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Magazine Chapter 13: Q. 13.50 (page 543)
In section\(13.2\) we considered two hypothetical examples to explain the logic behind one-way ANOVA. Now you are to further examine those examples.
a. Refer to Table \(13.1\) on page \(528\). Perform a one-way ANOVA on the data and compare your conclusion to that stated in the corresponding "what does it mean"? box. Use \(\alpha =0.05\).
b. Repeat part (a) for the data in Table \(13.2\) on page \(528\).
Short Answer
The solution is
Step by step solution
Step 1. Given information
The data given is
Step 2. Calculation
Calculate the SST, SSTR and SSE using given relation
\(SST=\sum x^{2}-\frac{(\sum x)^{2}}{n}\)
\(SST=7272-\frac{(270)^{2}}{24}=1197\)
\(SSTR=\frac{\sum (x_{i})^{2}}{n_{i}}-\frac{\sum (x)^{2}}{n}\)
\(SSTR=\frac{120^{2}}{6}+\frac{150^{2}}{6}-\frac{(270)^{2}}{12}=75\)
\(SSE=SST-SSTR=1122\)
Then,
\(df_{T}=k-1=4-1=3\)
\(df_{E}=n-k=24-4=20\)
\(MSTR=\frac{SSTR}{df_{T}}=\frac{75}{1}=75\)
\(MSE=\frac{SSE}{df_{E}}=\frac{1122}{10}=112.2\)
\(F=\frac{MSTR}{MSE}=\frac{75}{112.2}\approx 0.67\)
Then make an ANOVA table.
At the \(5%\) significance level data do not provide the sufficient evidence because p-value fail to reject null hypothesis.
\(P>0.05\Rightarrow\) Fail to Reject \(H_{0}\)
Program:
Query:
- First, we have defined the samples.
- Calculate the value of SST and SSTR.
- Then calculate the SSE.
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