Question

11. Twelve samples of water form a particular source have an average lead concentration of 12.5µg/L with a standard deviation of 2.0µg/L. 1 (a) Find a 95% confidence interval for the lead concentration in water from this source. (b) Can you conclude that the lead concentration is less than 14µg/L?

Answer 1

a) The 95% confidence interval for the lead concentration in water from this source is between 11.23 µg/L and 13.77 µg/L.

b) The upper bound of the confidence interval is less than 14 µg/L, which means that yes, we can conclude that the lead concentration is less than 14µg/L.

Explanation:

We have the standard deviation for the samples, which mean that the t-distribution will be used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 12 - 1 = 11

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.95)/(2) = 0.975`. So we have T = 2.201

The margin of error is:

`M = T(s)/(√(n)) = 2.201(2)/(√(12)) = 1.27`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12.5 - 1.27 = 11.23 µg/L.

The upper end of the interval is the sample mean added to M. So it is 12.5 + 1.27 = 13.77 µg/L.

The 95% confidence interval for the lead concentration in water from this source is between 11.23 µg/L and 13.77 µg/L.

(b) Can you conclude that the lead concentration is less than 14µg/L?

The upper bound of the confidence interval is less than 14 µg/L, which means that yes, we can conclude that the lead concentration is less than 14µg/L.