Question

How do I set this up to solve? Thanks.

A manufacturer monitors the amount of peanuts put in jars of a given food product. A control device is to be evaluated to determine if it reduces the variability in the amount of peanuts placed in jars. To that end, a sample of 11 jars were evaluated using normal processing without the device and another sample of 9 jars were evaluated with the control device.

data : normal : 8.06 8.64 7.97 7.81 7.93 8.57 8.39 8.46 8.28 8.02 8.39

control : 7.99 8.12 8.34 8.17 8.11 8.03 8.14 8.14 7.87

Does the control device reduce variability? Use a significance level of .05.

Answer 1

We need to find the variance for both the samples and conduct a test of variance. What we also want to see is that if variation is reduced or in other words if the variation in the normal is greater than the variation in the control.

I have used bot, the crirtical value approach and the p value approach. I have used online calculators to calculate the same.

The calculations have been given after the hypothesis testing.

S₁² = 0.08, n1 = 11, df1 = 11 - 1 = 10

S₂² = 0.017, n2 = 9, df2 = 9 - 1 = 8

The Hypothesis:

H0: : There is no difference in variation between the normal amount and the control amount .

Ha: : The variation in the normal is greater than that of the control. Therefore variation is reduced..

The Test Statistic:

F = s₁²/s₂² = 0.08/0.017 = 4.706

The Critical Value at = 0.05, df1 = 10, df2 = 8 is 3.347

The p value: At F = 4.706, df1 = 10 and df2 = 8 is 0.0191

The Decision Rule: If Fobserved is > F critical, Then reject H0.

If p value is < , Then Reject H0.

The Decision: Since Fobserved(4.706) is > Fcritical(3.347) , we Fail to reject H0.

Since pvalue (0.0191) is < (0.05, We Reject H0.

The Conclusion: There is suficient evidence at the 95% significance level to conclude that the variation in the control has reduced i.e the devise reduces variability.

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For Normal

#	X	Mean	(x - mean)2
1	8.06	8.229	0.028561
2	8.64	8.229	0.168921
3	7.97	8.229	0.067081
4	7.81	8.229	0.175561
5	7.93	8.229	0.089401
6	8.57	8.229	0.116281
7	8.39	8.229	0.025921
8	8.46	8.229	0.053361
9	8.28	8.229	0.002601
10	8.02	8.229	0.043681
11	8.39	8.229	0.025921
Total	90.52		0.797291

SS(Sum of squares)	0.797291
Variance = SS/n-1	0.080

S₁² = 0.08

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For Control

#	X	Mean	(x - mean)2
1	7.99	8.101	0.012321
2	8.12	8.101	0.000361
3	8.34	8.101	0.057121
4	8.17	8.101	0.004761
5	8.11	8.101	8.1E-05
6	8.03	8.101	0.005041
7	8.14	8.101	0.001521
8	8.14	8.101	0.001521
9	7.87	8.101	0.053361
Total	72.91		0.136089

SS(Sum of squares)	0.136089
Variance = SS/n-1	0.017

S₂² = 0.017