Answer: [tex](98.04,\ 98.36)[/tex]
Step-by-step explanation:
Given : Sample size of healthy adults: n= 106
Degree of freedom = df =n-1 = 105
Sample mean body temperature : [tex]\overline{x}=98.2[/tex]
Sample standard deviation : [tex]s= 0.62[/tex]
Significance level ; [tex]\alpha= 1-0.99=0.01[/tex]
∵ population standard deviation is unknown , so we use t- critical value.
Confidence interval for the population mean :
[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
Using t-distribution table , we have
Critical value for df = 105 and [tex]\alpha=0.01[/tex]= [tex]t_{\alpha/2, df}=t_{0.005 , 105}=2.623[/tex]
A 99% confidence interval estimate of the mean body temperature of all healthy humans will be :
[tex]98.2\pm (2.623)\dfrac{0.62}{\sqrt{106}}[/tex]
[tex]98.2\pm (2.623)(0.0602197234662)[/tex]
[tex]98.2\pm (0.157956334652)[/tex]
[tex]\approx98.2\pm 0.16[/tex]
[tex](98.2-0.16,\ 98.2+0.16)[/tex]
[tex](98.04,\ 98.36)[/tex]
Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans = [tex](98.04,\ 98.36)[/tex]
