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An analysis of variance can be performed whenever we have a continuous (interval or ratio level) dependent variable that is normally distributed in the population, a categorical variable with three or more levels or categories, and we are interested in testing hypothesis about three or more population means. The dependent variable must be continuous, and the independent variable must be categorical with at least three levels.
If we have a continuous dependent variable and a categorical independent variable with only two categories or levels, the correct statistical test would be a two sample z or t test, assuming that the hypothesis test involved the equality of two population means.
It is called the analysis of variance, because we make inferences about the differences among population means based upon a comparison of the variance that exists within each sample, relative to the variance that exists across the samples. More specifically, we examine the ratio of variance between the samples to the variance within the samples. The greater this ratio, the more across sample variance there is relative to within sample variance. Therefore, as this ratio becomes greater than 1, we are more inclined to believe that the samples were drawn from different populations with different population means.
As suggested in the answer to the last question, the two types of variability we use in the ANOVA F test are the variability between the samples and the variability within the samples.
The formulas for the three degrees of freedom are:
dfTotal = N - 1To check your arithmetic, make sure that dfTotal = dfBetween + dfWithin.
a. In order to determine the various sum of squares, we will reproduce the data. We will then do some preliminary calculations like square each x score, and summing the x scores within each group.
LEVEL OF STRESS
HIGH MEDIUM LOW
NUMBER OF TIMES PHYSICALLY USED PUNISHMENT LAST YEAR
x x2 x x2 x x2
4 16 2 4 3 9
6 36 4 16 1 1
12 144 5 25 2 4
10 100 3 9 0 0
5 25 0 0 2 4
9 81 3 9 2 4
8 64 2 4 4 16
11 121 5 25 1 1
10 100 5 25 0 0
8 64 4 16 1 1
S = 83 751 33 133 16 40
Now that we have the various sums and squares, we are ready to calculate the different sum of squares using our computational formulas. First, the total sum of squares:
Next, the between sum of squares:
Now we can obtain the within sum of squares by subtraction:
b. the correct degrees of freedom for this table are:
dfBetween = k - 1 = 3 - 1 = 2
dfWithin = N - k = 30 - 3 = 27
dfTotal = N - 1 = 30 - 1 = 29
You can see that dfBetween + dfWithin = dfTotal.
The ratio of sum of squares to degrees of freedom can now be determined:
SSBetween/dfBetween = 242.6/2 = 121.30
SSWithin/dfWithin = 100.6/27 = 3.72
The F ratio is: Fobt = 121.30/3.72 = 32.61.
Our completed summary table looks like the following:
Sum of Squares df SS/df F Between 242.6 2 121.30 32.61
Within 100.6 27 3.72
Total 343.2 29
c. We will go though each step of the hypothesis test:
Step 1: The null hypothesis is that the three population means are not different, the alternative hypothesis is that the three population means are different. The implication of the null hypothesis is that there is no difference in the frequency of physical punishment against children among women in the high, medium, and low stress groups.
H0: m1 = m2 = m3
H1: m1 ¹ m2 ¹ m3
Step 2: Our statistical test is the F test, and our sampling distribution is the F distribution.
Step 3: Our alpha level is .05. We go to the F distribution with 2 and 27 degrees of freedom. The critical value of F from this table is 3.35. The critical region is any Fobt that falls to the right of this value. Our decision rule will be to reject the null hypothesis if Fobt ³ 3.35.
Step 4: We have already calculated our test statistic. From the summary table above, you can see that Fobt = 32.61.
Step 5: Since our obtained value of F is greater than the critical value, and falls into the critical region, our decision is to reject the null hypothesis. We conclude that the population means are not equal, and that the frequency of using physical punishment against one's child does vary by the amount of stress the woman feels.
c. The significant F test only tells us that at least two of the three means are significantly different from one another, it does not tell us which means are different. To test the significance of particular pairs of means, we have to calculate Tukey's Honest Significant Difference Test. This test is based upon the CD or critical difference. The formula is:
where q is the value from the studentized range. You get this value from Table 5 of Appendix B. To obtain the value of q you need an alpha level (.05 in this problem), the number of dfwithin (equal to 27 in this case), and the number of groups (3 in this case). Going to the studentized table, you find the value of q to be equal to 3.49. To find the critical difference, you plug these values into your formula:
The critical difference for the mean comparisons, then, is 2.13. If the absolute value of the difference between a pair of means is greater than 2.13, we can reject the null hypothesis that the two means are equal. Let's find the difference between each pair of sample means in the problem:
High Stress 8.30
Medium Stress - 3.30
³5.00³
High Stress 8.30
Low Stress - 1.60
³6.70³
Medium Stress 3.30
Low Stress - 1.60
³1.70³
The difference between the High and Medium Stress groups is greater than 2.13, so these groups are significantly different in their frequency of physical punishment. The difference between the High and Low Stress groups is also greater than 2.13, so these groups are significantly different in their frequency of physical punishment. The Medium and Low Stress groups, however, are not significantly different, the difference between their two means is less than 2.13.
d. eta2 is the ratio of Between Sum of Squares to Total Sum of Squares: 
. The unbiased value of eta2 is equal to one minus the ratio of within group variance to total variance, where the total variance is defined as SSTotal/dfTotal :
. This tells us that there is a moderately strong relationship between a woman's feelings of stress and the frequency with which she uses physical punishment against her children. Specifically, about 70% of the variance in the frequency of physical punishment explained by the mother's feelings of stress.
a. the correct degrees of freedom for this table are:
dfBetween = k - 1 = 3 - 1 = 2
dfWithin = N - k = 45 - 3 = 42
dfTotal = N - 1 = 45 - 1 = 44
You can see that dfBetween + dfWithin = dfTotal.
The ratio of sum of squares to degrees of freedom can now be determined: SSBetween/dfBetween = 475.3/2 = 237.65 SSWithin/dfWithin = 204.5/42 = 4.87
The F ratio is: Fobt = 237.65/4.87 = 48.80.
Our completed summary table looks like the following:
Sum of Squares df SS/df F
Between 475.3 2 237.65 48.80
Within 204.5 42 4.87
Total 679.8 44
b. We will go though each step of the hypothesis test:
Step 1: The null hypothesis is that the three population means are not different, the alternative hypothesis is that the three population means are different.
H0: m1 = m2 = m3 H1: m1 ¹ m2 ¹ m3
Step 2: Our statistical test is the F test, and our sampling distribution is the F distribution.
Step 3: Our alpha level is .01. We go to the F distribution with 2 and 42 degrees of freedom. The critical value of F from this table is 5.18. The critical region is any Fobt that falls to the right of this value. Our decision rule will be to reject the null hypothesis if Fobt ³ 5.18.
Step 4: We have already calculated our test statistic. From the summary table above, you can see that Fobt = 48.80.
Step 5: Since our obtained value of F is greater than the critical value, and falls into the critical region, our decision is to reject the null hypothesis. We conclude that the population means are not equal.
c. The significant F test only tells us that at least two means are significantly different from one another, it does not tell us which means are different. To test the significance of particular pairs of means, we have to calculate Tukey's Honest Significant Difference Test. This test is based upon the CD or critical difference. The formula is:
where q is the value from the studentized range. You get this value from Table 5 of Appendix B. To obtain the value of q you need an alpha level (.01 in this problem), the number of dfwithin (equal to 42 in this case), and the number of groups (3 in this case). Going to the studentized table, you find the value of q to be equal to 4.37. To find the critical difference, you plug these values into your formula:
The critical difference for the mean comparisons, then, is 2.49. If the absolute value of the difference between a pair of means is greater than 2.49, we can reject the null hypothesis that the two means are equal. Let's find the difference between each pair of sample means in the problem:
"Get Tough" 125.2
"Moral Appeal" - 119.7
³5.5³
"Get Tough" 125.2
"Control" - 145.3
³20.1³
"Moral Appeal" 119.7
"Control" - 145.3
³25.6³
Each pair of differences is greater than 2.49. This means that the "moral appeal" states have a significantly lower level of drunk driving than the "get tough" states, the "get tough" states have a significantly lower level of drunk driving than "control" states, and the "moral appeal" states have significantly lower levels of drunk driving than the control states. It appears, then, that doing something about drunk drivers is better than doing little or nothing.
d. eta2 is the ratio of Between Sum of Squares to Total Sum of Squares: 
. The unbiased value of eta2 is equal to one minus the ratio of within group variance to total variance, where the total variance is defined as SSTotal/dfTotal :
. This tells us that there is a moderately strong relationship between the state's response to drunk driving and the rate of drunk driving in that state. Specifically, about 70% of the variance in levels of drunk driving is explained by the state's public policy.
a. the correct degrees of freedom for this table are:
dfBetween = k - 1 = 5 - 1 = 4
dfWithin = N - k = 250 - 5 = 245
dfTotal = N - 1 = 250 - 1 = 249
You can see that dfBetween + dfWithin = dfTotal.
The ratio of sum of squares to degrees of freedom can now be determined: SSBetween/dfBetween = 12.5/4 = 3.125 SSWithin/dfWithin = 616.2/245 = 2.51
The F ratio is: Fobt = 3.125/2.51 = 1.25
Our completed summary table looks like the following:
Sum of Squares df SS/df F
Between 12.5 4 3.125 1.25
Within 616.2 245 2.510
Total 628.7 249
b. We will go though each step of the hypothesis test:
Step 1: The null hypothesis is that the five population means are not different, the alternative hypothesis is that the five population means are different.
H0: m1 = m2 = m3 = m4 = m5 H1: m1 ¹ m2 ¹ m3 ¹ m4 ¹ m5
Step 2: Our statistical test is the F test, and our sampling distribution is the F distribution.
Step 3: Our alpha level is .05. We go to the F distribution with 4 and 245 degrees of freedom. The critical value of F from this table is 2.37. The critical region is any Fobt that falls to the right of this value. Our decision rule will be to reject the null hypothesis if Fobt ³ 2.37.
Step 4: We have already calculated our test statistic. From the summary table above, you can see that Fobt = 1.25.
Step 5: Since our obtained value of F is not greater than or equal to the critical value, and does not fall into the critical region, our decision is to not reject the null hypothesis. We conclude that different fear spots are not different in terms of their actual risk of victimization.
c. The value of eta2 is equal to 
. The unbiased estimate of eta2 is equal to 
There is virtually no relationship between a person's fear of a given geographical area and the actual frequency of criminal victimization in that area. About 2 percent of the variance in fear spots is explained by victimization levels.
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