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A) Find the correlation coefficient between X and Y. X = [45, 50, 56, 62, 65, 70], Y = [65, 70, 75, 80, 90, 100]

Mathematics

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2 parts
a
0.9664
b
107

Step-by-step solution

5 steps

Here is the solution for Question 16.

Part 16

Part a0.9664

Find the correlation coefficient between X and Y.

  1. List the given data for X (supply) and Y (price).

    X=[45,50,56,62,65,70]X = [45, 50, 56, 62, 65, 70]
    Y=[65,70,75,80,90,100]Y = [65, 70, 75, 80, 90, 100]
    The number of data points is n=6n=6.

  2. Calculate the sums needed for the correlation coefficient formula.

    • ∑X=45+50+56+62+65+70=348\sum X = 45+50+56+62+65+70 = 348
    • ∑Y=65+70+75+80+90+100=480\sum Y = 65+70+75+80+90+100 = 480
    • ∑XY=(45×65)+(50×70)+(56×75)+(62×80)+(65×90)+(70×100)\sum XY = (45 \times 65) + (50 \times 70) + (56 \times 75) + (62 \times 80) + (65 \times 90) + (70 \times 100) ∑XY=2925+3500+4200+4960+5850+7000=28435\sum XY = 2925 + 3500 + 4200 + 4960 + 5850 + 7000 = 28435
    • ∑X2=452+502+562+622+652+702\sum X^2 = 45^2 + 50^2 + 56^2 + 62^2 + 65^2 + 70^2 ∑X2=2025+2500+3136+3844+4225+4900=20630\sum X^2 = 2025 + 2500 + 3136 + 3844 + 4225 + 4900 = 20630
    • ∑Y2=652+702+752+802+902+1002\sum Y^2 = 65^2 + 70^2 + 75^2 + 80^2 + 90^2 + 100^2 ∑Y2=4225+4900+5625+6400+8100+10000=39250\sum Y^2 = 4225 + 4900 + 5625 + 6400 + 8100 + 10000 = 39250
  3. Apply the Pearson correlation coefficient formula.

    The formula is:

    r=n∑XY−(∑X)(∑Y)[n∑X2−(∑X)2][n∑Y2−(∑Y)2]r = \frac{n \sum XY - (\sum X)(\sum Y)}{\sqrt{[n \sum X^2 - (\sum X)^2][n \sum Y^2 - (\sum Y)^2]}}

    Substitute the calculated values:

    r=6(28435)−(348)(480)[6(20630)−(348)2][6(39250)−(480)2]r = \frac{6(28435) - (348)(480)}{\sqrt{[6(20630) - (348)^2][6(39250) - (480)^2]}} r=170610−167040[123780−121104][235500−230400]r = \frac{170610 - 167040}{\sqrt{[123780 - 121104][235500 - 230400]}} r=3570[2676][5100]r = \frac{3570}{\sqrt{[2676][5100]}} r=357013647600r = \frac{3570}{\sqrt{13647600}} r=35703694.2656r = \frac{3570}{3694.2656} r≈0.9664r \approx 0.9664

    The correlation coefficient between X and Y is 0.9664\boxed{0.9664}.

Part b107

Calculate the supply when the price of commodity is Rs. 150.

  1. Determine the regression line equation of X on Y (X=a+bYX = a + bY).

    First, calculate the slope bb:

    b=n∑XY−(∑X)(∑Y)n∑Y2−(∑Y)2b = \frac{n \sum XY - (\sum X)(\sum Y)}{n \sum Y^2 - (\sum Y)^2}

    Using values from part (a):

    b=35705100=0.7b = \frac{3570}{5100} = 0.7

    Next, calculate the means Xˉ\bar{X} and Yˉ\bar{Y}:

    Xˉ=∑Xn=3486=58\bar{X} = \frac{\sum X}{n} = \frac{348}{6} = 58 Yˉ=∑Yn=4806=80\bar{Y} = \frac{\sum Y}{n} = \frac{480}{6} = 80

    Now, calculate the y-intercept aa:

    a=Xˉ−bYˉa = \bar{X} - b\bar{Y} a=58−(0.7)(80)a = 58 - (0.7)(80) a=58−56a = 58 - 56 a=2a = 2

    The regression equation of X on Y is X=2+0.7YX = 2 + 0.7Y.

  2. Calculate the supply (X) when the price (Y) is Rs.

    Substitute Y=150Y = 150 into the regression equation:

    X=2+0.7(150)X = 2 + 0.7(150) X=2+105X = 2 + 105 X=107X = 107

    The supply when the price of the commodity is Rs. 150 is 107\boxed{107}.

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