Bonferroni Test Shown below are weights (kg) of poplar trees obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the table below. The data are from a study conducted by researchers at Pennsylvania State University and were provided by Minitab, Inc. Also shown are partial results from using the Bonferroni test with the sample data.

No Treatment	Fertilizer	Irrigation	Fertilizer and Irrigation
1.21	0.94	0.07	0.85
0.57	0.87	0.66	1.78
0.56	0.46	0.10	1.47
0.13	0.58	0.82	2.25
1.30	1.03	0.94	1.64

1. Use a 0.05 significance level to test the claim that the different treatments result in the same mean weight.

2. What do the displayed Bonferroni SPSS results tell us?

3. Use the Bonferroni test procedure with a 0.05 significance level to test for a significant difference between the mean amount of the irrigation treatment group and the group treated with both fertilizer and irrigation. Identify the test statistic and either the P-value or critical values. What do the results indicate?
4. 
   

The data set all have the same sample size. The given sample size \(n = 5\)\(\left( {{n_1} = {n_2} = {n_3} = {n_4} = 5} \right)\).

No.of samples (k)\( = 4\)

Significance level \(\alpha \)\( = 0.05\)

To test the claim that females in different treatments result in the same mean weight. of trees, the null hypothesis is formulated as,

\({H_0}:{\mu _1} = {\mu _2} = {\mu _3}\)

The corresponding alternative hypothesis is

\[{H_1}:\]At least one of the means is different from the others

Let,

\({\bar X_1},{\bar X_2}{\rm{,}}{\bar X_{3\;}}{\rm{and }}{\bar X_4}\)denotes the sample means.

\({n_1},{n_2}{\rm{,}}{n_3}{\rm{ and }}{{\rm{n}}_4}\)denotes the sample sizes.

\(s_1^2,s_2^2{\rm{,}}s_3^2{\rm{ and }}s_4^2\)denotes the sample variances.

Then, the sample means are 0.754, 0.776, 0.518, and 1.598.

The sample variances are 0.2417, 0.0596, 0.1662, 0.2589.

Let,

\(\bar \bar X\)denotes the mean of sample means.

\(s_{\bar X}^2\)denotes the variance of sample means.

\(\begin{aligned}{c}\bar \bar X = \frac{1}{n}\sum\limits_{i = 1}^n {{{\bar X}_i}} \\ = \frac{{0.754 + 0.776 + 0.518 + 1.598}}{4}\\ = 0.9115\end{aligned}\)

\(\begin{aligned}{c}s_{\bar X}^2 = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{{\bar X}_i} - \bar \bar X} \right)}^2}} \\ = \frac{1}{3}\left[ {{{\left( {0.754 - 0.9115} \right)}^2} + {{\left( {0.776 - 0.9115} \right)}^2} + {{\left( {0.518 - 0.9115} \right)}^2} + {{\left( {1.598 - 0.9115} \right)}^2}} \right]\\ = 0.2231\end{aligned}\)

Thus, the mean of the sample mean is 0.9115, and the variance of the sample mean is 0.2231.

Let \(ns_{\bar X}^2\)denotes the variance between sample means. Then,

\(\begin{aligned}{c}ns_{\bar X}^2 = 5\left( {0.2231} \right)\\ = 1.1155\end{aligned}\)

Thus, the variance between samples is 1.1155.

Let,

\(s_p^2\)denotes the variance within samples, then

\(\begin{aligned}{c}s_p^2 = \frac{1}{n}\sum\limits_{i = 1}^n {s_i^2} \\ = \frac{1}{4}\left[ {0.2417 + 0.0596 + 0.1662 + 0.2589} \right]\\ = 0.1816\end{aligned}\)

Thus, the variance within samples is 0.1816.

The F test statistic is

\[\begin{aligned}{c}F = \frac{{{\rm{Variance between samples}}}}{{{\rm{variance within samples}}}}\\ = \frac{{ns_{\bar X}^2}}{{s_p^2}}\\ = \frac{{1.1155}}{{0.1816}}\\ = 6.1426\end{aligned}\]

Thus, the value of F is 6.1426.

Let, \(k\)is the number of samples, and n is the sample size. Then,

Numerator degrees of freedom = \(k - 1\)

Denominator degrees of freedom= \(k\left( {n - 1} \right)\)

Here,

\(k - 1 = 3\)and \(k\left( {n - 1} \right) = 16\).

With \(\alpha = 0.05\), the critical value of F is

\({F_{\left( {3,16} \right)}} = {\rm{3}}{\rm{.2388}}\)\(\left( {{\rm{from F - distribution table}}} \right)\)

Here, \(F > {F_{\left( {3,16} \right)}}\), the null hypothesis is to reject. So, the there is insufficient evidence to support the claim that different treatments result in the same mean weight.

b.

The SPSS table displays that for the pair no treatment and fertilizer and irrigation, the p-value is 0.039. This p-value is small \(\left( {p < 0.05} \right)\)so, there is significant different between the mean amount of group treated with no treatment and the group treated with both fertilizer and irrigation.

c.

The Bonferroni test is a separate t-test to test the significant difference between the mean amount of irrigation treatment group (represented as 3) and the group treated with both fertilizer and irrigation (represented as 4).

The null and alternative hypothesis is,

\({H_0}:{\mu _3} = {\mu _4}\)

\({H_1}:{\mu _3} \ne {\mu _4}\)

The test statistic is

\(\begin{aligned}{c}t = \frac{{{{\bar X}_3} - {{\bar X}_4}}}{{\sqrt {{\rm{MS}}\left( {{\rm{error}}} \right) \times \left( {\frac{1}{{{n_3}}} + \frac{1}{{{n_4}}}} \right)} }}\\ = \frac{{0.518 - 1.598}}{{\sqrt {0.1816 \times \left( {\frac{1}{5} + \frac{1}{5}} \right)} }}\\ = - 4.0072\end{aligned}\)

The number of degrees of freedom is,

\(\begin{aligned}{c}N - K = 20 - 4\\ = 16\end{aligned}\)

Using table-A3, the critical value is obtained at the degrees of freedom (16), and the significance level (0.05) is 2.120.

Here, \(\left| t \right| > {t_\alpha }.\)So, the null hypothesis is to reject. Thus, there is a significant difference between the mean amount of irrigation treatment group and the group treated with both fertilizer and irrigation. Thus, the difference is significant and negative, and this implies that the mean weight of popular trees is greater if a combination of both fertilizers and irrigation is used as compared to using only irrigation.