Answer: [tex](188.24,\ 211.76 )[/tex].
Step-by-step explanation:
Given : Sample size : n= 64 , the sample is a large sample (n>30), so we can apply z-test.
Sample mean = [tex]\overline{x}=200[/tex]
Standard deviation : [tex]\sigma=48[/tex]
Level of confidence:[tex]1-\alpha=0.95[/tex]
[tex]\Rightarrow\ \alpha=0.05[/tex]
Then, critical z-value =[tex]z_{\alpha/2}=1.96[/tex]
The confidence interval to estimate the population mean is given by :_
[tex]\overline{x}\ \pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]=200\ \pm\ (1.96)\dfrac{48}{\sqrt{64}}\\\\=200\pm11.76=(200-11.76, 200+11.76)=(188.24,\ 211.76 )[/tex]
Hence, the 95% confidence interval to estimate the population mean can be computed as [tex](188.24,\ 211.76 )[/tex].
