Question

CONSIDER THE FOLLOWING 5 POINTS (1,-1) (2,0) (4.1) (5,2) and (6,1)

Determine the REGRESSION line y = m x + b.; b, What would the predicted value be for x = 3;c?
Is there a good correlation between the TWO VARIABLES? HOW?

Answer 1

Explanation:

To determine the regression line for the given points, we need to calculate the slope (m) and the y-intercept (b) of the line. We can use the formula for the slope of a line:

m = (nΣxy - ΣxΣy) / (nΣx^2 - (Σx)^2)

where n is the number of data points, Σxy is the sum of the product of x and y, Σx is the sum of x, Σy is the sum of y, and Σx^2 is the sum of the squares of x.

Using the given points, we can calculate:

n = 5

Σx = 18

Σy = 3

Σxy = 17

Σx^2 = 86

Plugging these values into the formula, we get:

m = (5 * 17 - 18 * 3) / (5 * 86 - 18^2) = 0.235

To find the y-intercept (b), we can use the formula:

b = 1/n * Σy - m/n * Σx

Plugging in the values, we get:

b = 1/5 * 3 - 0.235/5 * 18 = -0.294

Therefore, the regression line is:

y = 0.235x - 0.294

To find the predicted value for x = 3, we can plug in x = 3 into the equation:

y = 0.235(3) - 0.294 = -0.088

Therefore, the predicted value for x = 3 is -0.088.

To determine whether there is a good correlation between the two variables, we can calculate the correlation coefficient (r). The correlation coefficient measures the strength and direction of the linear relationship between two variables. A value of r between -1 and 1 indicates the strength and direction of the correlation, with 0 indicating no correlation and -1 or 1 indicating a perfect negative or positive correlation, respectively.

Using the given points, we can calculate:

r = [nΣxy - ΣxΣy] / sqrt[(nΣx^2 - (Σx)^2)(nΣy^2 - (Σy)^2)]

Plugging in the values, we get:

r = [5 * 17 - 18 * 3] / sqrt[(5 * 86 - 18^2)(5 * 14 - 3^2)] = 0.876

Since the correlation coefficient is close to 1, there is a strong positive correlation between the two variables. Therefore, we can conclude that there is a good correlation between the two variables.

Answer 2

To find the regression line y = mx + b, we first need to calculate the slope m and the y-intercept b using the following formulas:

m = (nΣ(xy) - ΣxΣy) / (nΣ(x^2) - (Σx)^2)

b = (Σy - mΣx) / n

where n is the number of points, Σxy is the sum of the products of x and y for all points, Σx and Σy are the sums of x and y values, and Σ(x^2) is the sum of the squares of x values.

Using these formulas, we get:

n = 5

Σxy = 39

Σx = 18

Σy = 1

Σ(x^2) = 86

m = (539 - 181) / (586 - 18^2) ≈ 0.372

b = (1 - 0.37218) / 5 ≈ -0.416

Therefore, the regression line is y = 0.372x - 0.416.

To find the predicted value for x = 3, we plug in x = 3 into the regression line and get:

y = 0.372(3) - 0.416 ≈ 0.108

So the predicted value for x = 3 is approximately 0.108.

To determine if there is a good correlation between the two variables, we can calculate the correlation coefficient r, which measures the strength and direction of the linear relationship between the variables. The formula for r is:

r = (nΣ(xy) - ΣxΣy) / sqrt((nΣ(x^2) - (Σx)^2)(nΣ(y^2) - (Σy)^2))

Using the values we calculated earlier, we get:

n = 5

Σxy = 39

Σx = 18

Σy = 1

Σ(x^2) = 86

Σ(y^2) = 6

r = (539 - 181) / sqrt((586 - 18^2)(5*6 - 1^2)) ≈ 0.932

Since r is close to 1, we can conclude that there is a strong positive linear correlation between the two variables.

Explanation: