Question

An environmental group is conducting a study to determine whether dolphins in a certain region are ingesting unhealthy levels of arsenic. A marine biologist classifies arsenic levels greater than 7.0 parts per million as unhealthy. The arsenic levels in a random sample of 28 dolphins in the region are measured and recorded, and the data are shown in the stemplot.

Key: 2|4 = 2.4 parts per million

Part A: What proportion of dolphins in the sample have arsenic levels classified by the marine biologist as unhealthy? (4 points)

Part B: The mean aresenic level of the 28 dolphins in the sample is 5.26 parts per million and the standard deviation is 1.49 parts per million. Construct and interpret a 99% confidence interval for the mean arsenic level of dolphins in the region. (6 points) (10 points)

Answer 1

Part A: What proportion of dolphins in the sample have arsenic levels classified by the marine biologist as unhealthy? (4 points)

The dolphins in the sample that have arsenic levels greater than 7.0 parts per million are classified by the marine biologist as unhealthy

greater than 7.0 parts per million are = 7.1 , 7.3, 7.5, 7.6, 7.7 (Last two stems)

= 5 dolphins.

Proportion = 5 / 28

Proportion = 0.179

Part B: The mean aresenic level of the 28 dolphins in the sample is 5.26 parts per million and the standard deviation is 1.49 parts per million. Construct and interpret a 99% confidence interval for the mean arsenic level of dolphins in the region. (6 points) (10 points)

.Since we do not have known population SD of arsenic levels, we will use the t-dist to caclulate the confidence intervals

(1- $\alpha$ )% confidence interval for mean is

SD (8+,-1,a/2 (m)

Critical value at $\alpha$ = 1-99% = 0.01

In-1,0/2 = t27,0.005

= 2.7707   Using t-dist table with df = 28 - 1 and p = 0.01 /2

Sub all the values

1.49、 (5.2642.7707 *

(4.48.6.0402)

We are 99% that the true population mean arsenic level lies within this range.