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A clinical trial was conducted to evaluate the effectiveness of three treatment creams (Cream 1 - 3) for fever blisters. Twenty-four patients were randomly divided into four groups of equal size. Each group of six (6) patients were then allocated to be treated with one of the four creams (Cream 1-3 and plus a control group labelled Cream 4). The data recorded are the number of days [Days] from initial appearance of the blisters until healing is complete. (Data source: Kleinbaum, D.G., Kupper, L.L. and Muller, K.E., 1988 ).


Note: Mainly need to be done by hand but may make use of R/RStudio when possible.
a) Comment on two main interesting features of the blister data based on the boxplot of the data below.
b) Suggest an appropriate analysis (or model) for this data for comparing the four treatment groups (creams).
c) Suppose we wish to perform a one-way ANOVA for the blister data. Using the descriptive statistics below to complete the ANOVA table shown on next page. Also show, in details, how each item, such as SST, is computed. (Note that you may need first to work out the overall/grand mean based on the treatment/Cream group means given below.)
Cream i (1-4) n mean sd
 1 6 5.000 1.265
 2 6 4.333 1.033
 3 6 5.167 1.472
 4 6 7.500 1.643 

d) Based on the results in your completed ANOVA table above in part c, test if there is any difference in average number of days until healing across the four cream groups. Remember to state null and alternative hypotheses, give the test statistic value and associated degrees of freedom, p-value or range of it, and draw your conclusion (assuming that all relevant assumptions are ok for this part).
e) If one wishes to carry out multiple pairwise comparisons of all the four creams (including the control group) at an overall 5% significance level, 
i) what is the total number of these comparisons
ii) what are the significance level for each individual comparison (test) if using Bonferroni approach and the relevant t critical value in this case? (Do NOT carry out the tests.)
f) If someone wants to find out if there is any difference in average number of days until healing between the three real treatment creams (Cream 1-3) as a whole and the control group (Cream 4), form an appropriate contrast to evaluate and test the difference. Remember to state null and alternative hypotheses, give the test statistic value and associated degrees of freedom, the evidence for rejecting or not rejecting the null hypothesis, and give a conclusion that includes a relevant 95% confidence interval. (Note: Means and sample sizes by Cream group are presented in part c.)


g) In the case that only the normality assumption of an ANOVA model is violated, explain which non-parametric procedure may be considered. (Do NOT carry out this procedure!)
h) What would you suggest to try if both normality and equal variance assumptions are violated Why.

Interesting features of the blister data

For all the three cream (1, 2 and 3) the minimum number of days to start witnessing the healing effect on the blisters is four days. This is indicated by the lower bases of all the box-plots scaling at 4 days on the y-axis. However, for the control treatment, it takes longer for the healing to be evident. This is shown by the base of the box-plot for cream 4 touching on seven days. But different from the first three creams, it takes less number of days for the members under control group to be all healed of the blisters. This is indicated by the range depicted by the box-plot (8-7=1). All the members in the control group get their blisters within one day..

The number of days taken for the blisters to heal under cream-1 and cream-3 are not normally distributed but skewed to the left.  This is indicated by the second quartile line which is position on the left side of the box-plot. The numbers of days for cream-2 are skewed to the right. This is indicated by the second quartile line which is position on the right side of the box-plot.  It is only under the control group that the number of days taken for the blisters to heal is normally distributed. This is shown by the median or second quartile line being at the middle.

  1. Appropriate analysis

The most appropriate analysis for this model is an analysis of variance (ANOVA). This is due to the test’s ability to determine the difference between group means for samples which are more than two (David & George, 2006).

  1. Hypothesis test

Null hypothesis: There is no difference in the mean number of days until healing across the four cream groups.

Versus

Alternative: There is a significant difference in the mean number of days until healing across the four cream groups.

  1.  
  2. There are 6 pairs to be compared.
  3. The level of significance for each comparison is 5% and the t-critical value is 0.025
  4. Hypothesis test

If we want to compare the mean number of days for the tree creams and the control group, we reduce the three samples for cream 1-3 into one sample by computing the average number of days for the three creams. The average number will form the test statistic value of the test.  At this point we will be having two samples. We therefore employ a test to test for difference in the number of days that the blisters take to heal between cream 1-3 and the control group (cream 4). The hypothesis is as below;

The test statistic value = (5 + 7.5 + 4.333 + 5.167)/4 = 5.5

The degrees of freedom is (24-1) = 23

H0: mean cream 1-3 = mean cream 4 = 5.5

Versus

H1: At least one mean is different

If a comparison is done between the level of significance and the p-value, it can be observed that the p-value is greater than the level of significance, (.74 > .05). The decision is therefore to accept the null hypothesis and reject the alternative. This leads the study to conclude that there is no significant difference in the average number of days taken for the blisters to heal between cream1-4 and control group which also known as cream 4.

  1. Alternative test incase violation of normality of ANOVA

Appropriate analysis for comparing the four treatment groups

Analysis of variance is one of the most robust tests to withstand violations of assumptions. It tolerates violations up to a certain extent above which it can have a great effect on the results. This happens especially in platykurtotic distributions. However there are other tests to be used instead of ANOVA incase assumption of normality is violated. The non-parametric test that can take place of ANOVA is a Kruskal-Wallis H test (Pallant, 2009).

  1. Mann-Whitney test is the most appropriate test to use instead of anova in case both normality and equal variance assumptions are violated. This is because the test is less sensitive to normality and equal variances.
  2. Assumptions of the analysis
  • There is homogeneity of variance among the three groups. This means that the variances of the three groups are equal.
  • The data in the three groups are normally distributed.
  • All the three groups are independent of each other.

To test for normality we can either use box-plot analysis or even conduct a Lavene’s test (Gravetter, 2011).

Applying box-plot for low dose we get the chart below;

As can be observed in figure one above, the second quartile or the median line is right at the middle of the box-plot. This is an indication that the data is normally distributed.

the data for high dose and the control group are almost normally distributed. This is shown by the median line cutting almost at the middle of the box-plot. If the line was above or below the middle, it would mean that the data is skewed.

To test for the homogeneity of variance among the groups, a Lavene’s test is employed (Warmbrod, 2001). This test uses F-test and maintains its null hypothesis that the variance is the same in all the groups. If in case the p-value is established to be less than .05, then the null hypothesis is rejected.

the null hypothesis that the error variance of the dependent variable is equal across groups.Since the p-value is less than .05, we accept the null hypothesis that there is homogeneity of variance across the groups.

H0: There is no significant difference in the average BMD among the three groups of rats.

Versus

              H1: There is a significant difference in the average BMD among the three groups of rats.

Since the p-value, .94 is greater than the level of significance, .05; we therefore reject the null hypothesis and accept the alternative that there is a significant difference in the average BMD among the three groups of rats.

  1. Pairwise comparison

Comparing the low dose and the control group we employ t-test since we are testing for the difference in means between two samples only. The t-test results are as shown in the table below;

Hypothesis

H0: There is no significant difference in the average BMD between the low dose and the control group.  

Versus

H1: There is a significant difference in the average BMD between the low dose and the control group.Since the p-value, .46 is greater than the level of significance, .05; we therefore accept the null hypothesis and reject the alternative that there is no significant difference in the average BMD between the low dose and the control group.

Another pairwise comparison was conducted using t-test between low-dose and high-dose group. The tests results were as follows below 

H0: There is no significant difference in the average BMD between the low dose and the control group.  

Versus

H1: There is a significant difference in the average BMD between the low dose and the control group.

Since the p-value, .004 is less than the level of significance, .05; we therefore reject the null hypothesis and accept the alternative that there is a significant difference in the average BMD between the low dose and the high-dose -group.

  1. Suggestion on the method to use
  2. The researcher can employ multiple analysis of variance (MANOVA).
  3. ii) The significance level will be 99% and the critical value will be .05.
  4. Testing whether test statistic value = 0.21 in the control group
  5. In this kind of test, we employ a one sample t-test as follows

Since the p-value, .01 is less than the level of significance, .05; we therefore reject the null hypothesis and accept the alternative that the mean BMD in the control group is higher than .21.

  1. We will use a non-parametric test such as Mann Whitney test which is not sensitive to inequality of variance in groups of data.
  • Since the value will be greater than the value being tested and the initial test was testing whether the value is .21, this will be a case of a one-tailed test and therefore the power of the test would be (.01x2=.02). This means that the interpretation of the result will remain the same; rejecting the null hypothesis and accepting the alternative.
  1. If the sample size increases, the power will increase, this is because the level of precision will go high due to the fact that a large population sample is associated with a well representation of the entire population.

References

Warmbrod, J. R. (2001). Conducting, interpreting, and reporting quantitative research. Research Pre-Session, New Orleans, Louisiana

Pallant, J. F. (2009). A step by step guide to data analysis using SPSS. Survival manual.

David, S., & George , P. (2006). Introduction to the practice of Statistics. Freeman.

Gravetter , F. J. (2011). Research methods for behavioral sciences.

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My Assignment Help. Analysis Of Clinical Trial For Fever Blisters [Internet]. My Assignment Help. 2021 [cited 18 July 2024]. Available from: https://myassignmenthelp.com/free-samples/stat680-applied-statistics/analysis-of-variance.html.

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