Question

(a) A study was conducted to determine whether the yield of olive oil is associated with the average temperature of the area. The data in the table below provide the average kilograms of olive oil per tree (y) and the average temperature (2), measured in degrees Celsius. The data correspond to areas taken for 12 different countries. Average temperature (2) 5 7.5 5 7 8 3 2 8 11 4 5 Olive oil yield (y) 10 20 15 17 25 5 2 13 30 3 20 co 10 The summary statistics for these data are: Sum of z data: 73.5 Sum of the squares of x data: 522.25 Sum of y data: 170 Sum of the squares of y data: 3246 Sum of the products of x and y data: 1239 i. Draw a scatter diagram of these data on the graph paper provided. Carefully label the diagram. ii. Calculate the sample correlation coefficient. Interpret its value. iii. Calculate and report the least squares line of y on r. Draw the line on the scatter diagram. iv. Based on the regression model above, what olive oil yield would you expect in an area with average temperature of 5 degrees Celsius? Would you trust this value? Justify your answer.

Answer 1

1)

30 28 26 24 22 20 18 16 olive oil yield (y) 14 12 10 * 4 2. 7 8 9 10 11 average temperature (x)

3)
х Y X*Y X2 y2 5 10 50 25 100 7.5 20 150 56.25 400 5 15 75 25 225 7 17 119 49 289 8 25 200 64 625 3 5 15 9 25 2 2 4 4 4 8 13 104 64 169 11 30 330 121 900 4 3 12 16 9 5 20 100 25 400 8 10 80 64 100 Sum = 73.5 170 1239 522.25 3246
from the above we calculate the mean and sum of squares
n 73.5 X Χ; = 6.125 12 n 170 Y Σ Yi 14.166666666667 n 12 2 n n 1 SSxx = ΣΧ? = 522.25 – 73.52 /12 = 72.0625 n 1=1 1=1 2 η n SSYY = ΣΥ? ΣΥ; = 3246 – 170? /12 = 837.66666666667 n i=1 1=1 SSXY - Σκη - (Σκ) (Σ) = 1239 - 73.5 x 170/12 = 197.75
here slope is denoted as 'm' and y intercept as 'n'

197.75 m = SSXY SSxx = 2.7441 72.0625 n= Y - i . m = 14.166666666667 6.125 x 2.7441 = -2.6412 Therefore, we find that the regression equation is: Y = -2.6412 + 2.7441X

31.40 26.40 21.40 26.40 11.40 6.40 1.40 0.20 -3.60 2.20 4.20 6.20 8.20 10.20 12.20 X Regression equation: Y = -2.6412 + 2.7441 * X
4) so according to the regression line when X=5

Then y =-2.6412 + 2.7441 × 5 = 11.0793

So the yield is 11.0793 , but the observed value is 10,15, and 20 respectively according to the question .

From the scatterness of the plot we can see there is much variation but according to our value it may happen that under some circumstances the yield is 11.0793 , and it is not unusual. .

3.) the correlation coefficient is positive and the strength of the correlation coefficient is moderately good but not too much strong .

1 Sample correlation me ay (Enkey In (8x²) (E2² Jn (EG) (Ey)² (12x1239) (73.5)(110) 12 x 522.25-73.5) (2x 3246 - (170) 2373 2378 0.8048 1864.75 X 10052 2948.299 2 C ol VI