Answer: (20.86, 22.52)
Step-by-step explanation:
Formula to find the confidence interval for population mean :-
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = sample mean.
z*= critical z-value
n= sample size.
[tex]\sigma[/tex] = Population standard deviation.
By considering the given question , we have
[tex]\overline{x}= 21.69[/tex]
[tex]\sigma=3.23[/tex]
n= 58
Using z-table, the critical z-value for 95% confidence = z* = 1.96
Then, 95% confidence interval for the amount of time spent on administrative issues will be :
[tex]21.69\pm (1.96)\dfrac{3.23}{\sqrt{58}}[/tex]
[tex]=21.69\pm (1.96)\dfrac{1.7}{7.61577}[/tex]
[tex]=21.69\pm (1.96)(0.223221)[/tex]
[tex]\approx21.69\pm0.83[/tex]
[tex]=(21.69-0.83,\ 21.69+0.83)=(20.86,\ 22.52)[/tex]
Hence, the 95% confidence interval for the amount of time spent on administrative issues = (20.86, 22.52)
