We have two factors: {factor1_name} and {factor2_name}.
\n{capitalise(factor1_name)} has two levels: {factor1_levels[0]} and {factor1_levels[1]}.
\n{capitalise(factor2_name)} has two levels: {factor2_levels[0]} and {factor2_levels[1]}.
\nFirst regard the different treatment combinations as a set of independent samples and analyse as for a one-way analysis with unrelated measurements. From this analysis, we obtain the Total Sum of Squares, the Between Treatments Sum of Squares ($\\mathit{BTSS}$) and Residual Sum of Squares ($\\mathit{RSS}$). Note that the degrees of freedom for this step are $\\var{n-4} = \\var{n}-4$ as there are $4$ treatments.
\nYou should obtain
\n| $\\overline{x_i}$ | $T_i$ | $\\sum x^2$ | $n_i$ | |
|---|---|---|---|---|
| {capitalise(factor1_levels[0])}, {factor2_levels[0]} | \n$\\var{mean1}$ | \n$\\var{t[0]}$ | \n$\\var{ssq[0]}$ | \n$\\var{num_samples}$ | \n
| {capitalise(factor1_levels[0])}, {factor2_levels[1]} | \n$\\var{mean2}$ | \n$\\var{t[1]}$ | \n$\\var{ssq[1]}$ | \n$\\var{num_samples}$ | \n
| {capitalise(factor1_levels[1])}, {factor2_levels[0]} | \n$\\var{mean3}$ | \n$\\var{t[2]}$ | \n$\\var{ssq[2]}$ | \n$\\var{num_samples}$ | \n
| {capitalise(factor1_levels[1])}, {factor2_levels[1]} | \n$\\var{mean4}$ | \n$\\var{t[3]}$ | \n$\\var{ssq[3]}$ | \n$\\var{num_samples}$ | \n
| \n | \n | $G = \\var{g}$ | \nSum of Squares = $\\var{ss}$ | \n$N=\\var{n}$ | \n
From this we obtain:
\n\\[ \\mathit{BTSS} = \\sum \\frac{T_i^2}{\\var{num_samples}} - \\frac{G^2}{\\var{n}} = \\frac{\\var{t[0]}^2}{\\var{num_samples}}+\\frac{\\var{t[1]}^2}{\\var{num_samples}}+\\frac{\\var{t[2]}^2}{\\var{num_samples}}+\\frac{\\var{t[3]}^2}{\\var{num_samples}}-\\frac{\\var{g}^2}{\\var{n}}=\\var{btss}\\]
\n\\[\\mathit{TSS} = \\sum \\sum x^2- \\frac{G^2}{\\var{n}}=\\var{ss}- \\frac{\\var{g}^2}{\\var{n}}=\\var{tss} \\]
\nboth to 2 decimal places.
\n\\[\\mathit{RSS} = \\mathit{TSS} - \\mathit{BTSS} = \\var{tss} - \\var{btss} = \\var{rss} \\]
\nNow ignore the {factor2_name} factor and calculate totals $T_i$ for each level of {factor1_name}. From these totals calculate a variance estimate for {factor1_name} using the same method as before. The degrees of freedom will be one fewer than the number of levels of {factor2_name} and is therefore $1$.
\n\n| {capitalise(factor1_levels[0])} ({factor2_levels[0]} and {factor2_levels[1]}) | \n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n$\\var{r1[8]}$ | \n$\\var{r1[9]}$ | \n$\\var{r2[0]}$ | \n$\\var{r2[1]}$ | \n$\\var{r2[2]}$ | \n$\\var{r2[3]}$ | \n$\\var{r2[4]}$ | \n$\\var{r2[5]}$ | \n$\\var{r2[6]}$ | \n$\\var{r2[7]}$ | \n$\\var{r2[8]}$ | \n$\\var{r2[9]}$ | \n
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| {capitalise(factor1_levels[1])} ({factor2_levels[0]} and {factor2_levels[1]}) | \n$\\var{r3[0]}$ | \n$\\var{r3[1]}$ | \n$\\var{r3[2]}$ | \n$\\var{r3[3]}$ | \n$\\var{r3[4]}$ | \n$\\var{r3[5]}$ | \n$\\var{r3[6]}$ | \n$\\var{r3[7]}$ | \n$\\var{r3[8]}$ | \n$\\var{r3[9]}$ | \n$\\var{r3[0]}$ | \n$\\var{r4[1]}$ | \n$\\var{r4[2]}$ | \n$\\var{r4[3]}$ | \n$\\var{r4[4]}$ | \n$\\var{r4[5]}$ | \n$\\var{r4[6]}$ | \n$\\var{r4[7]}$ | \n$\\var{r4[8]}$ | \n$\\var{r4[9]}$ | \n
You should produce the following data from this table:
\n| $T_i$ | $\\sum x^2$ | $n_i$ (number of observations) | |
|---|---|---|---|
| {capitalise(factor1_levels[0])} ({factor2_levels[0]} and {factor2_levels[1]}) | \n$\\var{f1t[0]}$ | \n$\\var{f1ssq[0]}$ | \n$\\var{2*num_samples}$ | \n
| {capitalise(factor1_levels[1])} ({factor2_levels[0]} and {factor2_levels[1]}) | \n$\\var{f1t[1]}$ | \n$\\var{f1ssq[1]}$ | \n$\\var{2*num_samples}$ | \n
| \n | $G = \\var{g}$ | \nSum of Squares = $\\var{f1ss}$ | \n$N = \\var{n}$ | \n
So we can calculate:
\n\\[\\text{Variance estimate for }\\var{factor1_name} = \\sum \\frac{T_i^2}{\\var{2*num_samples}}- \\frac{G^2}{\\var{n}}=\\frac{\\var{f1t[0]}^2}{\\var{2*num_samples}}+\\frac{\\var{f1t[1]}^2}{\\var{2*num_samples}}-\\frac{\\var{g}^2}{\\var{n}}=\\var{f1btss}\\]
\nRepeat step 2 with the factors switched, i.e. use the totals $T_i$ for the {factor2_name} factor levels, ignoring {factor1_name}. This gives a Between Treatments Sum of Squares. Again, the degrees of freedom will be one fewer than the number of levels of {factor1_name}, i.e. $2 - 1 = 1$.
\n| {capitalise(factor2_levels[0])} ({factor1_levels[0]} and {factor1_levels[1]}) | \n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n$\\var{r1[8]}$ | \n$\\var{r1[9]}$ | \n$\\var{r2[0]}$ | \n$\\var{r3[1]}$ | \n$\\var{r3[2]}$ | \n$\\var{r3[3]}$ | \n$\\var{r3[4]}$ | \n$\\var{r3[5]}$ | \n$\\var{r3[6]}$ | \n$\\var{r3[7]}$ | \n$\\var{r3[8]}$ | \n$\\var{r3[9]}$ | \n
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| {capitalise(factor2_levels[1])} ({factor1_levels[0]} and {factor1_levels[1]}) | \n$\\var{r2[0]}$ | \n$\\var{r2[1]}$ | \n$\\var{r2[2]}$ | \n$\\var{r2[3]}$ | \n$\\var{r2[4]}$ | \n$\\var{r2[5]}$ | \n$\\var{r2[6]}$ | \n$\\var{r2[7]}$ | \n$\\var{r2[8]}$ | \n$\\var{r2[9]}$ | \n$\\var{r4[0]}$ | \n$\\var{r4[1]}$ | \n$\\var{r4[2]}$ | \n$\\var{r4[3]}$ | \n$\\var{r4[4]}$ | \n$\\var{r4[5]}$ | \n$\\var{r4[6]}$ | \n$\\var{r4[7]}$ | \n$\\var{r4[8]}$ | \n$\\var{r4[9]}$ | \n
You should obtain the following data from this table:
\n| $T_i$ | $\\sum x^2$ | $n_i$ (number of observations) | |
|---|---|---|---|
| {capitalise(factor2_levels[0])} ({factor1_levels[0]} and {factor1_levels[1]}) | \n$\\var{f2t[0]}$ | \n$\\var{f2ssq[0]}$ | \n$\\var{2*num_samples}$ | \n
| {capitalise(factor2_levels[1])} ({factor1_levels[0]} and {factor1_levels[1]}) | \n$\\var{f2t[1]}$ | \n$\\var{f2ssq[1]}$ | \n$\\var{2*num_samples}$ | \n
| \n | $G = \\var{g}$ | \nSum of Squares = $\\var{f2ss}$ | \n$N = \\var{n}$ | \n
So we can calculate:
\n\\[ \\text{Variance estimate for }\\var{factor2_name} = \\sum \\frac{T_i^2}{\\var{2*num_samples}}- \\frac{G^2}{\\var{n}}=\\frac{\\var{f2t[0]}^2}{\\var{2*num_samples}}+\\frac{\\var{f2t[1]}^2}{\\var{2*num_samples}}-\\frac{\\var{g}^2}{\\var{n}}=\\var{f2btss}\\]
\nNow determine a Sum of Squares for Interaction by subtracting the sums of squares obtained for {factor1_name} (Step 2) and {factor2_name} (step 3) from the overall Between Treatments Sum of squares obtained in Step 1. The degrees of freedom is also obtained by subtraction and is 1.
\nThis gives:
\n\\[\\text{Variance estimate for the interaction}= \\var{btss}-\\var{f1btss}-\\var{f2btss} = \\var{interactionss}\\]
\nWe now have all the terms required to construct the ANOVA table and hence test the null hypothesis relating to each factor and to the interaction. Note that the $\\mathit{VR}$ values are obtained by dividing the $\\mathit{RMS}$ value into the $\\mathit{MS}$ values for the factors and the interaction.
\n| Source | df | SS | MS | VR | Decision |
|---|---|---|---|---|---|
| {capitalise(factor1_name)} | \n$1$ | \n$\\var{f1btss}$ | \n$\\var{f1btss}$ | \n$\\var{f1vr}$ | \n{dec(f1vr)} that {measurement_name} is independent of {factor1_name}. | \n
| {capitalise(factor2_name)} | \n$1$ | \n$\\var{f2btss}$ | \n$\\var{f2btss}$ | \n$\\var{f2vr}$ | \n{dec(f2vr)} that {measurement_name} is independent of {factor2_name}. | \n
| Interaction | \n$1$ | \n$\\var{interactionss}$ | \n$\\var{interactionss}$ | \n$\\var{ivr}$ | \n{dec(ivr)} that {factor1_name} and {factor2_name} are independent in terms of {measurement_name}. | \n
| Residual | \n$\\var{n-4}$ | \n$\\var{rss}$ | \n$\\var{mrs}$ | \n- | \n\n |
| Total | \n$\\var{n-1}$ | \n$\\var{precround(f1btss+f2btss+interactionss+rss,2)}$ | \n- | \n- | \n\n |
Now complete the following two factor ANOVA table from this data. Input the $\\mathit{MS}$ and $\\mathit{VR}$ data to 2 decimal places.
\n| Source | \ndf | \nSS | \nMS | \nVR | \n
|---|---|---|---|---|
| {capitalise(factor1_name)} | \n[[0]] | \n{f1btss} | \n[[1]] | \n[[2]] | \n
| {capitalise(factor2_name)} | \n[[3]] | \n{f2btss} | \n[[4]] | \n[[5]] | \n
| Interaction | \n[[6]] | \n{interactionss} | \n[[7]] | \n[[8]] | \n
| Residual | \n[[9]] | \n{rss} | \n[[10]] | \n- | \n
| Total | \n[[11]] | \n{precround(f1btss+f2btss+interactionss+rss,2)} | \n- | \n- | \n
Also input the mean values of the factors at their various levels:
\n| \n | $\\overline{x_i}$ | \n
|---|---|
| {capitalise(factor1_levels[0])}, {factor2_levels[0]} | \n[[12]] | \n
| {capitalise(factor1_levels[0])}, {factor2_levels[1]} | \n[[13]] | \n
| {capitalise(factor1_levels[1])}, {factor2_levels[0]} | \n[[14]] | \n
| {capitalise(factor1_levels[1])}, {factor2_levels[1]} | \n[[15]] | \n
Using the following $p$-values for the $F_{1,36}$ statistic find the appropriate significance levels for the factors as given by their $\\mathit{VR}$ value and then comment on the null hypotheses for each factor and the interaction.
\n| $10\\%$ | \n$5\\%$ | \n$1\\%$ | \n$0.1\\%$ | \n
| $2.86$ | \n$4.12$ | \n$7.41$ | \n$12.88$ | \n
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Individuals who are identified as having hyperhidrosis suffer from excess sweating. This can be a problem for sport climbers who rely on dry hands for gripping climbing holds. One way to alleviate the problem may be a hand anti-perspirant.
\nSweating can be quantified using a \"gravimetric measurement\" (GM), which measures the rate of sweat production in mg/min, by measuring the change in weight of an absorbant paper placed in the palm of an individual's hand.
\nThe following data represent the results of an experiment to compare the responses of individuals considered to be sufferers or non-sufferers, who have been given a hand anti-perspirant or placebo.
\n| {capitalise(factor1_name)} | \n{capitalise(factor2_name)} | \nGravimetric Measurement (GM), in mg/min | \n|||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| {capitalise(factor1_levels[0])} | \n{capitalise(factor2_levels[0])} | \n$\\var{r1[0]}$ | \n$\\var{r1[1]}$ | \n$\\var{r1[2]}$ | \n$\\var{r1[3]}$ | \n$\\var{r1[4]}$ | \n$\\var{r1[5]}$ | \n$\\var{r1[6]}$ | \n$\\var{r1[7]}$ | \n$\\var{r1[8]}$ | \n$\\var{r1[9]}$ | \n
| {capitalise(factor2_levels[1])} | \n$\\var{r2[0]}$ | \n$\\var{r2[1]}$ | \n$\\var{r2[2]}$ | \n$\\var{r2[3]}$ | \n$\\var{r2[4]}$ | \n$\\var{r2[5]}$ | \n$\\var{r2[6]}$ | \n$\\var{r2[7]}$ | \n$\\var{r2[8]}$ | \n$\\var{r2[9]}$ | \n|
| {capitalise(factor1_levels[1])} | \n{capitalise(factor2_levels[0])} | \n$\\var{r3[0]}$ | \n$\\var{r3[1]}$ | \n$\\var{r3[2]}$ | \n$\\var{r3[3]}$ | \n$\\var{r3[4]}$ | \n$\\var{r3[5]}$ | \n$\\var{r3[6]}$ | \n$\\var{r3[7]}$ | \n$\\var{r3[8]}$ | \n$\\var{r3[9]}$ | \n
| {capitalise(factor2_levels[1])} | \n$\\var{r4[0]}$ | \n$\\var{r4[1]}$ | \n$\\var{r4[2]}$ | \n$\\var{r4[3]}$ | \n$\\var{r4[4]}$ | \n$\\var{r4[5]}$ | \n$\\var{r4[6]}$ | \n$\\var{r4[7]}$ | \n$\\var{r4[8]}$ | \n$\\var{r4[9]}$ | \n
Carry out a two factor ANOVA on the data to test the following null hypotheses:
\n