In a 1868 paper, German physician Carl Wunderlich reported based on over a milli
ID: 3363829 • Letter: I
Question
In a 1868 paper, German physician Carl Wunderlich reported based on over a million body temperature readings that the mean body temperature for healthy adults is 98.6° F. However, it is now commonly believed that the mean body temperature of a healthy adult is less than what was reported in that paper. To test this hypothesis a researcher measures the following body temperatures from a random sample of healthy adults.98.4, 98.7, 98.7, 98.6, 97.2, 98.1, 98.1, 97.6
(a) Find the value of the test statistic. (b) Find the 10% critical value In a 1868 paper, German physician Carl Wunderlich reported based on over a million body temperature readings that the mean body temperature for healthy adults is 98.6° F. However, it is now commonly believed that the mean body temperature of a healthy adult is less than what was reported in that paper. To test this hypothesis a researcher measures the following body temperatures from a random sample of healthy adults.
98.4, 98.7, 98.7, 98.6, 97.2, 98.1, 98.1, 97.6
(a) Find the value of the test statistic. (b) Find the 10% critical value
Explanation / Answer
Given that,
population mean(u)=98.6
sample mean, x =98.175
standard deviation, s =0.5445
number (n)=8
null, Ho: =98.6
alternate, H1: <98.6
level of significance, = 0.1
from standard normal table,left tailed t /2 =1.415
since our test is left-tailed
reject Ho, if to < -1.415
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =98.175-98.6/(0.5445/sqrt(8))
to =-2.2077
| to | =2.2077
critical value
the value of |t | with n-1 = 7 d.f is 1.415
we got |to| =2.2077 & | t | =1.415
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -2.2077 ) = 0.03151
hence value of p0.1 > 0.03151,here we reject Ho
ANSWERS
---------------
null, Ho: =98.6
alternate, H1: <98.6
a.
test statistic: -2.2077
b.
critical value: -1.415
decision: reject Ho
p-value: 0.03151
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