There isn't enough evidence to suggest that the true average concentration in the surface water is different than the concentration in the bottom water.
Step-by-step explanation:
We want to test if the true average concentration in the surface water is different than the concentration in the bottom water.
For this, we obtain a new distribution that is the difference between zinc concentration at the bottom and at the top for the 6 samples
A B C D E F
Zinc on surface
0.411 0.238 0.390 0.410 0.605 0.609
Zinc on bottom
0.430 0.266 0.369 0.531 0.627 0.716
Difference (zinc at bottom - zinc at surface)
0.019 0.028 -0.021 0.121 0.022 0.107
We can the find the mean of the differences and the standard deviation of the sample of differences
Mean = xbar = (Σx)/N = (0.276/6) = 0.046
Standard deviation = σ = √{[Σ(x - xbar)²]/(N-1)}
We use (N - 1) because this distribution is that of a sample.
x = each variable
xbar = mean = 0.046
N = number of variables = 6
σ = 0.055612948132607 = 0.055613
For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, we want to test if the true average concentration in the surface water is different than the concentration in the bottom water.
Hence, the null hypothesis is that there isn't significant evidence to suggest that the true average concentration in the surface water is different than the concentration in the bottom water.
The alternative hypothesis is that there is significant evidence to suggest that the true average concentration in the surface water is different than the concentration in the bottom water.
Mathematically, if the mean of zinc concentration at the surface and the bottom are μ₁ and μ₂ respectively and the mean difference is μ = μ₂ - μ₁
The null hypothesis is represented as
H₀: μ = 0 or μ₂ = μ₁
The alternative hypothesis is represented as
Hₐ: μ ≠ 0 or μ₂ ≠ μ₁
To do this test, we will use the t-distribution because no information on the population standard deviation is known
So, we compute the t-test statistic
t = (x - μ₀)/σₓ
x = mean difference = 0.046
μ₀ = 0
σₓ = standard error = (σ/√n)
σ = 0.055613
n = Sample size = 6
σₓ = (0.055613/√6) = 0.0227
t = (0.046 - 0) ÷ 0.0227
t = 2.026 = 2.03
checking the tables for the p-value of this t-statistic
Degree of freedom = df = n - 1 = 6 - 1 = 5
Significance level = 0.05
The hypothesis test uses a two-tailed condition because we're testing in both directions.
p-value (for t = 2.03, at 0.05 significance level, df = 5, with a two tailed condition) = 0.098613
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.098613
0.098613 > 0.05
Hence,
p-value > significance level
This means that we fail to reject the null hypothesis & say that there isn't enough evidence to suggest that the true average concentration in the surface water is different than the concentration in the bottom water.
Hope this Helps!!!
