Question

In Exercise \(13.42-13.47\) we provide data from independent simple random samples from several populations. In each case,

a. compute SST, SSTR and SSE by using the computing formulas given in Formula \(13.1\) on page \(535\).

b. compare your results in part (a) for SSTR and SSE with those you obtained in Exercises \(13.24-13.29\) where you employed the defining formulas.

c. construct a one-way ANOVA table.

d. decide at the \(5%\) significance level, whether the data provide sufficient evidence to conclude that the means of the populations from which the samples were drawn are not all the same.

Short answer

Part a. \(SST=88\)

\(SSTR=36\)

\(SSE=52\)

Part b.

Part c. The null hypothesis is rejected that mean data provided a sufficient evidence to support the claim for the means of the population from which the sample were drawn are not all the same.

Step-by-step solution

Part a. Step 1. Given information

The data is given

Part a. Step 2. Calculation

Calculate the SST, SSTR and SSE using given relation

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

\(SST=808-\frac{(120)^{2}}{16}=88\)

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

\(SSTR=\frac{20^{2}}{4}+\frac{18^{2}}{3}+\frac{30^{2}}{5}+\frac{25^{2}}{5}-\frac{(27)^{2}}{3}=36\)

\(SSE=SST-SSTR=52\)

Program:

Query:

• First, we have defined the samples.
• Calculate the value of SST and SSTR.

• Then calculate the SSE.

Part b. Step 1. Calculation

Calculate the SST, SSTR and SSE using given relation

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

\(df_{E}=n-k=20-3=17\)

\(MST=\frac{SST}{df_{T}}=\frac{88}{2}=44\)

\(MSE=\frac{SST}{df_{E}}=\frac{36}{17}=2.2857\)

\(F=\frac{MST}{MSE}=\frac{52}{2.2857}\approx 5.25\)

After calculating these values put all into the table and get ANOVA table

Part c. Step 1. Calculation

The \(p-\)value is the probability value which obtaining by the test statistics, or a value more extreme. The \(P-\)value is the number in the row title of the \(F-\)distribution table which containing \(F-\)value in the row \(dfd=df_{E}=7\) and in the column \(dfn=df_{T}=2\)

So, the \(p-\)value lie between

\(0.025<P<0.050\)

And if the \(p-\)value is less than significance level then it will reject the null hypothesis.

\(P<0.05\Rightarrow\) Reject \(H_{0}\)