# EXERCISE 36 Analysis of Variance (ANOVA) I

EXERCISE36: Analysis of Variance (ANOVA) I

1. The researchers found a significant difference between the two groups (control and treatment) for change in mobility of the women with osteoarthritis (OA) over 12 weeks with the results of F(1, 22) = 9.619, p = 0.005. Discuss each aspect of these

TheF-value is so great at the 5% level of significance to propose asignificance difference between treatment and control groups. Thep-value 0.005 is less than 0.05 (0.005 &lt 0.05) therefore thisrecommends a rejection of the null hypothesis, meaning that thetreatment and control groups are found to be completely different.

1. State the null hypothesis for the Baird and Sands (2004) study that focuses on the effect of the GI with PMR treatment on patients’ mobility level. Should the null hypothesis be rejected for the difference between the two groups in change in mobility scores over 12 weeks? Provide a rationale for your answer.

Thenull hypothesis for the Baird and Sands (2004) is that the meanmobility scores for the two groups are similar. As stated before inquestion (1), since the p-value 0.005 is less than 0.05 (0.005 &lt0.05) this means I reject the null hypothesis. So the mean“difficulty with mobility score” for the two groups must bedifferent.

1. The researchers stated that the participants in the intervention group reported a reduction in mobility difficulty at week 12. Was this result statistically significant, and if so at what probability?

Yesthe result obtained in the research was statistically significant atprobability p &lt 0.001, going by the text.

1. If the researchers had set the level of significance or 〈 = 0.01, would the results of p = 0.001 still be statistically significant? Provide a rationale for your answer.

Yesthis still implies statistical significance because 0.001 &lt 0.01

1. If F (3, 60) = 4.13, p = 0.04, and 〈 = 0.01, is the result statistically significant? Provide a rationale for your answer. Would the null hypothesis be accepted or rejected?

Inthis situation the results obtained would be considered staticallysignificant since 0.04 &gt 0.01. In this case I would agree to thenull hypothesis and conclude that there is no difference betweentreatment and control groups.

1. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide a rationale for your answer.

Yes,actually predicting correlations between variables to say whether ornot there exists a correlation or relationship between groups is themain function of the ANOVA. Usually, if the p-value is smaller thanthe significance level there exists a difference between the groups,and if the p-value is bigger than the significance level there is nodifference between the groups.

1. If a study had a result of F (2, 147) = 4.56, p = 0.003, how many groups were in the study, and what was the sample size?

IfK = number of groups in the study, and df= K – 1 then K = df+1= 2+1 = 3. So there are 3 groups in the study. There are N – K numberof participants so N = K+147 = 3+147 = 150 participants.

1. The researchers state that the sample for their study was 28 women with a diagnosis of OA, and that 18 were randomly assigned to the intervention group and 10 were randomly assigned to the control group. Discuss the study strengths and/or weaknesses in this statement.

Inmy opinion, the researcher could have used a larger sample size. ButI concur with the researcher’s choice to put more participants intothe intervention group rather than the control group. That is becauseI would be more interested in the results from the intervention groupso a larger sample is suitable, but there is still enough in thecontrol group to play their part as well.

1. In your opinion, have the researchers established that guided imagery (GI) with progressive muscle relaxation (PMR) reduces pain and decreases mobility difficulties in women with OA?

Inmy opinion, the outcome of the research is significant enough todeserve further investigation. I think it may be useful to carry outa couple other related studies done by independent groups to certifythe outcome of the Baird &ampSands group, as there is usually thechances of some bias inflowing the study. By caring out severalindependent studies this will strengthen the argument that GI withPMR reduces plan and decreases mobility difficulty in women with OA.

1. The researchers stated that this was a 12-week longitudinal, randomized clinical trial pilot study with 28 women over 65 years of age with the diagnosis of OA. What are some of the possible problems or limitations that might occur with this type of study?

Sincethe study took quite a long time, researches will need to be carefulthat those in the control and intervention control groups do notdiscover which group they are in, as this possibility exists to agreater degree given the length of the trial. Also, a slightly largergroup of women may help to certify the results to a greater extent.

REFERENCE

EXERCISE36: Analysis of Variance (ANOVA) I

1. The researchers found a significant difference between the two groups (control and treatment) for change in mobility of the women with osteoarthritis (OA) over 12 weeks with the results of F(1, 22) = 9.619, p = 0.005. Discuss each aspect of these

TheF-value is so great at the 5% level of significance to propose asignificance difference between treatment and control groups. Thep-value 0.005 is less than 0.05 (0.005 &lt 0.05) therefore thisrecommends a rejection of the null hypothesis, meaning that thetreatment and control groups are found to be completely different.

1. State the null hypothesis for the Baird and Sands (2004) study that focuses on the effect of the GI with PMR treatment on patients’ mobility level. Should the null hypothesis be rejected for the difference between the two groups in change in mobility scores over 12 weeks? Provide a rationale for your answer.

Thenull hypothesis for the Baird and Sands (2004) is that the meanmobility scores for the two groups are similar. As stated before inquestion (1), since the p-value 0.005 is less than 0.05 (0.005 &lt0.05) this means I reject the null hypothesis. So the mean“difficulty with mobility score” for the two groups must bedifferent.

1. The researchers stated that the participants in the intervention group reported a reduction in mobility difficulty at week 12. Was this result statistically significant, and if so at what probability?

Yesthe result obtained in the research was statistically significant atprobability p &lt 0.001, going by the text.

1. If the researchers had set the level of significance or 〈 = 0.01, would the results of p = 0.001 still be statistically significant? Provide a rationale for your answer.

Yesthis still implies statistical significance because 0.001 &lt 0.01

1. If F (3, 60) = 4.13, p = 0.04, and 〈 = 0.01, is the result statistically significant? Provide a rationale for your answer. Would the null hypothesis be accepted or rejected?

Inthis situation the results obtained would be considered staticallysignificant since 0.04 &gt 0.01. In this case I would agree to thenull hypothesis and conclude that there is no difference betweentreatment and control groups.

1. Can ANOVA be used to test proposed relationships or predicted correlations between variables in a single group? Provide a rationale for your answer.

Yes,actually predicting correlations between variables to say whether ornot there exists a correlation or relationship between groups is themain function of the ANOVA. Usually, if the p-value is smaller thanthe significance level there exists a difference between the groups,and if the p-value is bigger than the significance level there is nodifference between the groups.

1. If a study had a result of F (2, 147) = 4.56, p = 0.003, how many groups were in the study, and what was the sample size?

IfK = number of groups in the study, and df= K – 1 then K = df+1= 2+1 = 3. So there are 3 groups in the study. There are N – K numberof participants so N = K+147 = 3+147 = 150 participants.

1. The researchers state that the sample for their study was 28 women with a diagnosis of OA, and that 18 were randomly assigned to the intervention group and 10 were randomly assigned to the control group. Discuss the study strengths and/or weaknesses in this statement.

Inmy opinion, the researcher could have used a larger sample size. ButI concur with the researcher’s choice to put more participants intothe intervention group rather than the control group. That is becauseI would be more interested in the results from the intervention groupso a larger sample is suitable, but there is still enough in thecontrol group to play their part as well.

1. In your opinion, have the researchers established that guided imagery (GI) with progressive muscle relaxation (PMR) reduces pain and decreases mobility difficulties in women with OA?

Inmy opinion, the outcome of the research is significant enough todeserve further investigation. I think it may be useful to carry outa couple other related studies done by independent groups to certifythe outcome of the Baird &ampSands group, as there is usually thechances of some bias inflowing the study. By caring out severalindependent studies this will strengthen the argument that GI withPMR reduces plan and decreases mobility difficulty in women with OA.

1. The researchers stated that this was a 12-week longitudinal, randomized clinical trial pilot study with 28 women over 65 years of age with the diagnosis of OA. What are some of the possible problems or limitations that might occur with this type of study?

Sincethe study took quite a long time, researches will need to be carefulthat those in the control and intervention control groups do notdiscover which group they are in, as this possibility exists to agreater degree given the length of the trial. Also, a slightly largergroup of women may help to certify the results to a greater extent.

REFERENCE