Answer: Correct critical value = 2.2622
Step-by-step explanation:
Confidence interval for population mean when population standard deviation is unknown:
[tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex] , where [tex]\overline{x}[/tex] = sample mean, n= sample size, s= sample standard deviation, [tex]t_{\alpha/2}[/tex] = two-tailed t value.
As per given: n= 10
degree of freedom : df = n-1=9
[tex]\overline{x}=242.6\\\\s=73.33\\\\\alpha=5\%=0.05[/tex]
Critical t-value : [tex]t_{\alpha/2,df}=t_{0.05/2,9}=t_{0.025,9}=2.2622[/tex]
So, the 95 percent confidence interval estimate for the mean :
[tex]242.6\pm (2.2622)\dfrac{73.33}{\sqrt{10}}\\\\=242.6\pm (2.2622)(23.19)\\\\\approx242.6\pm52.46\\\\=(242.6-52.46,\ 242.6+52.46)\\\\=(190.14,\ 295.06)[/tex]
The 95 percent confidence interval estimate for the mean:(190.14, 295.06)
