As per given , we have
n= 88
[tex]\overline{x}=36.5\text{ minutes}[/tex]
Population standard deviation : [tex]\sigma=8.6\text{ minutes}[/tex]
Critical value for 98% confidence interval : [tex]z_{\alpha/2}=2.33[/tex]n
Margin of error : [tex]E=z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]={2.33}\dfrac{8.6}{\sqrt{88}}\\\\=2.13605797717\approx2.14[/tex]
98% confidence interval : [tex]36.5\pm 2.14[/tex]
[tex](36.5-2.14,\ 36.5+2.14)=(34.36,\ 38.64)[/tex]
The 98% confidence interval (34.36, 38.64) found by normal distribution with the appropriate calculations for a standard deviation that is known is narrower than the interval (28.7,44.3) found by the t-distribution (population standard deviation is unknown) .
