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.
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).
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.
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.
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).
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.
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.
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.
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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