Example 3.10.1. If \(k=3\), then four possible Latin square design are

The model \[Y_{ijl} = \mu + \alpha _i +\beta _j + \tau _l +e_{ijl}\] \[i=1,2,\ldots ,k\qquad j=1,2,\ldots ,k\qquad l=1,2,\ldots ,k\]

• \(\mu = \) Overall mean
• \(\alpha _i=\) Effect of the \(i^{\text {th}}\) level of factor \(A\).
• \(\beta _j = \) Effect of the \(j^{\text {th}}\) level of factor \(B\).
• \(\tau _l=\) Effect of the \(l^{\text {th}}\) treatments.
• \(Y_{ijl} = \) Observation for the \(i^{\text {th}}\) level of factor \(A\) and \(j^{\text {th}}\) level of factor \(B\) with the \(l^{\text {th}}\) treatment.

\[\sum \alpha _i = 0=\sum \beta _j=\sum \tau _l,\qquad \varepsilon _{ijl}\thicksim ^{iid}N(0,\sigma ^2)\]

Advantages of Latin Design

1.

Allows greater reductions in variability of experimental errors.

2.

Treatment effects can be studied using a small number of experimental units.

Disadvantages

1.

The number of levels in each factor \((A\) and \(B)\) must be equal to the number of treatments.

2.

The assumption of no interaction between either Factor and treatment, as well as no interaction between the two factors.

3.

The randomisation is complex.

\begin {align*} Y_{ijl} - \overline {Y}... & = \overline {Y}_i..-\overline {Y}... + \overline {Y}._j.-\overline {Y}... + \overline {Y}.._l-\overline {Y}... + Y_{ijl}-\overline {Y}_i..-\overline {Y}_i..-\overline {Y}._j.-\overline {Y}..._l+2\overline {Y}... \end {align*}

\(\therefore \) squaring both sides and adding we get \begin {align*} SST & = \underbrace {SSA}_{SSR} + \underbrace {SSB}_{SSC} + SSTrt +SSE\\ K^2-1 & = K-1 + K-1 + K- 1 + (K-2)(K-1) \end {align*}

• \(SST =\displaystyle {\sum ^k_{i=1}\sum ^k_{j=1}\big (Y_{ij(l)}-\overline {Y}...\big )^2=\sum ^k_{i=1}\sum ^k_{j=1}Y^2_{ij(l)}-\frac {Y^2...}{k^2}}\)
  
• \(SSA = \displaystyle {k\sum ^k_{i=1}\big (\overline {Y}_i..-\overline {Y}...\big )^2=\sum ^k_{i=1}\frac {Y^2_i..}{k}-\frac {Y^2...}{k^2}}\)
  
• \(SSB = \displaystyle {k\sum ^k_{j=1}\big (\overline {Y}._j.-\overline {Y}...\big )^2=\sum ^k_{j=1}\frac {Y^2._j.}{k}-\frac {Y^2...}{k^2}}\)
  
• \(SSTrt = \displaystyle {k\sum ^k_{l=1}\big (\overline {Y}.._l-\overline {Y}...\big )^2=\sum ^k_{l=1}\frac {Y^2.._l}{k}-\frac {Y^2...}{k^2}}\)
  

\[SSE = SST - SSA - SSB - SSTrt\]

Example 3.10.2. An experiment was done on the effects of five different types of background music (Treatments \(T_1,\ldots )\) on productivity of bank tellers. A given type of music is played for one day and the productivity measured (observed). A day of the week and week of the experimental period are the two factors. (Blocking variables).

1.

Test at \(\alpha = 0.05\) level of significance if there is effect on productivity by background of music.

2.

Find \(95\%\) simultaneous C.Is for the five treatments.

3.

Find \(95\%\) simultaneous C.Is for all the simple contrast using all the three methods, Bonferoni, S-Method and Tukey’s method.

\[Y_{ij(l)} = \mu +\alpha _i + \beta _j + e_{ij(l)}\] \[i=1,2,\ldots ,5\qquad j=1,2,\ldots ,5\qquad l=1,2,\ldots ,5\] \[\sum \alpha _i=\sum \beta _j=\sum \tau _i =0\]

\begin {align*} Row\qquad Y.._1 & = 14 + 16 + 7 + 13 + 7 = 57\qquad \overline {Y}.._1 = 11.4\\ Y.._2 & = 21 + 34 + 32 + 25 + 21 = 133\qquad \overline {Y}.._2=26.6\\ Y.._3 & = 17 + 13 + 13 + 24 + 31 = 98\qquad \overline {Y}.._3 = 19.6\\ Y.._4 & = 18 + 15 + 29 + 31 + 26 = 119\qquad \overline {Y}.._4=23.8\\ Y.._5 & = 17 + 21 + 27 + 17 + 26 = 108\qquad \overline {Y}.._5=21.6\\ Y... & = 515 \end {align*}

1.

Source of	SS	df	MS	F
Variation
Weeks	82.0	4	20.5	1.306
Days	477.2	4	119.3	7.60
Treatments	664.5	4	166.1	10.58
Type of
Error	188.4	12
Total	1,412	24

\(f_{4,12}^{0.05}=3.26\)

2.

\(95\%\) simultaneous C.Is. Five C.Is

Bonferroni: \(\frac {\alpha }{5}=\frac {0.05}{5}=0.01\)

\[\overline {Y}.._l \pm t_{0.01,12}\sqrt {\text {MSE}\,\times \frac {1}{5}}\qquad l=1,2,3,4,5\]

S-method \[\overline {Y}.._l\pm \sqrt {5f^{0.05}_{5,12}}\sqrt {\text {MSE}\,\times \frac {1}{5}}\qquad l=1,2,3,4,5\]

3.

\(95\%\) simultaneous C.Is for a simple contrasts.

Bonferroni: \(\displaystyle {\binom {5}{2}=10}\) C.Is \[\frac {0.05}{10}=0.005\] \[\overline {Y}.._l-\overline {Y}.._{l'}\pm t_{0.0025,12}\sqrt {\text {MSE}\,\times \frac {2}{5}}\qquad l<l'\]

S-method \[\overline {Y}.._l-\overline {Y}.._{l'} \pm \sqrt {5 f_{5,12}^{0.05}}\sqrt {\text {MSE}\,\times \frac {2}{5}}\qquad l<l'\]

Tukey: \(\displaystyle {\overline {Y}.._l-\overline {Y}.._{l'}\pm q_{(0.95,5,12)}\sqrt {\text {MSE}}},\quad q_{0.95,5,12}=4.51\)

The estimates of the parameters in the ANOVA model \[Y_{ij} =\mu + \tau _i + \beta \big (X_{ij}-\overline {X}..\big ) +\varepsilon _{ij}\] \[\widehat {\mu } =\overline {Y}..\, , \quad \widehat {\tau }_j=\overline {Y}._j-\overline {Y}..\, ,\quad \widehat {\beta }=\frac {SSE_{XY}}{SSE_X}\] \[\widehat {\beta }=\frac {\displaystyle {\sum ^k_{j=1}\sum ^{n_j}_{i=1}\big (Y_{ij}-\overline {Y}._j\big )\big (X_{ij}-\overline {X}._j\big )}}{\displaystyle {\sum \sum \big (X_{ij}-\overline {X}._j\big )^2}}\] \[\widehat {\mu } = \overline {Y}._j-\widehat {\beta }\big (\overline {X}._j-\overline {X}..\big )\] \[adj\big (\overline {Y}._j\big ) = \overline {Y}._j - \widehat {\beta }\big (\overline {X}.j-\overline {X}..\big )\]

Simultaneous confidence intervals for treatment measures we use \(adj(\overline {Y}._j)\).

\(H_0:\beta = 0\) vs \(H_1: \beta \neq 0\), test statistic is given by \[F=\frac {\big (SSE_{XY}\big )^2}{SSE_{X}\big (MSE(adj)\big )}\thicksim f_{1,N-K-1}\quad \Bigg |\quad f_{1,v}=t^2_{\alpha }\] \[t_{\alpha } = \sqrt {f_{1,v}}\] \[T=\frac {SSE_{XY}}{\sqrt {SSE_X\big (MSE(adj)\big )}}\thicksim t_{N-K-1}\]

\(H_0:\beta >0\) vs \(H_a: \beta <0\)

Example 3.10.3.

\[\overline {X}._1=\frac {126}{5}\]

• \(SST_Y = \displaystyle {\sum ^3_{j=1}\sum ^5_{i=1}\big (Y_{ij}-\overline {Y}..\big )^2=346.40}\)
• \(SSTrt_Y =\displaystyle { \sum ^3_{j=1}5\big (\overline {Y}._j-\overline {Y}..\big )^2=140.40}\)
• \(SSE_Y = SST_Y-SSTrt_Y=206.00\)
  

• \(SST_X = \displaystyle {\sum ^3_{j=1}\sum ^5_{i=1}\big (X_{ij}-\overline {X}..\big )^2=261.73}\)
• \(SSTrt_X = \displaystyle {\sum ^3_{j=1}5\big (\overline {X}._j-\overline {X}..\big )^2=66.13}\)
• \(SSE_X = SST_X - SSTrt_X = 195.60\)
  

• \(SST_{XY} =\displaystyle {\sum ^5_{i=1}\sum ^3_{j=1}\big (X_{ij}-\overline {X}..\big )\big (Y_{ij}-\overline {Y}..\big )=282.60}\)
• \(SSTrt_{XY} = \displaystyle {5\sum ^3_{j=1}\big (\overline {X}._j-\overline {X}..\big )\big (\overline {Y}._j-\overline {Y}..\big ) = 96.00}\)
• \(SSE_{XY} = SST_{XY}-SSTrt_{XY}=186.60\)
  

Wrong analysis (One- way Anova)

\begin {align*} SST (adj) & = SST_Y - \frac {\big (SST_{XY}\big )^2}{SST_X}\\ & = 346.40-\frac {(282.40)^2}{261.73}\\ & = 41.27\\ \end {align*}

\begin {align*} SSE(adj) & = SSE_Y - \frac {\big (SSE_{XY}\big )^2}{SSE_X}\\ & = 206.00 - \frac {(186.60)^2}{195.60}\\ & = 27.99\\ \end {align*}

\[SSTrt(adj)= SST(adj)-SSE(adj)=13.208\]

We fail to Reject \(H_0:\beta = 0\).

\begin {align*} F & = \frac {\big (SSE_{XY}\big )^2}{SSE_X\big (MSE(adj)\big )}\\ & = \frac {(186.60)^2}{(2.54)(195.60)}\\ & = 70.08 \end {align*}

\(\displaystyle {f^{0.05}_{1,11}=4.84.\qquad }\) Reject \(H_0\).

\[\widehat {\beta }=\frac {SSE_{XY}}{SSE_X}=\frac {186.60}{195.60}=0.954\]

The adjusted means are
\(\displaystyle {\overline {Y}._1=\frac {207}{5}=41.40}\,,\qquad \displaystyle {\overline {Y}._2=\frac {216}{5}=43.20}\,,\qquad \displaystyle {\overline {Y}._3=\frac {180}{5}=36}\) \[\overline {Y}.. = 40.2\] \[adj(\overline {Y}._j)=\overline {Y}._j-\widehat {\beta }\big (\overline {X}._j-\overline {X}..\big )\]

\(\overline {X}._1=25.20\,, \qquad \overline {X}._2=26.00\,,\qquad \overline {X}._3=21.20\) \[\overline {X}.. = 24.13\] \begin {align*} adj(\overline {Y}._1) & = 41.40 - 0.954(25.20-24.13)=40.38\\ adj(\overline {Y}._2) & = 43.20 - 0.954(26.00-24.13)=41.42\\ adj(\overline {Y}._3) & = 36.00 - 0.954(21.20-24.13)=38.80\\ \end {align*}