Answer:
Point estimation of the mean = 17.6
Error = 4.6
95% CI: 13.0 ≤ μ ≤ 22.2
The distribution can be approximated to a normal because the sample size n=50 is bigger than 30.
Step-by-step explanation:
We have this information:
- Sample mean:
[tex]\bar{X}=17.598[/tex]
- Sample standard deviation
[tex]s=16.01712719[/tex]
- Sample size
[tex]n=50[/tex]
For now on, we round to one decimal place.
The best estimation for the population mean is the sample mean
[tex]\mu=\bar{X}=17.6[/tex]
The margin of error is equal to the t-value multiplied by the sample standard deviation and divided by the sample size.
The t value depends on the degrees of freedom and the width of the confidence interval. In this case it is a 95% CI and the degrees of freedom are 49. For this conditions, the t-value is t=2.01.
Then, the margin of error is
[tex]E=t_{49}*\sigma/\sqrt{n}=2.01*16.0/\sqrt{50}=4.553[/tex]
Then, the confidence interval can be constructed as:
[tex]\bar{X}-t\sigma/\sqrt{n}\leq\mu\leq\bar{X}-t\sigma/\sqrt{n}\\\\17.6-4.6\leq \mu \leq 17.6+4.6\\\\ 13.0\leq \mu \leq 22.2[/tex]