Question

The US Geological Survey, in cooperation with the Florida Department of Environment protection, investment the effects of waste disposal practices on ground water quality at five poultry farms in north-central Florida. At one site they drilled four monitoring wells, numbered \(1, 2, 3\) and \(4\). Over a period of \(9\) months, water samples were collected from the last three wells and analyzed for a variety of chemicals, including potassium, chlorides, nitrates and phosphorus. The concentrations in milligrams per litter are provided on the WeissStats site. For each of the four chemicals, decide whether the data provide sufficient evidence to conclude that a difference exists, in mean concentrations among the three wells. Use \(\alpha =0.01\).

Short answer

The solution is

Step-by-step solution

Step 1. Given information

The data is given

Step 2. Calculation

Calculate the SST, SSTR and SSE using given relation

\(SST=\sum x^{2}-\frac{(\sum x)^{2}}{n}\)

\(SST=1561154-\frac{(4228)^{2}}{12}=7.15\times 10^{4}\)

\(SSTR=\frac{\sum (x_{i})^{2}}{n_{i}}-\frac{\sum (x)^{2}}{n}\)

\(SSTR=\frac{1828^{2}}{4}+\frac{1225^{2}}{4}+\frac{1175^{2}}{4}-\frac{(4228)^{2}}{12}=6.60\times 10^{4}\)

\(SSE=SST-SSTR=5.45\times 10^{3}\)

Then,

\(df_{T}=k-1=3-1=2\)

\(df_{E}=n-k=12-2=10\)

\(MSTR=\frac{SSTR}{df_{T}}=\frac{6.60\times10^{4}}{2}=3.30\times 10^{4}\)

\(MSE=\frac{SSE}{df_{E}}=\frac{5.45\times 10^{3}}{10}=0.545\times 10^{3}\)

\(F=\frac{MSTR}{MSE}=\frac{3.30\times 10^{4}}{0.545\times 10^{3}}\approx 60.55\times 10^{3}\)

Then make an ANOVA table.

At the \(1%\) significance level data provide the sufficient evidence because p-value reject null hypothesis.

\(P<0.01\Rightarrow \) Reject \(H_{0}\)

Program:

Query:

• First, we have defined the samples.
• Calculate the value of SST and SSTR.
• Then calculate the SSE.
• Then calculate MSTR and MSE and F.

• Make an ANOVA table.