It is desired to determine whether there is less variability in the silver plating done by Company 1 than in that done by Company 2. If independent random samples of size 12 of the two companies' work yield s1 = 0.035 mil and s2=0.062 mil, test the null hypothesis sigma12 = sigma22 against the alternative hypothesis sigma12 < sigma22 at the 0.05 level of significance. The F statistic for this test is. The decision is to, at alpha = 0.05, reject H0.

The conclusion is that the data provides evidence to believe that the plating done by Company 1 is less variable than that done by Company 2. Enter 0 if this statement is FALSE or 1 otherwise.

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Mathematics

8 May, 00:47

It is desired to determine whether there is less variability in the silver plating done by Company 1 than in that done by Company 2. If independent random samples of size 12 of the two companies' work yield s1 = 0.035 mil and s2=0.062 mil, test the null hypothesis sigma12 = sigma22 against the alternative hypothesis sigma12 < sigma22 at the 0.05 level of significance. The F statistic for this test is. The decision is to, at alpha = 0.05, reject H0.

The conclusion is that the data provides evidence to believe that the plating done by Company 1 is less variable than that done by Company 2. Enter 0 if this statement is FALSE or 1 otherwise.

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Answers (1)

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Sanai Kennedy · Answer 1

The conclusion is True

Step-by-step explanation:

Solution:

- The standard deviation of company s1 = 0.035

- The standard deviation of company s2 = 0.062

- Null Hypothesis : s1^2 = s2^2

- Alternate hypothesis : s1^2 < s2^2

- Criteria to reject Null: M > M_a (12 - 1, 12 - 1)

- From Tables, M > 2.82

- M - statistics value M = s2^2 / s1^2

- M = (0.062 / 0.035) ^2 = 3.14

- M = 3.14 > 2.82 ... Criteria of rejection is met

- Null Hypothesis must be rejected at level of significance a = 0.05

- p value, 1 - 0.965 = 0.035 < 0.05 (Significance Level)

- The conclusion is True