Drinking tea appears to offer a strong boost to the immune system. In a study, w
ID: 3314664 • Letter: D
Question
Drinking tea appears to offer a strong boost to the immune system. In a study, we see that production of interferon gamma, a molecule that fights bacteria, viruses, and tumors, appears to be enhanced in tea drinkers. In the study, eleven healthy non-tea-drinking individuals were asked to drink five or six cups of tea a day, while ten healthy non-tea and non-coffee-drinkers were asked to drink the same amount of coffee, which has caffeine but not the L-theanine that is in tea. The groups were randomly assigned. After two weeks, blood samples were exposed to an antigen and production of interferon gamma was measured. The results are shown in Table 1 and are available in ImmuneTea. The question of interest is whether the data provide evidence that production is enhanced in tea-drinkers.
Tea 5 11 13 18 20 47 48 52 55 56 68
Coffee 0 0 3 11 15 16 21 21 38 52
the absolute tolerance is +/-0.005
Tea 5 11 13 18 20 47 48 52 55 56 68
Coffee 0 0 3 11 15 16 21 21 38 52
Find a standardized test statistic and use the t-distribution to find the p-value and make a conclusion using a 5% significance level.
Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places.
test statistic =
p-value =
A randomization test might be a more appropriate test to use in this case. Construct a randomization distribution for this test and use it to find a p-value and make a conclusion.
Click here to access StatKey.
Round your answer to three decimal places.
p-value =
the absolute tolerance is +/-0.005
Explanation / Answer
Given that,
mean(x)=35.7273
standard deviation , s.d1=22.3611
number(n1)=11
y(mean)=17.7
standard deviation, s.d2 =16.6936
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =35.7273-17.7/sqrt((500.01879/11)+(278.67628/10))
to =2.10527
| to | =2.10527
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 2.10527 & | t | = 1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.1053 ) = 0.03228
hence value of p0.05 > 0.03228,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.10527
critical value: 1.833
decision: reject Ho
p-value: 0.03228
data provide evidence that production is enhanced in tea-drinkers.
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