One way ANOVA:

One-way ANOVA (“ANOVA”) compares the means of two or more independent groups to see if there is statistical evidence of single-factor ANOVA with significantly different relevant population means.

Single Factor ANOVA:

Judge. Analysis of variance (ANOVA) is one of the most commonly used techniques in life and environmental sciences.

a.

The following table shows examples of specific problems and their totals.

We have

$k=5$

$n_1=4,n_2=3,n_3=5,n_4=5,n_5=3$

$T_1=20,T_2=18,T_3=30,T_4=25,\text{and}{T}_5=27$

$n=\sum n_j=4+3+5+5+3=20$

$\sum x_i=\sum T_j=20+18+30+25+27=120$

Summing the squares of all of the facts withinside the above desk yields.

$\sum x_i^2=(7{)}^2+(4{)}^2+(5{)}^2+\dots .+(9{)}^2+(11{)}^2=808$

Consequently

$SST=\sum x_i^2-\frac{{\left(\sum x_i\right)}^2}{n}$

$=808-\frac{(120{)}^2}{20}$

$=808-720$

$=88$.

$SSTR=\sum \frac{T_j^2}{n_j}-\frac{{\left(\sum x_i\right)}^2}{n}$

$=\frac{(20{)}^2}{4}+\frac{(18{)}^2}{3}+\frac{(30{)}^2}{5}+\frac{(25{)}^2}{5}+\frac{(27{)}^2}{3}-\frac{(120{)}^2}{20}$

$=756-720$

localid="1654245418445" $=36$

$SSE=SST-SSTR$

$=88-36$

$=52$

b.

Both results are the same. I'm using different versions of the calculation, but both return the same result.

c.

Therefore, the processing is mean squared

$MSTR=\frac{SSTR}{k-1}$

$=\frac{36}{5-1}$

$=9$

The error is the mean square

$MSE=\frac{SSE}{n-k}$

$=\frac{52}{20-5}$

$=3.47$

Thevalues for $F$statistics are:

$F=\frac{MSTR}{MSE}$

$=\frac{9}{3.47}$

$=2.60$

Therefore, one-way ANOVA table

d.

The nullhypothesis and the alternative hypothesis are:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$

localid="1654244325820" $H_1:Notallthemeansareequal$

Must be tested at the $5\%$significance level. That is, localid="1654243659068" $\alpha =0.05$.

The population under consideration is$5$, that is, $k=5$, and the number of observations is $20$, that is,$n=20$.

Therefore, the degrees of freedom of the$F$statistic are:

$df=(k-1,n-k)$

$=(5-1,20-5)$

$=(4,15)$

From Table VIII, the critical value at the significance level is $5\%$$F\underset{}{{}_{0.05}}=3.06.$

You can find by Referring to tables VIII $df=(4,15)$and$0.05<P<0.10$

Do not reject$H_0$ because the $P$-value is greater than the significance level.

The data do not provide sufficient evidence to conclude that thepopulation means from which the samples wereextracted are not allequal.