Rhino viruses typically cause common colds. In a test of the effectiveness of ec
ID: 3323096 • Letter: R
Question
Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 46 of the 52 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 95 of the 110 subjects developed rhinovirus infections. Use a .01 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below.
a) test the claim using a hypothesis test
identify the test statistic
z = ___
identify the p value
p = ___
what is the conclusion based on the hypothesis test?
the p value is (greater/less than) the significance level, so (reject/fail to reject) the null hypothesis
b) test the claim by constructing an appropriate confidence interval
the 99% confidence level is ___ < (p1-p2) < ___ (round to 3 decimal places as needed)
what is the conclusion based on the confidence interval?
Because the confidence interval limits (include/do not inclue) 0, there (does/does not) appear the be a significant difference between the two proportions. There (is/is not) evidence to support this effect
c) Based on the results, does echinacea appear to have any effect on the infection rate?
Explanation / Answer
p1 = 46/52 = 0.88
n1 = 52
p2 = 95/110 = 0.86
n2 = 110
Pooled sample proportion. Since the null hypothesis states that P1=P2, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.
p = (p1 * n1 + p2 * n2) / (n1 + n2)
where p1 is the sample proportion from population 1, p2 is the sample proportion from population 2, n1 is the size of sample 1, and n2 is the size of sample 2.
p = (0.88*52 + 0.86*110) /(110+52) = 0.87
Standard error. Compute the standard error (SE) of the sampling distribution difference between two proportions.
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
= sqrt( 0.87 * ( 1 - 0.87 ) * [ (1/52) + (1/110) ] ) = 0.056
where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.
Test statistic. The test statistic is a z-score (z) defined by the following equation.
z = (p1 - p2) / SE
where p1 is the proportion from sample 1, p2 is the proportion from sample 2, and SE is the standard error of the sampling distribution.
z = (0.88-0.86)/0.056 = 0.357
we check the p value from the z table as 0.6394
the p value is (greater) the significance level, so (fail to reject) the null hypothesis
b )
we know the z value for 99% CI is
2.58
we know that the Ci is given as
(p1-p2) +- z*SE
putting the values
(0.88 - 0.86) +- 2.58*0.056 , so the interval is
-0.124 and 0.164
Because the confidence interval limits (include) 0, there (does not) appear the be a significant difference between the two proportions. There (is not) evidence to support this effect
c) Based on the results, does echinacea appear to have any effect on the infection rate?
as the result is not significant , hence there is no sufficient evidence
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