Question

A geologist examines 6 seawater samples for lead concentration. The mean lead concentration for the sample data is 0.903 cc/cubic meter with a standard deviation of 0.0566 . Determine the 95% confidence interval for the population mean lead concentration. Assume the population is approximately normal.
Step 2 of 2 :

Construct the 95% confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean lead concentration of sea water samples

Number of samples. n = 6

Mean, u = 0.903 cc/cubic meter

Standard deviation, s = 0.0566

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

0.903 +/- 1.96 × 0.0566/√6

= 0.903 +/- 1.96 × 0.0566/2.44948974278

= 0.903 +/- 0.045

The lower end of the confidence interval is 0.903 - 0.045 =0.858

The upper end of the confidence interval is 0.903 + 0.045 =0.948

Therefore, with 95% confidence interval, the mean lead concentration of the sea water is between 0.858 cc/cubic meter and 0.948 cc/cubic meter