Question

An ecological group suspects tuna are consuming dangerous levels of copper in a region of the world. Dangerous levels of copper are classified when they reach beyond 5.7 parts per

million. A random sample of 21 tunas in the region were tested and the data is shown in the stemplot.
Copper Levels
20
2.
31
3578
4 01
45678
5 | 0233
5689
634
6
Key: 20 = 2.0 parts per million
Part A: What proportion of the tuna in the sample have dangerous levels of copper in their systems? (4 points)
Part B: Construct and interpret a 98% confidence interval for the mean copper levels of the tuna if the sample of 21 tuna has a mean copper level of 4.77 parts per million and standard
deviation of 11 16 parts per million. (6 points)

Answer 1

Explanation:

A) From the stem-leaf plot, we see that out of 21 tunas, 5 have dangerous levels of copper since the levels go beyond 5.7 parts per million. The required proportion is 5/21=0.2381

B) Given the sample mean is {x}=4.77, sample standard deviation s=1.16 and the sample size is n=21.

Since the population standard deviation is not known, we use t-distribution.

So the 98% CI for mean is

4.77 ± t{1-0.02 /2,20} x 1.16/sqrt(21) = (4.13, 5.41)}

We are sure with 98% confidence the true copper level (in parts per million) lies in the interval (4.13, 5.41)