SOLUTION: This is a three part question that I need help with please. Thank you! A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from

Question 387397: This is a three part question that I need help with please. Thank you!
A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample. The target accuracy is a variance in measurements of 1.2 or less. If the lab technician takes 16 measurements and the variance of the measurements in the sample is 2.2, does this provide enough evidence to reject the claim that the lab technician's accuracy is within the target accuracy?
A. State the null and alternative hypotheses
B. Compute the value of the appropriate test statistic.
C. At a (probability of type error 1)=0.01 level of significance, what is your conclusion? 1)Do not reject hypothesis. 2) Reject b/c there is evidence that the target is higher. 3) Reject b/c there is not enough evidence the targets true variance is larger. 4) Cannot determine.
Found 2 solutions by stanbon, abush:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample. The target accuracy is a variance in measurements of 1.2 or less.
If the lab technician takes 16 measurements and the variance of the measurements in the sample is 2.2, does this provide enough evidence to reject the claim that the lab technician's accuracy is within the target accuracy?
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A. State the null and alternative hypotheses
Ho: var^2 >= 1.2
Ha: var^2 < 1.2 (claim)
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B. Compute the value of the appropriate test statistic.
Chi-Sq = (16-1)2.2^2/1.2^2 = 15*3.36 = 50.42
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C. At a (probability of type error 1)=0.01 level of significance, what is your conclusion? 1)Do not reject hypothesis. 2) Reject b/c there is evidence that the target is higher. 3) Reject b/c there is not enough evidence the targets true variance is larger. 4) Cannot determine.
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For df=19, the right-tail reject critical value is 36.19
Since the ts is greater than 36.19 reject Ho.
The test results do not support the claim.
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Cheers,
Stan H.
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Answer by abush(5) (Show Source): You can put this solution on YOUR website!
A. Ho: 02 <= 1.2 H1: 02>1.2
B.x2={(n-1)s2}/o2 s2=sample variance o2=population variance
x2=(16-1)(2.2)/1.2 =27.5
C.Critical values would be 4.601 and 32.801. 27.5 falls in the non critical region so do not reject the null hypothesis
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