A sample of 106 healthy adults have a mean body temperature of 98.2, and a standard deviation of 0.62. Construct a 99% confidence interval estimate of the mean body temperature …

Question

asked May 7, 2020 70.3k views

1 Answer

Sunspawn · Answer 1 · 2020-05-13T05:54:38+0000

Answer:
$(98.04,\ 98.36)$

Explanation:

Given : Sample size of healthy adults: n= 106

Degree of freedom = df =n-1 = 105

Sample mean body temperature :
$\overline{x}=98.2$

Sample standard deviation :
s= 0.62

Significance level ;
$\alpha= 1-0.99=0.01$

∵ population standard deviation is unknown , so we use t- critical value.

Confidence interval for the population mean :

$\overline{x}\pm t_(\alpha/2, df)(s)/(√(n))$

Using t-distribution table , we have

Critical value for df = 105 and
$\alpha=0.01$ =
$t_(\alpha/2, df)=t_(0.005 , 105)=2.623$

A 99% confidence interval estimate of the mean body temperature of all healthy humans will be :

$98.2\pm (2.623)(0.62)/(√(106))$

$98.2\pm (2.623)(0.0602197234662)$

$98.2\pm (0.157956334652)$

$\approx98.2\pm 0.16$

$(98.2-0.16,\ 98.2+0.16)$

$(98.04,\ 98.36)$

Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans =
$(98.04,\ 98.36)$