Solution.

Solution.

1.

Set up the hypothesis
\(H_0:\) The type of housing and compound are independent.
\(H_1:\) The type of housing and compound are dependent.

2.

Level of significance \(\alpha =5\%\).

3.

Test statistic. Each expected frequency is \[e_{ij}=\frac {(\text {row total})\times (\text {column total})}{\text {grand total}}\] so, for instance, \(e_{11}=\frac {754\times 973}{4266}=171.97\).

	Detail	Data	Flats
Chelstone	171.97	301.88	280.14
Northmead	217.59	381.96	354.45
Kabwata	583.44	1024.16	950.41

Then \[\chi ^2=\sum \frac {(O-e)^2}{e} =\frac {(95-171.97)^2}{171.97}+\frac {(297-301.88)^2}{301.88}+\cdots +\frac {(861-950.41)^2}{950.41}=103.96\]

4.

Critical region. With \(r=3\) rows and \(c=3\) columns, \[v=(r-1)(c-1)=2\times 2=4\] and from tables \(\chi ^2_{4,\,0.05}=9.488\). We reject \(H_0\) if \(\chi ^2>9.488\).

5.

Conclusion. \(103.96>9.488\), so we reject \(H_0\): the type of housing does vary between the compounds.

Where the difference lies is worth reading off the table rather than leaving at “significant”. Chelstone has \(95\) detached houses against \(172\) expected and \(362\) flats against \(280\); Kabwata has \(703\) detached against \(583\). Those three cells alone contribute \(34.5+23.9+24.5=82.9\) of the total \(103.96\). The compounds differ mainly in the balance between detached houses and flats, not across the board.

Solution.

1.

Set up the hypothesis
\(H_0: \mu _1=\mu _2=\mu _3\)
\(H_1:\) at least two of the means differ

2.

Level of significance \(\alpha =5\%\)

3.

Test statistic. Here \(k=3\) groups, \(n_j=4\) each and \(n=12\), with grand total \(T=8+14+12=34\), so the correction factor is \[\frac {T^2}{n}=\frac {34^2}{12}=\frac {1156}{12}=96.333\] \begin {align*} SS_T &=\sum x_{ij}^2-\frac {T^2}{n}=\Big [3^2+2^2+1^2+\cdots +1^2\Big ]-96.333=118-96.333=21.667\\\\ SS_B &=\sum _{j=1}^{k}\frac {T_j^2}{n_j}-\frac {T^2}{n}=\frac {8^2+14^2+12^2}{4}-96.333=\frac {404}{4}-96.333=4.667\\\\ SS_W &=SS_T-SS_B=21.667-4.667=17.000 \end {align*}

Source	SS	df	MS	\(F\)
Between samples	4.667	\(k-1=2\)	\(4.667/2=2.333\)	\(2.333/1.889=1.235\)
Within samples	17.000	\(n-k=9\)	\(17.000/9=1.889\)
Total	21.667	\(n-1=11\)

Note that the degrees of freedom add up, \(2+9=11\), and so do the sums of squares. If they do not, there is an arithmetic slip somewhere above.

4.

Critical region. From tables \(F_{2,\,9,\,0.05}=4.26\), so we reject \(H_0\) if \(F>4.26\).

5.

Conclusion. \(F=1.235<4.26\), so we fail to reject \(H_0\). There is no evidence at the \(5\%\) level that the three fertilisers differ in mean yield.

Look at where the variation went. Of the total \(21.667\), only \(4.667\) lies between the fertilisers and \(17.000\) lies within them — plot-to-plot differences under the same fertiliser are far larger than the differences between fertilisers. That is what \(F\) compares, and it is why the test comes out as it does.

Proof. The trick is to add and subtract the group mean \(\overline {x}_i\), so that each deviation from the grand mean splits into two pieces: \[x_{ij}-\overline {\overline {x}}=\big (\overline {x}_i-\overline {\overline {x}}\big )+\big (x_{ij}-\overline {x}_i\big )\] Squaring and summing, \begin {align*} \sum _{i=1}^k\sum _{j=1}^{n_i}(x_{ij}-\overline {\overline {x}})^2 &=\sum _{i=1}^k\sum _{j=1}^{n_i}\Big [(\overline {x}_i-\overline {\overline {x}})+(x_{ij}-\overline {x}_i)\Big ]^2\\\\ &=\sum _{i=1}^k\sum _{j=1}^{n_i}\Big [(\overline {x}_i-\overline {\overline {x}})^2+(x_{ij}-\overline {x}_i)^2 +2(\overline {x}_i-\overline {\overline {x}})(x_{ij}-\overline {x}_i)\Big ]\\\\ &=\sum _{i=1}^k n_i(\overline {x}_i-\overline {\overline {x}})^2+\sum _{i=1}^k\sum _{j=1}^{n_i}(x_{ij}-\overline {x}_i)^2 +2\sum _{i=1}^k(\overline {x}_i-\overline {\overline {x}})\sum _{j=1}^{n_i}(x_{ij}-\overline {x}_i) \end {align*}

The identity only works because that last term is zero. In the inner sum \((\overline {x}_i-\overline {\overline {x}})\) is a constant, so it comes outside, and what is left is \[\sum _{j=1}^{n_i}(x_{ij}-\overline {x}_i)=0\] since deviations about their own mean always cancel. The whole cross-product term therefore vanishes, leaving \[SS_T = SS_B + SS_W\hspace {1cm}\blacksquare \]

Test whether there is any significance difference between the treatment and \(\alpha = 0.025\)

1.

Set up the hypothesis
\(H_0:\) \(\mu _1=\mu _2=\mu _3=\mu _4=\mu _5\)
\(H_1:\) At least two are different

2.

We set \(\alpha = 0.05\)
\(N=rc=5\times 5=25\) \[\text {Correction factor}\hspace {0.5cm}=\frac {g^2}{N}=\frac {133^2}{25}\hspace {1cm}\implies cf = 707.56\]

3.

Test statistic \begin {align*} SS_T & =\sum \sum x^2_{ij}-cf\\ &=\Big [5^2+9^2+3^2+2^2+\cdots +7^2\Big ]-707.56\\ &=863-707.56\\ \implies \hspace {1cm} SS_T &=155.5 \end {align*}

\begin {align*} SS_C & = \frac {1}{r}\sum t_i^2-cf\\ &=\frac {1}{5}\Big [26^2+39^2+20^2+14^2+34^2\Big ]-707.56\\ &=\frac {1}{5}[3949]-707.56\\\\ \implies \hspace {0.5cm }SS_C &=82.3 \end {align*}

\begin {align*} SS_T &=SS_C +SS_E\\\\ \implies \hspace {0.5cm} SS_E &= SS_T-SS_C\\ &=155.5-82.3 \\ &=73.2 \end {align*}

Solution. From tables \(t_{0.025,\,6}=2.447\), so \begin {align*} LSD &= 2.447\times \sqrt {1.57\left (\frac {1}{4}+\frac {1}{4}\right )}\\\\ &= 2.447\times \sqrt {0.785}\\\\ &= 2.447\times 0.886 = 2.168 \end {align*}

Comparing each difference against \(2.168\):

\(v_2-v_1=2.25>2.168\), so \(v_1\) and \(v_2\) differ significantly.

\(v_3-v_1=3.00>2.168\), so \(v_1\) and \(v_3\) differ significantly.

\(v_4-v_1=4.00>2.168\), so \(v_1\) and \(v_4\) differ significantly.

All three exceed the \(LSD\), so \(v_1\) differs from each of the others. Note that \(v_2-v_1\) only just clears the bar, at \(2.25\) against \(2.168\); a conclusion resting on that small a margin should be reported as such rather than stated flatly.