Question

During a routine check of the fluoride content of Gotham City\'s water supply, the following results were obtained from replicate analyses of a single sample: 0.511 mg/L, 0.487 mg/L, 0.511 mg/L, 0.487 mg/L, and 0.519 mg/L Determine the mean and 90% confidence interval for the average fluoride concentration in this sample.
Mean = 0.503mg/L
Find 90% confidence interval

Answer 1

Answer:

The mean is $x=0.503\frac{mg}{L}$

The 90% confidence interval is:

$i_{0.90}=[0.492\frac{mg}{L},0.514\frac{mg}{L}]$

Explanation:

1. First organize the data:

$x_{1}=0.487$

$x_{2}=0.487$

$x_{3}=0.511$

$x_{4}=0.511$

$x_{5}=0.519$

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.

$x=\frac{0.511+0.487+0.511+0.487+0.519}{5}$

$x=0.503\frac{mg}{L}$

3. Find 90% confidence interval.

The formula to find the confidence interval is:

$i_{0.90}=[x+/-z_{\frac{\alpha}{2}}*(\frac{d}{\sqrt{n}})]$ (Eq.1)

where x is the mean, d is the standard deviation and n is the sample size.

And

$1-\alpha=0.90$

$\alpha=0.10$

$\frac{\alpha}{2}=0.05$

$z_{0.05}=1.645$

4. Find the standard deviation

$d=\sqrt{\frac{(x_{1}-x)^{2}+(x_{2}-x)^{2}+(x_{3}-x)^{2}+(x_{4}-x)^{2}+(x_{5}-x)^{2}}{n-1}}$

$d=\sqrt{\frac{(0.487-0.503)^{2}+(0.487-0.503)^{2}+(0.511-0.503)^{2}+(0.511-0.503)^{2}+(0.519-0.503)^{2}}{4}}$

$d=\sqrt{\frac{(-0.016)^{2}+(-0.016)^{2}+(0.008)^{2}+(0.008)^{2}+(0.016)^{2}}{4}}$

$d=\sqrt{2.24*10^{-4}}$

$d=0.015$

5. Replace values in (Eq.1):

$i_{0.90}=[0.503+/-1.645*(\frac{0.015}{2.236})]$

For the addition:

$i_{0.90}=[0.503+1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.514$

For the subtraction:

$i_{0.90}=[0.503-1.645*(\frac{0.015}{2.236})]$

$i_{0.90}=0.492$

The 90% confidence interval is:

$i_{0.90}=[0.492,0.514]$