Answer: [tex]3.4779[/tex]
Step-by-step explanation:
Given : Sample size : [tex]n=10[/tex]
Sample mean : [tex]\overline{x}=73[/tex]
Standard deviation : [tex]\sigma =6[/tex]
Significance level : [tex]\alpha=1-0.99=0.01[/tex]
Critical value : [tex]t_{n-1,\alpha/2}=t_{9,0.005}=1.833[/tex]
Formula to find the margin of error for population mean :-
[tex] t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=(1.833)\dfrac{6}{\sqrt{10}}\\\\\approx3.4779[/tex]
Hence, the margin of error if 99% confidence is 3.4779 .
