The effectiveness of treating respiratory infections with herbal remedies was st
ID: 3326304 • Letter: T
Question
The effectiveness of treating respiratory infections with herbal remedies was studied. “Days of fever” was used to measure effects. Among 421 children treated with herbal remedies, the mean number of days with fever was 0.49, with a standard deviation of 1.32. Among 316 children given placebo, the mean was 0.70, with a standard deviation of 1.16 days. Use a 0.01 significance level to test the claim that herbal remedies affect the number of days with fever. Do not assume the population standard deviations are equal. Based on these results, do herbal remedies appear to be effective?
Explanation / Answer
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Herbal remedies do not affect the number of days with fever. That is the mean number of days with fever after treatment with herbal remedies is same as that after treatment with placebo. 1 - 2 = 0
Alternative hypothesis: Herbal remedies affect the number of days with fever. That is the mean number of days with fever after treatment with herbal remedies is different than that after treatment with placebo. 1 - 2 0
Note that these hypotheses constitute a two-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis as, we do not know the population statndard deviation and population standard deviations of the two treatments may not be equal.
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)]
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.
SE = sqrt[(1.322/421) + (1.162/316] = 0.0916
Degree of freedom is given as,
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (1.322/421 + 1.162/316)2 / { [ (1.322 / 421)2 / (420) ] + [ (1.162 / 316)2 / (315) ] }
= 717 (Rounding to nearest integer)
t = [ (x1 - x2) - d ] / SE = [ (0.49 - 0.70) - 0 ] / 0.0916 = -2.29
Since we have a two-tailed test, the P-value is the probability that a t statistic having 717 degrees of freedom is more extreme than -2.29; that is, less than -2.29 or greater than 2.29.
We use the t Distribution Calculator to find P(t < -2.29) = 0.0112, and P(t > 2.29) = 0.0112. Thus, the P-value = 0.0112 + 0.0112 = 0.0224.
Interpret results. Since the P-value (0.0224) is greater than the significance level (0.01), we fail to reject the null hypothesis and conclude that at 0.01 significance level, there is not significant evidence that the mean number of days with fever after treatment with herbal remedies is different than that after treatment with placebo.That is, there is not significant evidence that Herbal remedies affect the number of days with fever.
Based on these results, at 0.01 significance level, herbal remedies do not appear to be effective.
(Note - If the significance level is 0.05, then at 0.05 significance level, herbal remedies appear to be effective as p-value (0.0224) is less than 0.05).
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