# Population variations, ANOVA table, Chi Square test

1) The following hypotheses are given:

Ho=2 group 1= 2 group 2

Ha=2 group 1≠2 group 2

A random sample of eight observations from the first population resulted in a standard deviation of 12. A random sample of seven observations in the second population resulted in a standard deviation of 7. At the .02 significance level, is there a difference in the variation of the two populations?

Group 2=Exercise 3 on page 419

The following hypotheses are given:

Ho=2 group 1= 2 group 2

Ha=2 group 1≠2 group 2

A random sample of eight observations from the first population resulted in a standard deviation of 12. A random sample of seven observations in the second population resulted in a standard deviation of 7. At the .01 significance level, is there a difference in the variation of the two populations?

7) The following is sample information. Test the hypotheses that the treatment means are equal. Use the .05 significance level.

Treatment 1 Treatment 2 Treatment 3

8 3 3

6 2 4

10 4 5

9 3 4

a) State the null hypothesis and alternative hypothesis.

b) What is the decision rule?

c) Compute the SST, SSDE and SS Total.

d) Complete an ANOVA table.

e) State your decision regarding the null hypothesis.

f) Give a real life example of when you could use ANOVA.

The other two groups will do p 428, # 8

8) The following is sample information. Test the hypothesis at the .05 significance level that the treatment means are equal.

Treatment 1 Treatment 2 Treatment 3

9 13 10

7 20 9

11 14 15

9 13 14

12 15

10

a) State the null hypothesis and alternative hypothesis.

b) What is the decision rule?

c) Compute the SST, SSDE and SS Total.

d) Complete an ANOVA table.

e) State your decision regarding the null hypothesis.

f) Give a real life example of when you could use ANOVA.

1) In a particular Chi Square goodness of fit test there are 4 categories and 200 observations. Use the .05 significance level.

a) How many degrees of freedom are there?

b) What is the critical value of Chi Square (you will need to look at the table).

2) The null hypothesis and the alternate are:

Ho: The cell categories are equal.

Ha: The cell categories are not equal.

Category fo

A 10

B 20

C 30

a) State the decision rule, using the .05 significance level.

b) Compute the value of Chi Square.

c) What is your decision regarding Ho?

d) Give a real life example of when you'd use Chi Square.

2) In a particular Chi Square good ness of fit test, there are 6 categories and 500 observations. Use the .01 significance level.

a) How many degrees of freedom are there?

b) What is the critical value of Chi Square (you will need to look at the table).

4) The null hypothesis and alternate are:

Ho: The cell categories are equal.

Ha: The cell categories are not equal.

Category fo

A 10

B 20

C 30

D 20

a) State the decision rule, using the .05 significance level.

b) Compute the value of Chi Square.

c) What is your decision regarding Ho?

d) Give a real life example of when you'd use Chi Square.

https://brainmass.com/statistics/chi-squared-test/population-variations-anova-table-chi-square-test-287077

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Q.(1):

The following hypotheses are given:

Null Hypothesis H0 :

Alternative Hypothesis H1 :

A random sample of eight observations from the first population resulted in a standard deviation of 12. A random sample of seven observations in the second population resulted in a standard deviation of 7. At the .02 significance level, is there a difference in the variation of the two populations?

Solution:

We have n1 = 8 , S1 = 12 and n2 = 7 , S2 = 7 ,, = 0.02

Degrees of freedom for numerator = df1 = 8 - 1 = 7

Degrees of freedom for denominator = df2 = 7 - 1 = 6

The Upper Critical value (FU) = F(df = 7, 6, = 0.02 ) = 6.2508 ( from table)

The Lower Critical value (FL)

Decision Rule :

Reject the Null Hypothesis (H0), if the calculated value of F-statistic lies outside the range of ( 0.1794 , 6.2508 ). Otherwise, accept the Null Hypothesis.

Computing Test Statistic:

Test Statistic

Decision:

Since the computed value of the F-statistic of 2.9388 lies within the range of ( 0.1794 , 6.2508 ), we can not reject the Null hypothesis at = 0.02 significance level.

Therefore, we conclude at = 0.02 level of significance that, the variances are equal.

Q.(2):

A random sample of eight observations from the first population resulted in a standard deviation of 12. A random sample of seven observations in the second population resulted in a standard deviation of 7. At the .01 significance level, is there a difference in the variation of the two populations?

Solution :

Null Hypothesis H0 :

Alternative Hypothesis H1 :

We have n1 = 8 , S1 = 12 and n2 = 7 , S2 = 7 ,, = 0.02

Degrees of freedom for numerator = df1 = 8 - 1 = 7

Degrees of freedom for denominator = df2 = 7 - 1 = 6

The Upper Critical value (FU) = F(df = 7, 6, = 0.01 ) = 8.26 ( from table)

The Lower Critical value (FL)

Decision Rule:

Reject the Null Hypothesis (H0), if the calculated value of F-statistic lies outside the range of ( 0.1391 , 8.26 ). Otherwise, accept the Null Hypothesis.

Decision :

Since the computed value of the F-statistic of 2.9388 lies within the range of ( 0.1391 , 8.26 ), we can not reject the Null hypothesis at = 0.01 significance level.

Therefore, we conclude at = 0.01 level of significance that, the variances are equal.

Q.(7):

The following is sample information. Test the hypotheses that the treatment means are equal. Use the .05 significance level.

Treatment 1 Treatment 2 Treatment 3

8 3 3

6 2 4

10 4 5

9 3 4

a) State the ...

#### Solution Summary

The expert examines population variations, ANOVA tables and Chi Square Tests.