Question

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 of all healthy humans. Group of answer choices ( 98.04, 98.36 ) ( 97.54, 97.95 ) ( 97.84, 98.12 ) ( 97.82, 98. 15 ) ( 97.95, 98.15 )

Answer 1

Answer: $(98.04,\ 98.36)$

Step-by-step 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 :

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

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 :

$98.2\pm (2.623)\dfrac{0.62}{\sqrt{106}}$

$98.2\pm (2.623)(0.0602197234662)$

$98.2\pm (0.157956334652)$

$\approx98.2\pm 0.16$

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

$(98.04,\ 98.36)$

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