Question

A simple random sample from a population with a normal distribution of 106106 body temperatures has x overbarxequals=98.1098.10degrees Upper F°F and sequals=0.610.61degrees Upper F°F. Construct anan 8080​% confidence interval estimate of the standard deviation of body temperature of all healthy humans.

Answer 1

Answer: (97.084, 99.116)

Step-by-step explanation:

As per given , we have

sample size :n= 106

degree of freedom: df = 105      (df=n-1)

$\overline{x}=98.10\\ s=0.61$

Significance interval : $\alpha: 1-0.80=0.20$

Since , population standard deviation is unknown , so we use t-test .

Using t-value table ,

$t_{df,\ \alpha/2}=t_{(105,\ 0.10)}=1.290$

Now , 80% Confidence interval will be :

$\overline{x}\pm t_{df,\ \alpha/2}\dfrac{s}{\sqrt{n}}$

$98.10\pm (1.290)\dfrac{0.61}{\sqrt{0.6}}$

$\approx98.10\pm 1.016$

$=(98.10- 1.016,\ 98.10+ 1.016)=(97.084,\ 99.116)$

Thus , The 80​% confidence interval estimate of the standard deviation of body temperature of all healthy humans = (97.084, 99.116).