4) The table below shows the study times and test scores for a number of students.

http://www.psychstat.missouristate.edu/introbook/sbk17m.htm

To find the line that best relates study time to the test score, we can use linear regression. Linear regression is a statistical method that helps us find the relationship between two variables by fitting a straight line to the data.

a) To find the line, we need to calculate the regression equation. The general equation for a straight line is y = mx + b, where y is the dependent variable (test score in this case), x is the independent variable (study time), m is the slope of the line, and b is the y-intercept.

First, let's calculate the means of study time (x) and test scores (y):

Mean of x = (9 + 16 + 21 + 26 + 33)/5 = 21
Mean of y = (59 + 61 + 64 + 65 + 73)/5 = 64.4

Now, let's calculate the deviations from the mean for both x and y. Deviation from the mean is the difference between each individual data point and the mean:

Deviation from mean of x = (9 - 21), (16 - 21), (21 - 21), (26 - 21), (33 - 21) = -12, -5, 0, 5, 12
Deviation from mean of y = (59 - 64.4), (61 - 64.4), (64 - 64.4), (65 - 64.4), (73 - 64.4) = -5.4, -3.4, -0.4, 0.6, 8.6

Next, we need to calculate the product of the deviations from the mean for each data point:

Product = (Deviations from mean of x) * (Deviations from mean of y)

Product = (-12 * -5.4), (-5 * -3.4), (0 * -0.4), (5 * 0.6), (12 * 8.6) = 64.8, 17, 0, 3, 103.2

Now, let's calculate the sum of the products:

Sum of Product = 64.8 + 17 + 0 + 3 + 103.2 = 188

Next, we need to calculate the sum of the squared deviations from the mean for both x and y:

Sum of Squared Deviation for x = (-12)^2 + (-5)^2 + (0)^2 + (5)^2 + (12)^2 = 274
Sum of Squared Deviation for y = (-5.4)^2 + (-3.4)^2 + (-0.4)^2 + (0.6)^2 + (8.6)^2 = 99.72

Now, let's calculate the slope (m) of the regression line:

m = (Sum of Product) / (Sum of Squared Deviation for x)
m = 188 / 274 ≈ 0.685

Next, let's calculate the y-intercept (b) of the regression line:

b = Mean of y - m * Mean of x
b = 64.4 - 0.685 * 21 ≈ 49.28

Therefore, the regression line equation is:
y = 0.685x + 49.28

b) Now, let's calculate the correlation coefficient (r) and r^2.

The correlation coefficient (r) is the measure of the strength and direction of the linear relationship between two variables. We can use the formula for r:

r = (Sum of Product) / (√(Sum of Squared Deviation for x) * √(Sum of Squared Deviation for y))

r = 188 / (√274 * √99.72) ≈ 0.837

To calculate r^2 (coefficient of determination), we square the correlation coefficient:

r^2 = 0.837^2 ≈ 0.701

Therefore, r ≈ 0.837 and r^2 ≈ 0.701.

c) To determine whether r is statistically significant using a 5% significance level, we need to calculate the degrees of freedom (df) and compare it with the critical value.

The degrees of freedom (df) for correlation coefficient are given by (n - 2), where n is the number of data pairs. In this case, n = 5.

df = 5 - 2 = 3

Next, we need to find the critical value for a 5% significance level with 3 degrees of freedom. Using statistical tables or software, you can find that the critical value for a 5% significance level with 3 degrees of freedom is approximately 0.878.

Since the correlation coefficient (r = 0.837) is smaller than the critical value (0.878), we can conclude that the correlation is not statistically significant at the 5% significance level.