Mathematics: 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

14 on edge i used calculator and evrything its 14Step-by-step explanation:...

It is desired to determine whether there is less

Nowellkwaku

20.01.2022

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

0,0(0 оценок)