Answer: 56.9 years to 63.1 years.
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 for [tex]\alpha[/tex].
Given: n= 24
degree of freedom = n- 1= 23
[tex]\overline{x}[/tex]= 60 years
s= 7.4 years
[tex]\alpha=0.05[/tex]
Two tailed t-critical value for significance level of [tex]\alpha=0.05[/tex] and degree of freedom 23:
[tex]t_{\alpha/2}=2.0687[/tex]
A 95% confidence interval on the true mean age:
[tex]60\pm (2.0686){\dfrac{7.4}{\sqrt{24}}}\\\\\approx60\pm3.1\\\\=(60-3.1,\ 60+3.1)\\\\=(56.9,\ 63.1)[/tex]
Hence, a 95% confidence interval on the true mean age. : 56.9 years to 63.1 years.
