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

The mean is
`x=0.503(mg)/(L)`

The 90% confidence interval is:

`i_(0.90)=[0.492(mg)/(L),0.514(mg)/(L)]`

Step-by-step 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=(0.511+0.487+0.511+0.487+0.519)/(5)`

`x=0.503(mg)/(L)`

3. Find 90% confidence interval.

The formula to find the confidence interval is:

`i_(0.90)=[x+/-z_{(\alpha)/(2)}*((d)/(√(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`

`(\alpha)/(2)=0.05`

`z_(0.05)=1.645`

4. Find the standard deviation

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

`d=\sqrt{((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{((-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*((0.015)/(2.236))]`

For the addition:

`i_(0.90)=[0.503+1.645*((0.015)/(2.236))]`

`i_(0.90)=0.514`

For the subtraction:

`i_(0.90)=[0.503-1.645*((0.015)/(2.236))]`

`i_(0.90)=0.492`

The 90% confidence interval is:

`i_(0.90)=[0.492,0.514]`