Question

In: Statistics and Probability

Test the given claim. Assume that a simple random sample is selected from a normally distributed...

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 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? - 43​, 79​, - 22​, - 72​, - 41​, 10​, 17​, 51​, - 6​, - 51​, - 108​, - 108

What are the null and alternative​ hypotheses?

A. H0​: sigmaless than 32.2 ft H1​: sigmaequals 32.2 ft

B. H0​: sigmaequals 32.2 ft H1​: sigmaless than 32.2 ft

C. H0​: sigmagreater than 32.2 ft H1​: sigmaequals 32.2 ft

D. H0​: sigmaequals 32.2 ft H1​: sigmagreater than 32.2 ft

E. H0​: sigmaequals 32.2 ft H1​: sigmanot equals 32.2 ft

F. H0​: sigmanot equals 32.2 ft H1​: sigmaequals 32.2 ft

Find the test statistic. X2 = __ ​(Round to two decimal places as​ needed.)

Determine the critical​ value(s). The critical​ value(s) is/are __. ​(Use a comma to separate answers as needed. Round to two decimal places as​ needed.)

Solutions

Expert Solution

Step 1:

Ho: = 32.2

Ha: > 32.2

D. H0​: sigmaequals 32.2 ft H1​: sigmagreater than 32.2 ft

Step 2: test statistics

sample mean = sum of all terms / no of terms = -294/12 = 24.5

sample sd = s

data data-mean (data - mean)2
-43 -18.5 342.25
79 103.5 10712.25
-22 2.5 6.25
-72 -47.5 2256.25
-41 -16.5 272.25
10 34.5 1190.25
17 41.5 1722.25
51 75.5 5700.25
-6 18.5 342.25
-51 -26.5 702.25
-108 -83.5 6972.25
-108 -83.5 6972.25

= 35.87

Chi square critical = CHISQ.INV.RT(probability,df) = CHISQ.INV.RT(0.95, 11) = 4.575

critical = 4.57

As the ( 35.87) is greater than critical, we reject the Null hypothesis.  

Hence we have sufficient evidence to believe that new production method has errors with a standard deviation greater than 32.2​ ft.

This makes the new production method worst than the existing. Yes the company should take necessary actions.


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