Answer:
[tex](832.156, \ 847.844)[/tex]
Step-by-step explanation:
Given data :
Sample standard deviation, s = 15
Sample mean, [tex]\overline x = 840[/tex]
n = 23
a). 98% confidence interval
[tex]$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$[/tex]
[tex]$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}[/tex]
[tex]$t_{(n-1 , \alpha/2)} \frac{s}{\sqrt n}$[/tex]
[tex]$t_{(n-1, a\pha/2)}=t_{(22,0.01)} = 2.508$[/tex]
∴ [tex]$E = 2.508 \times \frac{15}{\sqrt{23}}$[/tex]
[tex]$E = 7.844$[/tex]
So, 98% CI is
[tex]$(\overline x - E, \overline x + E)$[/tex]
[tex](840-7.844 , \ 840+7.844)[/tex]
[tex](832.156, \ 847.844)[/tex]