PSYC 3050 Exam 3 Calculations and Correlation Analysis

PSYC 3050
Exam 3 Calculations
Exam 3 Calculations
Name of Student
Name of University
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PSYC 3050
Exam 3 Calculations
Relationship 1
The relationship between hours studied and GPA:
1. What is the predictor variable?
The predictor variable is the variable named “study” denoting the number of hours studied by
the student.
2. What is the criterion variable?
The criterion variable is the variable named “GPA” denoting the GPA of the student.
3. Correlation: Use the steps of hypothesis testing to evaluate this hypothesis.
a. State the hypotheses.
The hypothesis states that the GPA scores of the student is affected by the hours studied by
the student. The null and alternate hypothesis can be therefore written as:
H0 : ρ=0 against H1: ρ ≠0
b. Determine the critical value of your statistic for rejecting the null based on the given alpha
level/sample size.
The critical value is given by the inverse of t-statistic with n-2 or 27 degrees of freedom
and alpha value equal to 0.05. The critical value was determined to be 2.0518.
c. Calculate the statistic.
The statistic of interest to test the hypothesis is given by |r √ n−2|
√ 1−r2 . The Pearson’s
correlation coefficient, r, given by ∑ ¿¿ ¿, was computed as 0.7539. The data has 29 sample
points. Thus the statistic was computed to have the value 5.9628.
d. Based on this statistic alone, would you reject or accept the null hypothesis?
The test suggests rejection of null hypothesis at 5% level of significance since the statistic
has value greater than the critical value.
e. Calculate the effect. What does this mean?
The value of 1−r2 is 0.4316, which is the effect that the predictor variable has on the
criterion variable. It denotes the proportion of variation in criterion variable as explained by
the predictor variable.
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PSYC 3050
Exam 3 Calculations
4. Regression: Address the following items
a. Calculate the slope of the regression line.
The slope, given by b=r × s y
sx
, where sx is the standard deviation of the predictor X and sy
is the sum of squared deviation from mean for the criterion variable was found to be
0.1024.
b. Calculate the intercept of the regression line.
The intercept a given by y−b x was found to be 1.632 where y is the mean of the criterion
variable and x is the mean of predictor.
c. Insert the calculated values for the slope and intercept into the equation for the
regression line.
The regression line was thus obtained as: y=1.632+ 0.1024 x
d. Determine the predicted GPA for students who study:
i. 4
Putting X as 4 in the equation specified in part c, the predicted GPA was found to be 2.042.
ii. 8
Putting X as 8 in the equation specified in part c, the predicted GPA was found to be 2.4519.
iii. 12
Putting X as 12 in the equation specified in part c, the predicted GPA was found to be 2.862.
iv. 16
Putting X as 16 in the equation specified in part c, the predicted GPA was found to be 3.272.
v. 20
Putting X as 20 in the equation specified in part c, the predicted GPA was found to be 3.681.
e. Determine the sum of squares for the error in predictions (standard error of the estimate).
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