Answer: (15.47, 24.53)
Step-by-step explanation:
We know that the confidence interval for population mean is given by :_
[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]
, where n= sample size.
[tex]\sigma[/tex] = standard deviation.
[tex]\overline{x}[/tex]= sample mean.
z*= Critical value.
Given : n= 450
[tex]\overline{x}=20[/tex]
[tex]\sigma=49[/tex]
Critical value for 95% confidence = z*=1.96 [From z-value table]
Then, the 95% confidence interval will be :-
[tex]20\pm (1.96)\dfrac{49}{\sqrt{450}}[/tex]
[tex]\approx 20\pm (1.96)(2.31)[/tex]
[tex]\approx 20\pm 4.53[/tex]
[tex]=(20-4.53,\ 20+4.53)=(15.47,\ 24.53)[/tex]
Hence, the 95% confidence interval for the mean change in score μ μ in the population of all high school seniors. : (15.47, 24.53)
