4. Trace metals in drinking water affect the flavor, and unusually high concentrations can be haz- ardous. A study collected water samples from six river locations and measured the concentration
of zinc (in mg/L). Water samples were collected from both the surface water and the bottom
water at each location. Does the data suggest that the true average concentration in the surface
water is different than the concentration in the bottom water? Assume that the zinc concentra-
tions are normally distributed. Using a level of significance of a = 0.05 test the claim that the
concentrations are different. (30 points)
Location
A
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
B

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!!!

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

answer is 144

step-by-step explanation: