Human oral normal body temperature is believed to be 98.6 degree F, but there is

Question

Human oral normal body temperature is believed to be 98.6 degree F, but there is evidence that it actually should be 98.2 degree F. From a sample of 52 healthy adults, the mean oral temperature was 98.285 with a standard deviation of 0.625 degrees. State any necessary assumptions about the underlying distribution of the data. (a) What are the null and alternative hypotheses? (b) Test the null hypothesis at alpha = 0.05. (c) What is the P-value for this test? (d) How does a 95% confidence interval answer the same question?

Explanation / Answer

Given that,
population mean(u)=98.2
sample mean, x =98.285
standard deviation, s =0.625
number (n)=52
null, Ho: <98.2
alternate, H1: >98.2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.675
since our test is right-tailed
reject Ho, if to > 1.675
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =98.285-98.2/(0.625/sqrt(52))
to =0.981
| to | =0.981
critical value
the value of |t | with n-1 = 51 d.f is 1.675
we got |to| =0.981 & | t | =1.675
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.9807 ) = 0.16568
hence value of p0.05 < 0.16568,here we do not reject Ho
ANSWERS
---------------
null, Ho: =98.2
alternate, H1: >98.2
test statistic: 0.981
critical value: 1.675
decision: do not reject Ho
p-value: 0.16568
e.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=98.285
Standard deviation( sd )=0.625
Sample Size(n)=52
Confidence Interval = [ 98.285 ± t a/2 ( 0.625/ Sqrt ( 52) ) ]
= [ 98.285 - 2.008 * (0.087) , 98.285 + 2.008 * (0.087) ]
= [ 98.111,98.459 ]

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.