Question

Proportion at 77mm (%)	Average Rainfall	Hand Feeding
67	100	1
75	150	0
80	148	0
72	70	1
91	210	0
69	120	1
55	50	1
77	167	0
84	230	0
92	189	0
58	40	1
69	93	1
74	133	0
72	80	1
66	108	1

A farmer who specialises in the production of carpet wool where the sheep are shorn twice per year is seeking a 75-mm-length clip from his Tukidale sheep. He believes that the proportion of sheep at each clip meeting this standard varies according to average rainfall during the six-month growing period and whether additional hand feeding of high protein sheep nits occurs during the period (because of a shortage of grass cover in the paddocks). Hand feeding is measured as 1 and no hand feeding as 0. Showing ALL formulas and working; a) Predict the proportion at 75mm if the rainfall is 180 mm and there is no hand feeding, a construct a 95% confidence interval estimate and 95% prediction interval.

e) Calculate the coefficients of partial determination and interpret their meaning.

f) Add an interaction term to the model and, at the 0.05 level of significance, determine whether it makes a significant contribution to the model.

Answer 1

Load the data into Excel.

Go to Data>Megastat.

Select the option Correlation/Regression and go to Regression.

Select average rainfall, hand feeding as the independent variable, x.

Select Proportion at 77mm (%) as the dependent variable, y.

Select the predicted value option and type the value 180,0.

Click OK.

The output will be as follows:

	R²	0.795
	Adjusted R²	0.761	n	15
	R	0.892	k	2
	Std. Error	5.134	Dep. Var.	Proportion at 77mm (%)
ANOVA table
Source	SS	df	MS	F	p-value
Regression	1,225.2756	2	612.6378	23.24	.0001
Residual	316.3244	12	26.3604
Total	1,541.6000	14
Regression output					confidence interval
variables	coefficients	std. error	t (df=12)	p-value	95% lower	95% upper
Intercept	55.9584
Average Rainfall	0.1478	0.0448	3.297	.0064	0.0501	0.2454
Hand Feeding	-2.1664	4.9300	-0.439	.6682	-12.9079	8.5751
Predicted values for: Proportion at 77mm (%)
			95% Confidence Interval		95% Prediction Interval
Average Rainfall	Hand Feeding	Predicted	lower	upper	lower	upper	Leverage
180	0	82.554	78.301	86.807	70.586	94.521	0.145

a) Predict the proportion at 75mm if the rainfall is 180 mm and there is no hand feeding, a construct a 95% confidence interval estimate and 95% prediction interval.

The proportion at 75mm from the output is 82.554.

The 95% confidence interval estimate from the output is between 78.301 and 86.807.

The 95% prediction interval estimate from the output is between 70.586 and 94.521​​​​​​​.

e) Calculate the coefficients of partial determination and interpret their meaning.

The regression equation is:

Proportion at 77mm (%) = 55.9584 + 0.1478*Average Rainfall - 2.1664*Hand Feeding

When average rainfall and hand feeding are kept constant, Proportion at 77mm is constant.

When average rainfall is increased by one mm and hand feeding is kept constant, Proportion at 77mm increases by 0.1478.

When average rainfall is kept constant and hand feeding is increased by one, Proportion at 77mm decreases by 2.1664.

f) Add an interaction term to the model and, at the 0.05 level of significance, determine whether it makes a significant contribution to the model.

We get the model as:

Proportion at 77mm (%)	Average Rainfall	Hand Feeding	Average Rainfall & Hand Feeding
67	100	1	100
75	150	0	0
80	148	0	0
72	70	1	70
91	210	0	0
69	120	1	120
55	50	1	50
77	167	0	0
84	230	0	0
92	189	0	0
58	40	1	40
69	93	1	93
74	133	0	0
72	80	1	80
66	108	1	108

Load the data into Excel.

Go to Data>Megastat.

Select the option Correlation/Regression and go to Regression.

Select average rainfall, hand feeding and average rainfall & hand feeding as the independent variable, x.

Select Proportion at 77mm (%) as the dependent variable, y.

The output will be as follows:

	R²	0.795
	Adjusted R²	0.739	n	15
	R	0.892	k	3
	Std. Error	5.358	Dep. Var.	Proportion at 77mm (%)
ANOVA table
Source	SS	df	MS	F	p-value
Regression	1,225.8630	3	408.6210	14.24	.0004
Residual	315.7370	11	28.7034
Total	1,541.6000	14
Regression output					confidence interval
variables	coefficients	std. error	t (df=11)	p-value	95% lower	95% upper
Intercept	54.9628
Average Rainfall	0.1534	0.0613	2.501	.0294	0.0184	0.2885
Hand Feeding	-0.5198	12.6077	-0.041	.9679	-28.2692	27.2296
Average Rainfall & Hand Feeding	-0.0136	0.0948	-0.143	.8888	-0.2222	0.1951

Since the p-value for the interaction term to the model is 0.8888 which is greater than the 0.05 level of significance, we fail to reject the null hypothesis. Therefore, it does not make a significant contribution to the model.