Chapter 26, Problem 26.39E

(a)

To determine

To find out how many observations per group does your analysis use.

Answer to Problem 26.39E

The number of observations per group is $12$ observations.

Explanation of Solution

In the question, it is given that for the table $26.2$ , do a one-way ANOVA that uses all $36$ observations with the fertilizer type nitrogen level as the only factor. Thus, we have the data as:

	Nitrogen 0	Nitrogen 28	Nitrogen 160
Phosphorus	0.29	0.21	0.18
	0.25	0.24	0.2
	0.27	0.21	0.19
	0.24	0.22	0.19
	0.24	0.19	0.16
	0.2	0.17	0.34
	0.64	0.41	0.31
	0.54	0.37	0.36
	0.53	0.5	0.37
	0.52	0.43	0.26
	0.41	0.39	0.17
	0.43	0.44	0.27

Thus, as we can see that there are three groups namely nitrogen $(0,28,160)$ levels. And in these the number of observations per group is $12$ observations.

(b)

To determine

To explain what do you conclude from the F statistic and P-value.

Answer to Problem 26.39E

We have sufficient evidence to conclude that at least one mean is different from the other mean.

Explanation of Solution

In the question, it is given that for the table $26.2$ , do a one-way ANOVA that uses all $36$ observations with the fertilizer type nitrogen level as the only factor. Thus, we have the data as:

	Nitrogen 0	Nitrogen 28	Nitrogen 160
Phosphorus	0.29	0.21	0.18
	0.25	0.24	0.2
	0.27	0.21	0.19
	0.24	0.22	0.19
	0.24	0.19	0.16
	0.2	0.17	0.34
	0.64	0.41	0.31
	0.54	0.37	0.36
	0.53	0.5	0.37
	0.52	0.43	0.26
	0.41	0.39	0.17
	0.43	0.44	0.27

Now, let us do the one-way analysis for this data by Excel. In Excel go to the data tab and then click on the data analysis option. Then the dialogue box will appear from which select the ANOVA one factor and then another dialogue box will appear and in that select the data and fill the entries and then click ok. Then the result will be as:

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	0.1014	2	0.0507	3.565097	0.039646	3.284918
Within Groups	0.4693	33	0.01422121
Total	0.5707	35

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected and also we can see from the result that the F statistics value is greater thenF critical value, so we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we have sufficient evidence to conclude that at least one mean is different from the other mean.

(c)

To determine

To find out which pairwise differences of means for the three nitrogen levels are significant at the overall $5\%$ level.

Answer to Problem 26.39E

  $\text{Nitrogen 0, 160}$ aresignificant at the overall $5\%$ level and the means for the two differ from each other.

Explanation of Solution

In the question, it is given that for the table $26.2$ , do a one-way ANOVA that uses all $36$ observations with the fertilizer type nitrogen level as the only factor. Thus, we have the data as:

	Nitrogen 0	Nitrogen 28	Nitrogen 160
Phosphorus	0.29	0.21	0.18
	0.25	0.24	0.2
	0.27	0.21	0.19
	0.24	0.22	0.19
	0.24	0.19	0.16
	0.2	0.17	0.34
	0.64	0.41	0.31
	0.54	0.37	0.36
	0.53	0.5	0.37
	0.52	0.43	0.26
	0.41	0.39	0.17
	0.43	0.44	0.27

Thus, to find the confidence interval for mean we will use the calculator $\text{TI}-89$ . In the STAT Intsmenu, choose $4:2-$ SampTInt. You must specify if you are using data stored in two lists or if you will enter the means, standard deviations, and sizes of both samples. You must also indicate whether to pool the variances (when in doubt, say no) and specify the desired level of confidence. So, we have the confidence level $5\%$ significance as:

  $\begin{array}{l}\text{Confidence interval(Nitrogen 0, 28)}=(-0.05,0.18)\\ \text{Confidence interval(Nitrogen 0, 160)}=(0.026,0.234)\\ \text{Confidence interval(Nitrogen 28, 160)}=(-0.021,0.151)\end{array}$

Thus, as we can see that in the group $\text{Nitrogen 0, 160}$ does not contain zero in the confidence interval thus, the means for the two differ from each other.

(d)

To determine

To explain do you think these pairwise comparisons are useful for these data.

Explanation of Solution

In the question, it is given that for the table $26.2$ , do a one-way ANOVA that uses all $36$ observations with the fertilizer type nitrogen level as the only factor. Thus, we have the confidence level $5\%$ significance as:

  $\begin{array}{l}\text{Confidence interval(Nitrogen 0, 28)}=(-0.05,0.18)\\ \text{Confidence interval(Nitrogen 0, 160)}=(0.026,0.234)\\ \text{Confidence interval(Nitrogen 28, 160)}=(-0.021,0.151)\end{array}$

As we can see that these pairwise comparisons are useful for these data because they can independently represent the data for the study and we can specify that which means for the groups are significant and which are not significant to each other.