Question

2. An experiment was conducted to test the effects of nitrogen fertilizer on lettuce production. Five rates of ammonium nitrate were applied to four replicate plots in a completely randomized design. The data are the number of heads of lettuce harvested from the plot. Use α =0.05

1. Write the statistical effects model and completely specify each component. State the necessary assumptions for the analysis.
2. Perform the appropriate analysis, identify what hypotheses are being tested, the test statistic, critical value and p-value. State your conclusions and make any specific recommendations for use of the fertilizers.
3. Use Tukey’s adjustment and perform pairwise comparisons for the different level of fertilizers.

4. Treatment (lb N/acre)	Heads of lettuce/plot
   0	104	114	90	140
   50	134	130	144	174
   100	146	142	152	156
   150	147	160	160	163
   200	131	148	154	163

Answer 1

(a) The model is:

Heads of lettuce/plot = 112.00 + 0.0 Treatment (lb N/acre)_0 + 33.5 Treatment (lb N/acre)_50 + 37.0 Treatment (lb N/acre)_100 + 45.5 Treatment (lb N/acre)_150 + 37.0 Treatment (lb N/acre)_200

(b) The hypothesis being tested is:

Null hypothesis	All means are equal
Alternative hypothesis	Not all means are equal
Significance level	α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor	Levels	Values
Treatment (lb N/acre)	5	0, 50, 100, 150, 200

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Treatment (lb N/acre)	4	4995	1248.7	5.61	0.006
Error	15	3338	222.5
Total	19	8333

Since the p-value (0.006) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that there are effects of nitrogen fertilizer on lettuce production.

(c) Using the Tukey Method and 95% Confidence

Treatment (lb N/acre)	N	Mean	Grouping
150	4	157.50	A
200	4	149.00	A
100	4	149.00	A
50	4	145.50	A
0	4	112.0		B

Means that do not share a letter are significantly different.

Tukey Simultaneous Tests for Differences of Means

Difference of Levels	Difference of Means	SE of Difference	95% CI	T-Value	Adjusted P-Value
50 - 0	33.5	10.5	(0.9, 66.1)	3.18	0.042
100 - 0	37.0	10.5	(4.4, 69.6)	3.51	0.023
150 - 0	45.5	10.5	(12.9, 78.1)	4.31	0.005
200 - 0	37.0	10.5	(4.4, 69.6)	3.51	0.023
100 - 50	3.5	10.5	(-29.1, 36.1)	0.33	0.997
150 - 50	12.0	10.5	(-20.6, 44.6)	1.14	0.785
200 - 50	3.5	10.5	(-29.1, 36.1)	0.33	0.997
150 - 100	8.5	10.5	(-24.1, 41.1)	0.81	0.925
200 - 100	0.0	10.5	(-32.6, 32.6)	0.00	1.000
200 - 150	-8.5	10.5	(-41.1, 24.1)	-0.81	0.925

Individual confidence level = 99.25%