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katen-ka-za [31]
3 years ago
14

Local environmentalists want to test if there is a difference in the pH of three different streams (called A, B, and C) feeding

into Lake Travis. They took a random sample of 15 measurements from each stream and, after confirming all relevant assumptions, ran a one-way ANOVA on the data and got a significant (p<0.05) result.
a. Write the appropriate null and alternative hypotheses for this analysis.
b. List one potential confounding variable.
c. In a post-hoc analysis, they got the following p values. Which pairs of streams have significantly different mean pH? Show work for full credit.

Stream A vs. B: p = 0.003
Stream A vs. C: p= 0.029
Stream A vs. C: p= 0.041
Mathematics
1 answer:
VashaNatasha [74]3 years ago
8 0

Answer:

Step-by-step explanation:

Hello!

The variable Y: pH of the water of a stream that feeds into Lake Travis (stream A, stream B, and stream C)

This is a one way ANOVA with one factor (ph of the water) and three levels (stream A, stream B, and stream C).

a)

The parameters of interest are the population mean of pH of each stream, then the hypotheses for this ANOVA are

H₀: μ_{A}=μ_{B}=μ_{C}=μ

H₁: At least one population mean is not equal to the others.

b)

There are many factors that may change the pH of water, for example:

1) Polution

2) Soil components (different concentrations of minearls) and pH of the soil

c)

The null hypothesis was rejected, so they made a post-hoc analysis to see wich of the population means is different from the others:

Using the same α: 0.05 as before

The deicsion rule for the p-value approach is

If p-value ≤ α, reject the null hypothesis.

If p-value ≥ α, do not reject the null hypothesis

1.

Stream A vs. B: p = 0.003

H₀: μ_{A}=μ_{B}

H₁: μ_{A}≠μ_{B}

p-value: 0.003 < α: 0.05 ⇒ Reject null hypothesis. The average pH of stream A is different than the average pH of stream B.

2.

Stream A vs. C: p= 0.029

H₀: μ_{A}=μ_{C}

H₁:  μ_{A}≠μ_{C}

p-value: 0.029 < α: 0.05 ⇒ Reject null hypothesis. The average pH of stream A is different than the average pH of stream C.

3.

Stream A vs. C: p= 0.041

H₀: μ_{A}=μ_{C}

H₁:  μ_{A}≠μ_{C}

p-value: 0.041 < α: 0.05 ⇒ Reject null hypothesis. The average pH of stream A is different than the average pH of stream C.

I hope this helps!

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Step-by-step explanation:

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Since the sample result is not significant, it suggests that the rate of 24.5% is a good estimate for the percentage of people that have sleepwalked.

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