The Correlation Coefficient

Week 4 Assignment

1. For each correlation coefficient below, calculate what proportion of variance is shared by the two correlated variables:

  1. r = 0.25 r2 = 0.0625
  2. r = 0.33 r2 = 0.1089
  3. r = 0.90 r2 = 0.81
  4. r = 0.14 r2 = 0.6584

2. For each coefficient of determination below, calculate the value of the correlation coefficient:

  1. r2 = 0.54 sqrt(r2) = 0.73
  2. r2 = 0.13 sqrt(r2) = 0.36
  3. r2 = 0.29 sqrt(r2) = 0.54
  4. r2 = 0.07 sqrt(r2) = 0.26

3. Suppose a researcher regressed surgical patients’ length of stay (dependent variable) in the hospital on a scale of functional ability measured 24 hours after surgery. Given the following, solve for the value of the intercept constant and write out the full regression equation:

Mean length of stay = 6.5days; mean score on scale = 33; slope = -0.10

Y = a + bX      Y = mean length of stay         X = mean score on scale         b = slope

6.5 = a + (-0.10)(33)

a = 9.8

Y = 9.8 -0.1X

4. Using the regression equation calculated in Exercise 3, compute the predicted value of Y (length of hospital stay) for patients with the following functional ability scores:

  1. X = 42 Y = 9.8 – 0.1(42) = 6
  2. X = 68 Y = 9.8 – 0.1(68) = 3
  3. X = 23 Y = 9.8 – 0.1(23) = 5
  4. X = 10 Y = 9.8 – 0.1(10) = 8

5. Use the regression equation below for predicting graduate GPA for the three presented cases.

Y′ = -1.636 + 0.793(undergrad GPA) + 0.004(GREverbal) – 0.0009(GREquant)

+0.009(Motivation)

Subject

undergrad GPA

GREverbal

GREquant

Motivation

1

2.9

560

540

55

2

3.2

550

590

65

3

3.4

600

550

70

Subject 1: Y′ = -1.636 + 0.793(2.9) + 0.004(560) – 0.0009(540) + 0.009(55) = 2.9127

Subject 2: Y′ = -1.636 + 0.793(3.2) + 0.004(550) – 0.0009(590) + 0.009(65) = 3.1556

Subject 3: Y′ = -1.636 + 0.793(3.4) + 0.004(600) – 0.0009(550) + 0.009(70) = 3.5952

6. Using the following information for R2, k, and N, calculate the value of the F statistic for testing the overall regression equation and determine whether F is statistically significant at the 0.05 level:

F = (R2/k)/[(1-R2)/(N-k-1)]

R2 = 0.13, k = 5, N = 120 F = (0.13/5) / [(1-0.13)/(120-5-1)] = 4

3.4 > tabled F = 2.45; significant – reject the null hypothesis.

R2 = 0.53, k = 5, N = 30 F = (0.53/5) / [(1- 0.53)/(30-5-1)] = 6

5.6 > tabled F = 2.84; significant- reject the null hypothesis.

R2 = 0.28, k = 4, N = 64 F = (0.53/5) / [(1- 0.53)/(30-5-1)] = 6

5.6 > tabled F = 2.76; significant- reject the null hypothesis.

R2 = 0.14, k = 4, N = 64 F = (0.14/ 4) / [(1-0.14)/(64-4-1)] = 3

2.3< tabled F = 2.76; not significant- retain the null hypothesis.

7. According to the University of Chicago, as men age, their cholesterol level goes up. A new drug (XAB) is being tested to determine if it can lower cholesterol in aging males and at what dose.  The data for the first test subject is below:

            Dose (mg)                               2            3        5          6          8          10 

Cholesterol level (mg/dL)      310       124      201      110      52        20

a. Plot the data and include a regression line in StatCrunch.  Copy and paste your graph into your Word document for full credit.

Simple linear regression results:

Dependent Variable: Cholesterol level (mg/dL)

Independent Variable: Dose (mg) 

Cholesterol level (mg/dL) = 305.75 - 29.926471 Dose (mg)

Sample size: 6

R (correlation coefficient) = -0.85274561

R-sq = 0.72717508

Estimate of error standard deviation: 61.710186

Parameter estimates:

Parameter

Estimate

Std. Err.

Alternative

DF

T-Stat

P-value

Intercept

305.75

57.724593

≠ 0

4

5.2967025

0.0061

Slope

-29.926471

9.1653284

≠ 0

4

-3.2651826

0.0309

Analysis of variance table for regression model:

Source

DF

SS

MS

F-stat

P-value

Model

1

40600.245

40600.245

10.661417

0.0309

Error

4

15232.588

3808.1471

   

Total

5

55832.833

     

b. What is the correlation coefficient r and what does it mean in this case? The correlation coefficient r is -0.8527 and this represents the magnitude and nature of the relationship between variables. Because this value is a negative, it indicates that high values of one variable is associated with low values on the other variable. The higher the value, the stronger the relationship. In this case, we have a moderately strong negative relationship.

c. What is the coefficient of determination and what does it mean in this case? The coefficient of determination is 0.7272 and it means that 72.72% of the variation is explained by the linear correlation.

d. Is there a statistically significant correlation between dose and cholesterol level in this case? Because the F statistic is 10.66 is greater than its critical value of 6.61 and the p value is less than 0.05, there is a statistically significant correlation between dose and cholesterol level in this case.

e. What is the predicted cholesterol level for a person taking a dose of 4 mg?  What about if they are not taking the drug at all (0 mg)?

Cholesterol level (mg/dL) = 305.75 - 29.926471 Dose (mg)

Y = 305.75 – 29.93(4) = 186.03

The predicted cholesterol level for a person taking a dose of 4 mg is 186.04 mg/dL. For a person not taking the drug, the predicted cholesterol level would be 305.75 mg/dL.

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