Gven
A survey of 22 randomly selected students finds that they save a mean of $195 per semester.
Assume the data comes from a normal distribution and the sample standard deviation is $10 per month.
To find the 90% confidence interval to estimate the population mean, Round your answers to two decimal places.
Explanation:
It is given that,
[tex]\begin{gathered} n=22 \\ \bar{x}=195 \\ s=10 \end{gathered}[/tex]Then, the confidence interval is determined by,
[tex]\begin{gathered} CI=\bar{x}\pm z\times\frac{s}{\sqrt{n}} \\ =195\pm1.6449\times\frac{10}{\sqrt{22}} \\ =195\pm3.507 \end{gathered}[/tex]That implies,
[tex]\begin{gathered} 195-3.507<\mu<195+3.507 \\ 191.493<\mu<198.507 \\ 191.50<\mu<198.51 \end{gathered}[/tex]Hence, the confidence interval is,
[tex]191.50\lt\mu\lt198.51[/tex]