In: Statistics and Probability
Please explain in manual written format and EXCEL function.
Mercury is a heavy metal that can cause severe health problems in even small concentrations. Fish and shellfish efficiently concentrate mercury into their flesh, so it is important to monitor seafood for its mercury content.
An extensive study conducted in 1980 concluded that the mean mercury level in oysters from the White Bear Estuary was .020 parts per million (ppm) with a population standard deviation , σ, of .022 ppm. In 2012, a sample of 48 oysters from the same estuary exhibited a mean mercury concentration of .004 ppm.
Can you conclude that the 2012 mercury concentration is lower than in 1980 at α = .01 level of significance?
Given:
In 1980 Mean=0.020 standard deviation=0.022
In 2012 Mean=0.004 standard deviation= ? sample size=48
Assuming standard deviation and sample size same in year 1980 & 2012. We have to check the following assumption
H0: Mean of mercury level in oysters from the White Bear Estuary in 1980 = Mean of mercury level in oysters from the White Bear Estuary in 2012
Vs
H1: Mean of mercury level in oysters from the White Bear Estuary in 1980 != Mean of mercury level in oysters from the White Bear Estuary in 2012
Output of Test using Minitab software
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Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 48 0.0200 0.0220 0.0032
2 48 0.0040 0.0220 0.0032
Difference = μ (1) - μ (2)
Estimate for difference: 0.01600
99% lower bound for difference: 0.00537
T-Test of difference = 0 (vs >): T-Value = 3.56 P-Value
= 0.000 DF = 94
Both use Pooled StDev = 0.0220
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Result: Here p-value is equal to zero, so we accept H0.
From the given data and assumptions made in analysis we conclude that the average of mercury level in oysters from the white bear estuary in 1980 is same as that in 2012.
Thanks.