Question

QUESTION 4 a) A farmer wanted to find the relationship between the amount of fertilizer used and the yield of corn. He selected seven acres of his land on which he used different amounts of fertlizer to grow corn. The following table gives the amount (in kilograms) of fertilizer used and the yield (in bushels) of corn for each of the seven acres. Fertilizer used 120 80 100 70 Yield of corn 142 112 132 96 119 104 136 75 110 i) Calculate the correlation coefficient and describe the relationship between the measured variables. (6 marks) i) Plot the graph and derive the regression equation for the above data. (6 marks) ii) Predict the yield of corn for 105 kilograms of fertilizer applied. (2 marks) CzeTheit b) The Health Club poll asked adults whether they felt genetically modified food was safe to at. The result showed 35% felt it was safe, 52% felt it was not safe and 13% had no opinion. A random sample of 120 adults was asked the same question at a local county fair. A total of 40 people felt that genetically modified food was safe, 60 people felt it was not safe and 20 had no opinion. At the 0.01 level of significance, is there sufficient evidence to conclude that the proportions differ from those reported in the survey? (6 marks)

Answer 1

( a )

Let x = Fertilizer used , y = yield of corn

Find X⋅Y , X² and Y² as it was done in the table below.

X	Y	X⋅Y	X⋅X	Y⋅Y
120	142	17040	14400	20164
80	112	8960	6400	12544
100	132	13200	10000	17424
70	96	6720	4900	9216
88	119	10472	7744	14161
75	104	7800	5625	10816
110	136	14960	12100	18496

There is strong positive correlation between the measured variables

II)

( iii )

so regression equation from the above graph is

y = 0.9027x + 37.22

given x = 105

then

y = 0.9027 ( 105 ) + 37.22

y = 132.0035 ~ 132

y =132

( b )

Null hypothesis ( H₀ ) : The proportion does not differ from those reported in the survey

Alternative hypothesis ( H₁ ) : The proportion differ from those reported in the survey

The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
Category 1	40	120*0.35=42	(40-42)2/42 = 0.095
Category 2	60	120*0.52=62.4	(60-62.4)2/62.4 = 0.092
Category 3	20	120*0.13=15.6	(20-15.6)2/15.6 = 1.241
Sum =	120	120	1.429

( 5 ) Conclusion : At alpha = 0.01 L.O.S there is no sufficient evidence to conclude that the proportion differ from those reported in the survey