A psychology student wanted to examine the value of predicting GPAs based on IQ scores. She recruited undergraduate students enrolled in an introductory psychology class and administered the IQ test (M = 100.260, SD = 12.985) and requested their GPAs from the registrar’s office (M = 2.456, SD = .861). Consistent with many previous reports, the student found that IQ was significantly correlated with GPA, r(86) = .497, p < .001.
Find a) sY-Ŷ. , b) covXY., and c) the equation for the regression line. Round to three decimal places.
Final Part of the problem:
Find a) sY-Ŷ. , b) covXY., and c) the equation for the regression line. Round to three decimal places.
To find the values of sY-Ŷ, covXY, and the equation for the regression line, we need to calculate them step by step.
a) sY-Ŷ (standard deviation of the predicted values):
- The standard deviation of the predicted values represents the average distance between the observed data points (GPAs) and their corresponding predicted values on the regression line.
- To calculate sY-Ŷ, we first need to calculate the predicted values (Ŷ) based on the regression equation.
- The regression equation is given by Ŷ = bX + a, where Y represents the dependent variable (GPA), X represents the independent variable (IQ), b is the slope of the regression line, and a is the y-intercept.
- In this case, the equation for the regression line is Ŷ = b * IQ + a, where b is the correlation coefficient (r) multiplied by the standard deviation of Y (GPA) divided by the standard deviation of X (IQ), and a is the mean of Y (GPA) minus b multiplied by the mean of X (IQ).
- Let's calculate the predicted values (Ŷ) using the given values:
Ŷ = r * (SDY/SDX) * X + (MY - r * MX * SDY/SDX)
Ŷ = .497 * (.861/12.985) * X + (2.456 - .497 * 100.260 * .861/12.985)
Ŷ = .033 * X + 1.960
- Now, we can calculate the standard deviation of the predicted values (sY-Ŷ):
sY-Ŷ = √((Σ(Ŷ - Ŷmean)^2) / (n - 2)), where n is the sample size.
We don't have the individual predicted values here, so we need to calculate the mean of the predicted values (Ŷmean) and the sum of squares of the differences between each predicted value and the mean (Σ(Ŷ - Ŷmean)^2).
sY-Ŷ = √(Σ(Ŷ - Ŷmean)^2 / (n - 2))
- We also know that the mean of the predicted values (Ŷmean) is equal to the mean of Y (GPA), which is 2.456 in this case.
- Since we don't have the individual predicted values, we substitute from the regression equation to find Ŷ - Ŷmean:
Ŷ - Ŷmean = Ŷ - MY = (.033 * X + 1.960) - 2.456
Ŷ - Ŷmean = .033 * X - .496
- As we don't have the individual values of X, we can't calculate each Ŷ - Ŷmean term. However, we can substitute it into the formula for sY-Ŷ:
sY-Ŷ = √((Σ(.033 * X - .496)^2) / (n - 2))
- We need to know the sample size (n) from the given information. However, the sample size (n) is not mentioned in the given question. So, we cannot calculate sY-Ŷ without the sample size.
b) covXY (covariance between X (IQ) and Y (GPA)):
- The covariance between two variables measures the extent to which they vary together.
- In this case, we need to find the covariance between X (IQ) and Y (GPA).
- The formula to calculate the covariance is:
covXY = Σ[(X - Xmean)(Y - Ymean)] / (n - 1), where n is the sample size.
- Since we don't have the individual values of X and Y, we can't calculate the covariance without the sample size.
c) The equation of the regression line:
- We have already calculated it in part a).
- The equation for the regression line is Ŷ = .033 * X + 1.960, where Ŷ is the predicted value of Y (GPA) based on X (IQ).
Without the sample size (n) provided in the given information, we cannot calculate sY-Ŷ (standard deviation of the predicted values) or covXY (covariance between X and Y).