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

-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

$k=5$

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

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

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

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

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

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

Step 4- Explanation (part b)

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" $=36$

Step 5- Explanation (part c)

$SSE=SST-SSTR$

$=88-36$

$=52$

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

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

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

$=9$

The error is the mean square

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

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

$=3.47$

Thevalues for $F$statistics are:

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

$=\frac{9}{3.47}$

$=2.60$

Step 8- Explanation (part f)

Therefore, one-way ANOVA table

Step 9- Explanation (part g)

d.

The nullhypothesis and the alternative hypothesis are:

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

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.