Please answer with details Please answer with details The im

Please answer with details

Please answer with details The impurity of fertilizer bags is Sigma^2 =6, n=30. (1) Test H(0): Sigm^2=6, H(1):Sigma^2 > 6 for s^2= 10, alpha =0.05: (2) Test H(0): Sigma^2 =6, H(1):Sigma^2

Solution

(1) It is a right-tailed test.

The degree of freedom =n-1=30-1=29

Given a=0.05, the critical value of chisquare with 0.95 and df=29 is 42.56 (from chisquare table)

The rejection region is if X^2>42.56, we reject the null hypothesis.

The test statistic is

Chisquare= (n-1)*s^2/o^2

=29*10/6 =48.33

Since 48.33 is larger than 42.56, we reject the null hypothesis.

So we can conclude that o^2>6

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(2)It is a left-tailed test.

The degree of freedom =n-1=30-1=29

Given a=0.025, the critical value of chisquare with 0.025 and df=29 is 16.05 (from chisquare table)

The rejection region is if X^2<16.05, we reject the null hypothesis.

The test statistic is

Chisquare= (n-1)*s^2/o^2

=29*3/6 =14.5

Since 14.5 is less than 16.05, we reject the null hypothesis.

So we can conclude that o^2<6

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(3)It is a two-tailed test.

The degree of freedom =n-1=30-1=29

Given a=0.01, the critical values of chisquare with 0.005 and df=29 is 13.12 (from chisquare table)

chisquare with 0.995 and df=29 is 52.34 (from chisquare table)

The rejection regions are if X^2>52.34 or X^2<13.12, we reject the null hypothesis.

The test statistic is

Chisquare= (n-1)*s^2/o^2

=29*12/6 =58

Since 58 is larger than 52.34, we reject the null hypothesis.

So we can conclude that o^2 is not equal to 6