Answer: [tex](4.24,\ 5.36 )[/tex]
Step-by-step explanation:
The confidence interval to estimate the population mean is given by :_
[tex]\overline{x}\ \pm\ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
Given : Sample size : n= 60 , the sample is a large sample (n>30), so we can apply z-test.
Sample mean = [tex]\overline{x}=4.8[/tex]
Standard deviation : [tex]\sigma=2.2[/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]
Then, a 95% confidence interval estimate of the population mean will be
[tex]=4.8\ \pm\ (1.96)\dfrac{2.2}{\sqrt{60}}\\\\=4.8\pm0.56=(4.8-11.76, 4.8+11.76)=(4.24,\ 5.36 )[/tex]
Hence, the 95% confidence interval to estimate the population mean = [tex](4.24,\ 5.36 )[/tex].
