Consider a treatment design consisting of five humidity level percentages (30, 40, 50, 60, and 70). Each of the five treatments is assigned randomly to three climate-controlled grow rooms in a completely randomized experimental design. The resulting fungus crop yields (lbs) are shown in the table below (Fungus Data):

We can see that the factor humidity has levels that are equally spaced. Therefore, we can use the orthogonal contrast coefficients to fit a polynomial to the response, fungus yields.  With \(k=5\) levels, we can only fit up to a quartic term. The orthogonal polynomial contrast coefficients for the example are shown in Table 10.1. Here, \(r=3\) denotes the number of replicates, \(p=1,\dots,4\) denotes the term order, and \(i = 1,\dots,5\) denotes the factor level. 

As mentioned before, one can easily find the orthogonal polynomial coefficients \((g_{pi})\) for a different order of polynomials using pre-documented tables for equally spaced intervals. However, let us try to understand how the coefficients are obtained. 

First note that the five values of \(x\) are 30, 40, 50, 60, and 70. Therefore, \(\bar{x} = 50\) and the spacing \(d = 10\). This means that the five values of \(\dfrac{x - \bar{x}}{d}\) are -2, -1, 0, 1, and 2.

Linear coefficients:

Quadratic coefficients:

Cubic coefficients:

Quartic coefficients:

Each set of coefficients corresponds to a contrast among the treatments. Notice the sum of coefficients is equal to zero (e.g. the quartic coefficients \((1, -4, 6, -4, 1)\) sum to zero, etc). Using orthogonal polynomial contrasts, we can partition the treatment sums of squares into a set of additive sums of squares corresponding to orthogonal polynomial contrasts. The contrast computations are identical to those we learned in Lesson 2.5. The resulting partitions can be used to sequentially test the significance of the linear, quadratic, cubic, and quartic terms in the model and find the polynomial order appropriate for the data. 

Table 10.1 shows how to obtain the sums of squares for each component \( (SSP_p) \) and compute the estimates of the \(\alpha_p\) coefficients for the orthogonal polynomial equation. Using the results in Table 10.1, we have estimated the orthogonal polynomial equation as:

\(\hat{y}_i = 17.36 + 1.20 g_{1i} - 1.03 g_{2i} + 0.11 g_{3i} + 0.09g_{4i}\)

Table 10.2 summarizes how the treatment sums of squares are partitioned and their test results. 

To test whether any of the polynomials are significant (i.e. \(H_0: \alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0\)), we can use the global F-test where the test statistic is equal to 35.48. We see that the p-value is almost zero and therefore we can conclude that at the 5% level, at least one of the polynomials is significant. Using the orthogonal polynomial contrasts we can determine which of the polynomials are useful. Table 10.2 shows that for this example only the linear and quadratic terms are useful. Removing the insignificant terms, the estimated orthogonal polynomial equation becomes:

\(\hat{y}_i = 17.36 + 1.20 g_{1i} - 1.03 g_{2i}\)

The polynomial relationship expressed as a function of y and x in actual units of the observed variables is more informative than when expressed in units of the orthogonal polynomial.

We can obtain the polynomial relationship using the actual units of observed variables by back-transforming using the formulas seen in Section 10.2. The necessary quantities to back-transform are \(\lambda_1=\lambda_2=1, d=10, \bar{x}=50\) and t = 5. Substituting these values we obtain

\(\begin{align}  
\hat{y}&=17.36+1.20g_{1i}-1.03g_{2i}\\  
&=17.36+1.20(1)\left( \dfrac{x-50}{10} \right)-1.03(1)\left( \left( \dfrac{x-50}{10} \right)^2-\dfrac{5^2-1}{12} \right)\\  
\end{align}\)

which simplifies to

\(\hat{y}=-12.23+1.15x-0.01x^2\)