Answer:
The mean is [tex]x=0.503\frac{mg}{L}[/tex]
The 90% confidence interval is:
[tex]i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}][/tex]
Explanation:
1. First organize the data:
[tex]x_{1}=0.487[/tex]
[tex]x_{2}=0.487[/tex]
[tex]x_{3}=0.511[/tex]
[tex]x_{4}=0.511[/tex]
[tex]x_{5}=0.519[/tex]
As there are 5 data, the sample size (n) is n=5
2. Calculate the mean x:
The mean is calculated adding up all the data and divide them between the sample size.
[tex]x=\frac{0.511+0.487+0.511+0.487+0.519}{5}[/tex]
[tex]x=0.503\frac{mg}{L}[/tex]
3. Find 90% confidence interval.
The formula to find the confidence interval is:
[tex]i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})][/tex] (Eq.1)
where x is the mean, d is the standard deviation and n is the sample size.
And
[tex]1-\alpha=0.90[/tex]
[tex]\alpha=0.10[/tex]
[tex]\frac{\alpha}{2}=0.05[/tex]
[tex]z_{0.05}=1.645[/tex]
4. Find the standard deviation
[tex]d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}[/tex]
[tex]d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}[/tex]
[tex]d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}[/tex]
[tex]d=\sqrt{2.24*10^{-4}}[/tex]
[tex]d=0.015[/tex]
5. Replace values in (Eq.1):
[tex]i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})][/tex]
For the addition:
[tex]i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})][/tex]
[tex]i_{0.90}=0.514[/tex]
For the subtraction:
[tex]i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})][/tex]
[tex]i_{0.90}=0.492[/tex]
The 90% confidence interval is:
[tex]i_{0.90}=[0.492,0.514][/tex]
