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.

hihi


Want latest solution of this assignment

Want to order fresh copy of the Sample Template Answers? online or do you need the old solutions for Sample Template, contact our customer support or talk to us to get the answers of it.