Answer:
The test statistic to test the null hypothesis equals 1.059
Step-by-step explanation:
From the given information; we have:
Treatment Observations
A 20 30 25 33
B 22 26 20 28
C 40 30 28 22
The objective is to find the test statistic to test the null hypothesis; in order to do that;we must first run through a series of some activities.
Let first compute the sum of the square;
Total sum of squares (TSS) = Treatment sum of squares [tex](T_r SS)[/tex] + Error sum of squares (ESS)
where:
(TSS) = [tex]\sum \limits ^v_{i=1} \sum \limits ^{n_i}_{j-1}(yij- \overline {y}oo)^2[/tex] with (n-1) df
[tex](T_r SS)[/tex] [tex]= \sum \limits ^v_{i=1} n_i( \overline yio- \overline {y}oo)^2[/tex] with (v-1) df
[tex](ESS) = \sum \limits ^v_{i=1} \sum \limits ^{n_i}_{j-1}(yij- \overline {y}io)^2[/tex] with (n-v) df
where;
v= 3
[tex]n_i=[/tex]4
i = 1,2,3
n =12
[tex]y_{ij}[/tex] is the [tex]j^{th[/tex] observation for the [tex]i^{th[/tex] treatment
[tex]\overline{y}io[/tex] is the mean of the [tex]i^{th[/tex] treatment i = 1,2,3 ; j = 1,2,3,4
[tex]\overline y oo[/tex] is the overall mean
From the given data
[tex]\overline y oo = \dfrac{1}{12} \sum \limits ^3_{i=1} \sum \limits ^{4}_{j=1}(yij)^2= 27[/tex]
[tex]TSS = \dfrac{1}{12} \sum \limits ^3_{i=1} \sum \limits ^{4}_{j=1}(yij- 27)^2 = 378[/tex]
[tex]T_r SS= \sum \limits^3_{i=1}4 (\overline y io - \overline yoo)^2[/tex]
[tex]=4(27-27)^2+4(24-27)^2+4(30-27)^2 = 72[/tex]
Total sum of squares (TSS) = Treatment sum of squares [tex](T_r SS)[/tex] + Error sum of squares (ESS)
(TSS) = 378 - 72
(TSS) = 306
Now; to the mean square between treatments (MSTR); we use the formula:
MSTR = TrSS/df(TrSS)
MSTR = 72/(3 - 1)
MSTR = 72/2
MSTR = 36
The mean square within treatments (MSE) is:
MSE = ESS/df(ESS)
MSE = 306/(12-3)
MSE = 306/(9)
MSE = 34
The test statistic to test the null hypothesis is :
[tex]T = \dfrac{ \dfrac{TrSS}{\sigma^2}/(v-1) }{ \dfrac{ESS}{\sigma^2}/(n-v) } = \dfrac{MSTR}{MSE} \ \ \ \approx \ \ T(\overline {v-1}, \overline {n-v})[/tex]
[tex]T = \dfrac{36}{34}[/tex]
T = 1.059
