In a random sample of 10 on-line statistics students, information was collected about their minutes studied and their chapter test scores. X is the minutes studied, and Y is the test score.
x : {20,30,45,60,80,90,45,120,90,70} y: {50,45,55,70,80,90,80,100,95,85}
Is this relationship positive, negative or no relation?
Can you plot the points on a scatter diagram? That should give you the answer.
In a random sample of 10 on-line statistics students, information was collected about their minutes studied and their chapter test scores. X is the minutes studied, and Y is the test score.
x : {20,30,45,60,80,90,45,120,90,70} y: {50,45,55,70,80,90,80,100,95,85}
Is this relationship positive, negative or no relation?
Positive
To determine the relationship between the minutes studied (X) and the test scores (Y), you can calculate the correlation coefficient. The correlation coefficient is a statistical measure that quantifies the strength and direction of the relationship between two variables.
To calculate the correlation coefficient, you need to follow these steps:
Step 1: Calculate the average (mean) of the minutes studied and the test scores. Let's denote them as X̄ (mean of X) and Ȳ (mean of Y).
X̄ = (20 + 30 + 45 + 60 + 80 + 90 + 45 + 120 + 90 + 70) / 10 = 63
Ȳ = (50 + 45 + 55 + 70 + 80 + 90 + 80 + 100 + 95 + 85) / 10 = 80
Step 2: Calculate the deviation of each data point from their respective means. Denote these deviations as dx and dy.
dx = (20 - 63, 30 - 63, 45 - 63, 60 - 63, 80 - 63, 90 - 63, 45 - 63, 120 - 63, 90 - 63, 70 - 63)
= (-43, -33, -18, -3, 17, 27, -18, 57, 27, 7)
dy = (50 - 80, 45 - 80, 55 - 80, 70 - 80, 80 - 80, 90 - 80, 80 - 80, 100 - 80, 95 - 80, 85 - 80)
= (-30, -35, -25, -10, 0, 10, 0, 20, 15, 5)
Step 3: Multiply the deviations together for each data point. Denote this product as dxdy.
dxdy = (-43 * -30, -33 * -35, -18 * -25, -3 * -10, 17 * 0, 27 * 10, -18 * 0, 57 * 20, 27 * 15, 7 * 5)
= (1290, 1155, 450, 30, 0, 270, 0, 1140, 405, 35)
Step 4: Calculate the sum of the products of deviations, denoted as Σdxdy.
Σdxdy = 1290 + 1155 + 450 + 30 + 0 + 270 + 0 + 1140 + 405 + 35
= 4765
Step 5: Calculate the sum of the squared deviations for X and Y, denoted as Σ(dx^2) and Σ(dy^2).
Σ(dx^2) = (-43)^2 + (-33)^2 + (-18)^2 + (-3)^2 + 17^2 + 27^2 + (-18)^2 + 57^2 + 27^2 + 7^2
= 3718
Σ(dy^2) = (-30)^2 + (-35)^2 + (-25)^2 + (-10)^2 + 0^2 + 10^2 + 0^2 + 20^2 + 15^2 + 5^2
= 1655
Step 6: Calculate the square root of the product of Σ(dx^2) and Σ(dy^2). Denote this as sqrt(Σ(dx^2) * Σ(dy^2)).
sqrt(Σ(dx^2) * Σ(dy^2)) = sqrt(3718 * 1655) ≈ 2033.86
Step 7: Calculate the correlation coefficient (r) using the formula:
r = Σdxdy / sqrt(Σ(dx^2) * Σ(dy^2))
r = 4765 / 2033.86 ≈ 2.342 (rounded to three decimal places)
Since the correlation coefficient (r) is positive (2.342), it indicates a positive relationship between the minutes studied (X) and the test scores (Y). This means that as the minutes studied increase, the test scores also tend to increase.