Question

Listed below are the amounts of mercury​ (in parts per​ million, or​ ppm) found in tuna sushi sampled at different stores. The sample mean is 0.836 ppm and the sample standard deviation is 0.253 ppm. Use technology to construct a 90​% confidence interval estimate of the mean amount of mercury in the population.

Answer 1

Answer:

• The 90%  inverval estimate of the mean amount of mercury in the population is [0.420 ppm, 1.252 ppm].

Explanation:

1) Data:

• μ = 0.836 ppm

• σ = 0.253 ppm

• z = ?

• 90% confidence interval estimate of the mean = ?

2)  Finding Zo for Pr = 90%

The symmetry of the standard normal distribution implies that, for the 90% confidence interval, 5% of the values (area) are above Zo and 5% are below = - Zo.

Using technology (statistic software) or a table of the standard normal probability, you find that for Pr ≥ 0.05 means Zo   ≥ 1.6045.

And, by symmetry, Pr ≤ 0.05 means Zo ≤ -1.6045.

3) Calculate the limits of the interval:

• Formula:  z = ( X - μ) / σ

• Upper limit:  X = zσ + μ = 1.645 (0.253 ppm) + 0.836 ppm = 1.252 ppm.

• Lower limit: X = zσ + μ = -1.645 (0.253ppm) + 0.836 =  0.420 ppm.

Then, the 90%  inverval estimate of the mean amount of mercury in the population is [0.420 ppm, 1.252 ppm] ← answer