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.

•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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%