Question

Checking Your Progress – Correlation & Regression Researchers investigated the relationship between amount of study time statistics class and mid-semester quiz scores. The data appear below: 1 28 95 2 25 95 3 3 58 4 10 75 5 0 44 6 15 83 7 20 91 8 24 87 9 7 65 10 8 70 Find the correlation between hours of study and quiz scores, and test it for significance. Then complete a simple linear regression analysis using hours of study to predict quiz scores. Sum of squares for study hours: Sum of squares for quiz scores: Sum of the cross products: Covariance: Value for Pearson r: Critical value for this Pearson r (two-tailed, alpha .05): The p-value associated with this Pearson r: Should you reject the null hypothesis? Coefficient of determination: What percentage of quiz score variance is explained by study hours? Is there a significant relationship between study hours and quiz scores? What is the regression constant for use in the regression equation to predict quiz scores on the basis of hours of study? ____________ ____________ ____________ ____________ ____________ ____________ ____________ ____________ ____________ ____________ ____________ ____________ What is the regression coefficient? The complete regression equation is: ____________________________________ What quiz score would you predict for a student who studied 17 hours? ____________ Use SPSS to make a scatterplot with QUIZ Scores as the dependent (or criterion) variable and Hours of Study as the independent (or predictor) variable. Edit it for APA style. Export it to Word and provide a figure number and caption. Attach it as the last page.

Answer 1

X Values
∑ = 140
Mean = 14
∑(X - M_x)² = SS_x = 872

Y Values
∑ = 763
Mean = 76.3
∑(Y - M_y)² = SS_y = 2622.1

X and Y Combined
N = 10
Covariance ∑(X - M_x)(Y - M_y) = 1445

R Calculation
r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

r = 1445 / √((872)(2622.1)) = 0.9556

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

The sample size is n=10, so then the number of degrees of freedom is df=n−2=10−2=8

The corresponding critical correlation value rc​ for a significance level of α=0.05, for a two-tailed test is rc​=0.632

Observe that in this case, the null hypothesis H0 :ρ=0 is rejected if ∣r∣>rc​=0.632.

Based on the sample correlation provided, we have that ∣r∣=0.9556>rc​=0.632, from which is concluded that the null hypothesis is rejected.We can conclude that there is significant correlation between two variables.

Coefficient of determination is r squared=(0.9556)=0.9132

Sum of X = 140
Sum of Y = 763
Mean X = 14
Mean Y = 76.3
Sum of squares (SS_X) = 872
Sum of products (SP) = 1445

Regression Equation = ŷ = bX + a

b = SP/SS_X = 1445/872 = 1.65711

a = M_Y - bM_X = 76.3 - (1.66*14) = 53.10046

ŷ = 1.65711X + 53.10046

Constant=53.10

coefficient of regression = 1.66

Required regression equation is  ŷ = 1.65711X + 53.10046

X=17 hours

ŷ = 1.65711X + 53.10046

ŷ = 1.65711*17 + 53.10046

ŷ = 81.27

ŷ = 81 score will be obtained if study for 17 hours