a. Given that:
- The mean age of the 36 respondents is 40 years.
- The standard deviation of ages is 10 years.
You need to construct a 95% confidence interval estimate of the mean age of the population from which the sample was selected.
Then, you need to use this formula:
[tex]x\pm z\cdot\frac{\sigma}{\sqrt{n}}[/tex]
Where "x" is the sample mean, "z" is the confidence level value, "n" is the sample size, and σ is the standard deviation.
In this case:
[tex]\begin{gathered} x=40 \\ \sigma=10 \\ n=36 \end{gathered}[/tex]
Therefore, by substituting values into the formula:
[tex]40\pm\frac{10}{\sqrt{36}}[/tex]
You get these two values:
[tex]40+\frac{10}{\sqrt{36}}\approx41.67\text{ }[/tex][tex]40-\frac{10}{\sqrt{36}}\approx38.33[/tex]
b. If you assume that:
[tex]\sigma=6[/tex]
You get the following values by substituting them into the formula:
[tex]40+\frac{6}{\sqrt{36}}=41[/tex][tex]40-\frac{6}{\sqrt{36}}=39[/tex]
Hence, the answers are:
a.
[tex](38.33,41.67)[/tex]
b.
[tex](39,41)[/tex]
