The baseline characteristics of different treatment groups are often included in
ID: 3230514 • Letter: T
Question
The baseline characteristics of different treatment groups are often included in journal articles. In one study, subjects treated with raw garlic had LDL 49 while 48 subjects lateral measurements with a mean of 151 and a standard deviation of 15, given placebos had LDL measurements with a mean of 149 and a standard deviation of 14 based on data from "Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia, " by Gardner et al., Archives of Internal Medicine, Vol. 167). Use a 0.05 significance level to cholesterol levels of sub- test the claim that there is no difference between the mean L groups appear to be about jests treated with raw garlic and given placebos. Do both the same?Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0, i.e., there is no difference between the mean LDL cholestrol levels of subjects treated with raw garlic and subjects given placebos.
Alternative hypothesis: 1 - 2 0, i.e., there is difference between the mean LDL cholestrol levels of subjects treated with raw garlic and subjects given placebos.
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(152/49) + (142/48)] = 6.22619
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (152/49 + 142/48)2 / { [ (152 / 49)2 / (48) ] + [ (142 / 48)2 / (47) ] }
DF = 75.2585757092 / (0.43927009579 + 0.35475768321) = 94.7807843751 or 95 (rounding to the nearest whole number)
t = [ (x1 - x2) - d ] / SE = [ (151 - 149) - 0 ] / 6.22619 = 0.32122373393 or 0.3212
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 95 degrees of freedom is more extreme than 0.3212; that is, less than -0.3212 or greater than 0.3212.
We use the t Distribution Calculator to find P(t < 0.3212)
The P-Value is 0.748764.
Interpret results. Since the P-value (0.748764) is more than the significance level (0.05), we can accept the null hypothesis. That is there is no difference between the mean LDL cholestrol levels of subjects treated with raw garlic and subjects given placebos.
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