Question

Short answer

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

Step-by-step solution

Step 1- Introduction

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.

Step 2- Information

a.

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

Step -3 Explanation (part a)

We have

$k=4$

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

$T_1=12,T_2=35,T_3=15,\text{and}{T}_4=18$

$n=\sum n_j=3+5+5+3=16$

$\sum x_i=\sum T_j=12+35+15+18=80$

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

$\sum x_i^2=(6{)}^2+(3{)}^2+(3{)}^2+\dots .+(4{)}^2+(6{)}^2=438$

Step 4- Explanation (Part b)

Consequently

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

$=474-\frac{(80{)}^2}{16}$

$=474-400$

$=74$

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

$=\frac{(12{)}^2}{3}+\frac{(35{)}^2}{5}+\frac{(15{)}^2}{5}+\frac{(18{)}^2}{3}-\frac{(80{)}^2}{16}$

$=446-400$

$=46$

Step 5-Explanation (part C)

$SSE=SST-SSTR$

$=74-46$

$=28$

Step 6-Explanation (part d)

b.

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

Step 7- Explanation (part e)

c.

Therefore, the processing is mean squared

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

$=\frac{46}{4-1}$

$=15.33$

The error is the mean square

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

$=\frac{28}{16-4}$

$=2.33$

The value of $F$- Statistic is

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

$=\frac{15.33}{2.33}$

$=6.58$

Step 7- Explanation (part f)

Therefore, one-way ANOVA table

Step 9- Explanation (Part g)

d.

The nullhypothesis and the alternative hypothesis are:

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

$H_1:Notallthemeansareequal$

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

The population under consideration is $4$, that is, $k=4$, and the number of observations is $16$, that is , $n=16$.

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

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

$=(4-1,16-4)$

$=(3,12)$

From Table VIII, the critical value at the $5\%$significance level is $F_{0.05}=3.49$

Step 10- Conclusion

You can find by Referring to tables VIII $df=(3,12)$and $0.005<P<0.01$

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 the population means from which the samples were extracted are not all equal.