Question

Rhino viruses typically cause common colds. In a test of the effectiveness of? echinacea, 31 of the 38 subjects treated with echinacea developed rhinovirus infections. In a placebo? group, 93 of the 109 subjects developed rhinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rhinovirus infections.

a. Test the claim using a hypothesis test.
Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis? test?

Identify the test statistic.
z= _?(Round to two decimal places as? needed.)
Identify the? P-value.
?P-value= ?(Round to three decimal places as? needed.)
What is the conclusion based on the hypothesis? test?
The? P-value is
?
less than
greater than
the significance level of alphaequals0.01?, so
?
reject
fail to reject
the null hypothesis. There
?
is
is not
sufficient evidence to support the claim that echinacea treatment has an effect.

Test the claim by constructing an appropriate confidence interval.
The 99?% confidence interval is _ <(p 1 - p 2)< _
?(Round to three decimal places as? needed.)
What is the conclusion based on the confidence? interval?
Because the confidence interval limits
?
do not include
include
?0, there
?
does
does not
appear to be a significant difference between the two proportions. There
?
is not
is
evidence to support the claim that echinacea treatment has an effect.

Based on the? results, does echinacea appear to have any effect on the infection? rate?
A.
Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate.
B.
Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate.
C.
Echinacea does not appear to have a significant effect on the infection rate.
D.
The results are inconclusive.

Answer 1

we have p1 =31/38 and p2 = 93/109, n1 =38 and n2 = 109

Null hypothesis:-

Alternate hypothesis-

It is two tailed becaue we have to determine whether there is any significant difference or not.

z statistic =

setting the values, we get

z statistic =

this gives

z statistic =

Using normal distribution table, we can get the p value corresponding to z score of -0.52

P value = 0.6031

(A) At 0.01, p value is insignificant because p value is greater than 0.01

Fail to reject the null hypothesis because there is not suffiicient evidence to support the claim.

(B) Confidence interval =

here z = 2.576 for 99% confidence level

setting the values, we get

this gives = (-0.2214,0.1466)

So, the confidence interval includes 0, means fails to reject the null hypothesis

Because the confidence interval limits
include? 0, there
does not
appear to be a significant difference between the two proportions. There
is not
evidence to support the claim that echinacea treatment has an effect.

option C is correct "Echinacea does not appear to have a significant effect on the infection rate." because there is no significant difference