Answer:
Confidence interval 99%
[tex]4.0\leq\mu \leq 5.4[/tex]
Step-by-step explanation:
In this case, we have to estimate a confidence interval (CI) with an unknown standard deviation of the population (we only known the s.d. of the sample).
The estimated standard deviation is
[tex]s_M=s/\sqrt{N}=2.9/\sqrt{125}=2.9/ 11.18=0.26[/tex]
For a 99% CI and degrees of freedom df=124, the t-statistic is t=2.616.
So the confidence interval is:
[tex]M-t*s_M\leq \mu\leq M+t*s_M\\\\4.7-2.616*0.26\leq\mu \leq 4.7+2.616*0.26\\\\4.7-0.68\leq\mu \leq 4.7+0.68\\\\4.0\leq\mu \leq 5.4[/tex]