Answer: Â [tex](9.90,\ 10.00)[/tex]
Step-by-step explanation:
Confidence interval for population mean:
[tex]\overline{x}\pm z^c\dfrac{\sigma}{\sqrt{n}}[/tex]
where , [tex]\overline{x}[/tex] = sample mean , [tex]\sigma[/tex] = population standard deviation, n= sample size, [tex]z^c[/tex] = Critical z-value.
As per given, [tex]n= 35,\ \ \ \overline{x} =9.95,\ \ \ \ \sigma=0.15[/tex]
We will find 95% confidence interval for which [tex]z^c=1.96[/tex]
Required confidence interval:
[tex]9.95\pm(1.96)\dfrac{0.15}{\sqrt{35}} \\\\= 9.95\pm (0.049695)\\\\=( 9.95-0.049695,\ 9.95+0.049695)\\\\=(9.900305,\ 9.999695)\approx(9.90,\ 10.00)[/tex]
Required confidence interval : [tex](9.90,\ 10.00)[/tex]
