Question

Periodically, a town water department tests the the drinking water of homeowners for contaminants such as lead. The lead levels in water specimens collected for a sample of 10 residents of the town had a mean of 2.4mg/L and a standard deviation of 1.2 mg/L. Complete parts a through c.

a. Construct a 90 % confidence interval for the mean lead level in water specimens from the town.

b. Interpret the interval in terms of this application. Choose the correct answer below.

A. The water department can be 90 % confident that the mean lead level in drinking water for all residents in the town is outside this interval.

B. The water department can be 90 % confident that the mean lead level in drinking water for all residents in the town is 2.4 mg/L.

C. The water department is confident that 90 % of the mean lead level in drinking water for all residents in the town is described by this interval.

D. The water department can be 10 % confident that the mean lead level in drinking water for all residents in the town is within this interval.

E. The water department can be 90 % confident that the mean lead level in drinking water for all residents in the town is within this interval.

c. What is meant by the phrase "90 % confidence interval"?

Answer 1

The 90% confidence interval for the mean lead level in the town's water specimens is (1.703, 3.097) mg/L. The water department can be 90% confident that the mean lead level for all residents is within this interval. A 90% confidence interval indicates that if many samples were taken, about 90% of the constructed intervals would capture the true population mean.

Step-by-step explanation:

To construct a 90% confidence interval for the mean lead level in water specimens from the town, we use the sample mean (μ = 2.4 mg/L) and the standard deviation (s = 1.2 mg/L) alongside the sample size (n = 10). The critical value for a 90% confidence level with 9 degrees of freedom (n-1) can be found in a t-distribution table, typically around 1.833. The confidence interval is calculated using the formula:

CI = μ ± (t * (s / √n))

Substituting the values we have:

CI = 2.4 ± (1.833 * (1.2 / √10))

This gives us the interval:

CI = 2.4 ± 0.697

Therefore, the confidence interval is (1.703, 3.097) mg/L.

The correct interpretation of this interval, in terms of this application, is:

E. The water department can be 90% confident that the mean lead level in drinking water for all residents in the town is within this interval.

Lastly, the phrase "90% confidence interval" means that if we were to take many samples and construct a confidence interval from each, approximately 90% of these intervals would contain the true population mean. It's a measure of how certain we are that the interval includes the true mean.