ greater, does the new production method appear to be better or worse than the old​ method? Should the company take any​ action?

Answer 1

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the new production method used to manufacture aircraft altimeters has a standard deviation greater than 32.2 ft (δ>32.2)

A random sample of n= 12 aircraft altimeters was taken and the measurement errors of each altimeter were recorded.

The variable of interest is X: measuring error of an aircraft altimeter that was manufactured using the new method.

The parameter of interest is the population variance, if the variance is greater than the variance of the other method, this will mean that the measurement errors of the altimeters will be more dispersed ( heterogeneous)

To study the population variance you have to use the Chi-Square distribution.

The hypotheses are:

H₀: δ ≤ 1036.84 (δ≤32.2)

H₁: δ² > 1036.84 (δ>32.2)

α: 0.05

$X^2_{H_0}= \frac{(n-1)S^2}{Sigma^2} ~X^2_{n-1}$

∑X= 280

∑X²= 43416

S²= 1/(n-1)[∑X²-(∑X)²/n]= 1/11[43416-(280)²/12]= 3352.97

$X^2_{H_0}= \frac{(11*3352.97)}{1036.84}= 35.57$

This test is one tailed  to the right, this means that, using the critical value approach, that you will reject the null hypothesis to high values of X²$X^2_{n-1;1-\alpha }= X^2_{11;0.95}= 19.675$

Decision rule:

If $X^2_{H_0}$ ≥ 19.675, the decision is to reject the null hypothesis

If $X^2_{H_0}$ < 19.675, the decision is to not reject the null hypothesis.

The calculated value is greater than the critical value, the decision is to reject the null hypothesis.

Then using a 5% significance level, you can conclude that the population variance of the error in measurement of the aircraft altimeters manufactured using the new method is greater than 1036.84 ft², at the same level you can say that the standard deviation of the error in measurement of the aircraft altimeters manufactured using the new method is greater than 32.2 ft.

It seems that the altimeters s manufactured using the new method is less efficient than the ones manufactured using the old method. To avoids aircraft accidents, like crashing, it is essential to have accurate altitude measurements. The company should dispose of all altimeters manufactured using the new method and change the manufacturing method to avoid possible accidents.

I hope this helps!

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Full text

Test the given claim. Assume that a simple random sample is selected from a normally distributed population. Use either the P value method or the traditional method of testing hypotheses. Company A uses a new production method to manufacture aircraft altimeters. A simple random sample of new altimeters resulted in errors listed below. Use a 0.05 level of significance to test the claim that hte new production method has errors with a standard deviation greater than 32.2 ft which was the standard deviation for the old production method. If it appears that the standard deviation is greater does the new production method appear to be better or worse than the old method? Should the company take any action?

-40, 78,-24,-70,-40,15,19,54,-9,-51,-106,-106

What are the null and alternative hypotheses?

A. standard deviation =32.2 ft standard deviation greater than 32.2 ft

B. standard deviation greather than 32.2 ft standard deviation= 32.2 ft

C. standard deviation= 32.2 ft standard deviation is not equal to 32.2. ft

D. standard deviation not equal to 32.2. ft standard deviation is equal to 32.2 ft

E. standard deviation is less than 32.2 ft stanard deviation equal 32.2 ft

F. standard deviation is equal to 32.2 ft standard deviation less than 32.2 ft

Find the test statistic x^2= round two decimal places

Determine the critical value= use a comma to seperate

Since the statistic is ____ the critical values are _____H_0. there is ______evidence to support a claim.

The variation appears to be ______ than in the past so the new method could be _______ because there will be more ______ in altimeters. Therefore the company _______ take immediate action to reduce the variation.