Question
A counselor asked college students who had jobs how many hours they worked each week. The counselor then compared the number of hour each student worked and the student's overall grade average. The table shows the data.
1. What is the value of the correlation coefficient (`r`-value)? Do not round
Hours Worked Overall Grade
16 83
31 72
29 70
26 83
15 89
19 89
12 99
2. Based on the correlation coefficient of the data, what type of linear association exists between the hours worked and overall grade average?
strong negative weak negative weak positive strong positive
1. What is the value of the correlation coefficient (`r`-value)? Do not round
Hours Worked Overall Grade
16 83
31 72
29 70
26 83
15 89
19 89
12 99
2. Based on the correlation coefficient of the data, what type of linear association exists between the hours worked and overall grade average?
strong negative weak negative weak positive strong positive
Answers
GPT-4o mini
To calculate the correlation coefficient (r-value) for the data provided, we need to follow these steps:
1. Calculate the mean of the hours worked and the overall grades.
2. Determine the deviations from the mean for both variables.
3. Calculate the covariance of the two variables.
4. Find the standard deviation of both variables.
5. Use these values to calculate the correlation coefficient \( r \).
Let's break down the calculations:
### Data Summary
- **Hours Worked**: 16, 31, 29, 26, 15, 19, 12
- **Overall Grades**: 83, 72, 70, 83, 89, 89, 99
### Step 1: Calculate the means
- Mean hours worked \( \bar{x} \):
\[
\bar{x} = \frac{16 + 31 + 29 + 26 + 15 + 19 + 12}{7} = \frac{ 16 + 31 + 29 + 26 + 15 + 19 + 12 }{7} = \frac{ 148 }{7} \approx 21.14
\]
- Mean overall grade \( \bar{y} \):
\[
\bar{y} = \frac{83 + 72 + 70 + 83 + 89 + 89 + 99}{7} = \frac{ 83 + 72 + 70 + 83 + 89 + 89 + 99 }{7} = \frac{ 585 }{7} \approx 83.57
\]
### Step 2: Deviations from the means
Calculate deviations for each data point:
| Hours Worked (x) | Overall Grade (y) | \( x - \bar{x} \) | \( y - \bar{y} \) |
|------------------|-------------------|-------------------|-------------------|
| 16 | 83 | 16 - 21.14 = -5.14| 83 - 83.57 = -0.57|
| 31 | 72 | 31 - 21.14 = 9.86 | 72 - 83.57 = -11.57|
| 29 | 70 | 29 - 21.14 = 7.86 | 70 - 83.57 = -13.57|
| 26 | 83 | 26 - 21.14 = 4.86 | 83 - 83.57 = -0.57|
| 15 | 89 | 15 - 21.14 = -6.14| 89 - 83.57 = 5.43|
| 19 | 89 | 19 - 21.14 = -2.14| 89 - 83.57 = 5.43|
| 12 | 99 | 12 - 21.14 = -9.14| 99 - 83.57 = 15.43|
### Step 3: Covariance of x and y
Calculate \( (x - \bar{x})(y - \bar{y}) \) and the mean of that product.
\[
\sum (x - \bar{x})(y - \bar{y}) = (-5.14)(-0.57) + (9.86)(-11.57) + (7.86)(-13.57) + (4.86)(-0.57) + (-6.14)(5.43) + (-2.14)(5.43) + (-9.14)(15.43)
\]
\[
= 2.93 - 114.06 - 106.70 - 2.77 - 33.33 - 11.62 - 140.94 = -406.29
\]
The covariance is:
\[
\text{Cov}(x,y) = \frac{-406.29}{7 - 1} = -67.71
\]
### Step 4: Standard deviation of \( x \) and \( y \)
Calculate standard deviation for \( x \):
\[
s_x = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}
\]
Calculate \( (x - \bar{x})^2 \):
\[
(-5.14)^2 + (9.86)^2 + (7.86)^2 + (4.86)^2 + (-6.14)^2 + (-2.14)^2 + (-9.14)^2
= 26.58 + 97.27 + 61.78 + 23.58 + 37.58 + 4.58 + 83.27 = 334.12
\]
\[
s_x = \sqrt{\frac{334.12}{6}} \approx \sqrt{55.69} \approx 7.46
\]
Calculate standard deviation for \( y \):
\[
s_y = \sqrt{\frac{\sum (y - \bar{y})^2}{n - 1}}
\]
Calculate \( (y - \bar{y})^2 \):
\[
(-0.57)^2 + (-11.57)^2 + (-13.57)^2 + (-0.57)^2 + (5.43)^2 + (5.43)^2 + (15.43)^2
= 0.32 + 133.74 + 184.59 + 0.32 + 29.50 + 29.50 + 238.00 = 615.97
\]
\[
s_y = \sqrt{\frac{615.97}{6}} \approx \sqrt{102.66} \approx 10.13
\]
### Step 5: Calculate the correlation coefficient \( r \)
\[
r = \frac{\text{Cov}(x,y)}{s_x s_y}
= \frac{-67.71}{(7.46)(10.13)}
\approx \frac{-67.71}{75.53}
\approx -0.895
\]
### Final Answers
1. The value of the correlation coefficient \( r \) is approximately **-0.895** (do not round).
2. Based on the correlation coefficient of the data, the type of linear association that exists between the hours worked and overall grade average is **strong negative**.
Thus, as the hours worked increase, the overall grade average tends to decrease.
1. Calculate the mean of the hours worked and the overall grades.
2. Determine the deviations from the mean for both variables.
3. Calculate the covariance of the two variables.
4. Find the standard deviation of both variables.
5. Use these values to calculate the correlation coefficient \( r \).
Let's break down the calculations:
### Data Summary
- **Hours Worked**: 16, 31, 29, 26, 15, 19, 12
- **Overall Grades**: 83, 72, 70, 83, 89, 89, 99
### Step 1: Calculate the means
- Mean hours worked \( \bar{x} \):
\[
\bar{x} = \frac{16 + 31 + 29 + 26 + 15 + 19 + 12}{7} = \frac{ 16 + 31 + 29 + 26 + 15 + 19 + 12 }{7} = \frac{ 148 }{7} \approx 21.14
\]
- Mean overall grade \( \bar{y} \):
\[
\bar{y} = \frac{83 + 72 + 70 + 83 + 89 + 89 + 99}{7} = \frac{ 83 + 72 + 70 + 83 + 89 + 89 + 99 }{7} = \frac{ 585 }{7} \approx 83.57
\]
### Step 2: Deviations from the means
Calculate deviations for each data point:
| Hours Worked (x) | Overall Grade (y) | \( x - \bar{x} \) | \( y - \bar{y} \) |
|------------------|-------------------|-------------------|-------------------|
| 16 | 83 | 16 - 21.14 = -5.14| 83 - 83.57 = -0.57|
| 31 | 72 | 31 - 21.14 = 9.86 | 72 - 83.57 = -11.57|
| 29 | 70 | 29 - 21.14 = 7.86 | 70 - 83.57 = -13.57|
| 26 | 83 | 26 - 21.14 = 4.86 | 83 - 83.57 = -0.57|
| 15 | 89 | 15 - 21.14 = -6.14| 89 - 83.57 = 5.43|
| 19 | 89 | 19 - 21.14 = -2.14| 89 - 83.57 = 5.43|
| 12 | 99 | 12 - 21.14 = -9.14| 99 - 83.57 = 15.43|
### Step 3: Covariance of x and y
Calculate \( (x - \bar{x})(y - \bar{y}) \) and the mean of that product.
\[
\sum (x - \bar{x})(y - \bar{y}) = (-5.14)(-0.57) + (9.86)(-11.57) + (7.86)(-13.57) + (4.86)(-0.57) + (-6.14)(5.43) + (-2.14)(5.43) + (-9.14)(15.43)
\]
\[
= 2.93 - 114.06 - 106.70 - 2.77 - 33.33 - 11.62 - 140.94 = -406.29
\]
The covariance is:
\[
\text{Cov}(x,y) = \frac{-406.29}{7 - 1} = -67.71
\]
### Step 4: Standard deviation of \( x \) and \( y \)
Calculate standard deviation for \( x \):
\[
s_x = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}
\]
Calculate \( (x - \bar{x})^2 \):
\[
(-5.14)^2 + (9.86)^2 + (7.86)^2 + (4.86)^2 + (-6.14)^2 + (-2.14)^2 + (-9.14)^2
= 26.58 + 97.27 + 61.78 + 23.58 + 37.58 + 4.58 + 83.27 = 334.12
\]
\[
s_x = \sqrt{\frac{334.12}{6}} \approx \sqrt{55.69} \approx 7.46
\]
Calculate standard deviation for \( y \):
\[
s_y = \sqrt{\frac{\sum (y - \bar{y})^2}{n - 1}}
\]
Calculate \( (y - \bar{y})^2 \):
\[
(-0.57)^2 + (-11.57)^2 + (-13.57)^2 + (-0.57)^2 + (5.43)^2 + (5.43)^2 + (15.43)^2
= 0.32 + 133.74 + 184.59 + 0.32 + 29.50 + 29.50 + 238.00 = 615.97
\]
\[
s_y = \sqrt{\frac{615.97}{6}} \approx \sqrt{102.66} \approx 10.13
\]
### Step 5: Calculate the correlation coefficient \( r \)
\[
r = \frac{\text{Cov}(x,y)}{s_x s_y}
= \frac{-67.71}{(7.46)(10.13)}
\approx \frac{-67.71}{75.53}
\approx -0.895
\]
### Final Answers
1. The value of the correlation coefficient \( r \) is approximately **-0.895** (do not round).
2. Based on the correlation coefficient of the data, the type of linear association that exists between the hours worked and overall grade average is **strong negative**.
Thus, as the hours worked increase, the overall grade average tends to decrease.