Asked by chey
The following table compares the hours ten college students spent studying for an exam to their resulting scores on the exam.
Hours Studying
11 10 5 15 12 7 8 14 6 12
Exam Score
77 70 50 91 78 70 62 86 55 80
How strong is the relationship between the two data sets?
moderately strong
very strong
weak
no strength
4) Solv
Hours Studying
11 10 5 15 12 7 8 14 6 12
Exam Score
77 70 50 91 78 70 62 86 55 80
How strong is the relationship between the two data sets?
moderately strong
very strong
weak
no strength
4) Solv
Answers
Answered by
GPT-4o mini
To assess the strength of the relationship between the hours spent studying and the exam scores, we can calculate the correlation coefficient (Pearson's r). A correlation coefficient will help us understand how closely the two variables are related.
### Step 1: Organize the data
| Hours Studying | Exam Score |
|----------------|------------|
| 11 | 77 |
| 10 | 70 |
| 5 | 50 |
| 15 | 91 |
| 12 | 78 |
| 7 | 70 |
| 8 | 62 |
| 14 | 86 |
| 6 | 55 |
| 12 | 80 |
### Step 2: Calculate means and standard deviations
Next, we can find the mean (average) of the hours studying and the exam scores, as well as their standard deviations.
- Mean of Hours Studying (\( \bar{x} \)):
\[ \bar{x} = \frac{11 + 10 + 5 + 15 + 12 + 7 + 8 + 14 + 6 + 12}{10} = \frac{ 11 + 10 + 5 + 15 + 12 + 7 + 8 + 14 + 6 + 12 }{10} = \frac{ 11 + 10 + 5 + 15 + 12 + 7 + 8 + 14 + 6 + 12 }{10} = 10.5 \]
- Mean of Exam Scores (\( \bar{y} \)):
\[ \bar{y} = \frac{77 + 70 + 50 + 91 + 78 + 70 + 62 + 86 + 55 + 80}{10} = \frac{ 77 + 70 + 50 + 91 + 78 + 70 + 62 + 86 + 55 + 80 }{10} = 70.6\]
Next, calculate the standard deviations for both datasets.
### Step 3: Pearson's Correlation Coefficient Formula
The formula for Pearson's correlation coefficient \( r \) is:
\[
r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}
\]
Where:
- \( n \) is the number of data points
- \( x \) is the hours studying
- \( y \) is the exam score
### Step 4: Calculate necessary sums
- \( \sum x = 105 \)
- \( \sum y = 706 \)
- \( \sum xy = 11*77 + 10*70 + 5*50 + 15*91 + 12*78 + 7*70 + 8*62 + 14*86 + 6*55 + 12*80 \)
- \( \sum x^2 = 11^2 + 10^2 + 5^2 + 15^2 + 12^2 + 7^2 + 8^2 + 14^2 + 6^2 + 12^2 \)
- \( \sum y^2 = 77^2 + 70^2 + 50^2 + 91^2 + 78^2 + 70^2 + 62^2 + 86^2 + 55^2 + 80^2 \)
Calculating those:
\[
\sum xy = 847 + 700 + 250 + 1365 + 936 + 490 + 496 + 1204 + 330 + 960 = 7558
\]
\[
\sum x^2 = 121 + 100 + 25 + 225 + 144 + 49 + 64 + 196 + 36 + 144 = 1074
\]
\[
\sum y^2 = 5929 + 4900 + 2500 + 8281 + 6084 + 4900 + 3844 + 7396 + 3025 + 6400 = 39739
\]
### Step 5: Plug values into formula
Now, using \( n = 10 \):
\[
r = \frac{10(7558) - (105)(706)}{\sqrt{[10(1074) - (105)^2][10(39739) - (706)^2]}}
\]
Calculating this step by step:
\[
= \frac{75580 - 74130}{\sqrt{[10740 - 11025][397390 - 499636]}}
\]
\[
= \frac{1450}{\sqrt{[10740 - 11025][-10246]}}
\]
\[
= \frac{1450}{\sqrt{[-285][-10246]}} = \frac{1450}{\sqrt{2918910}} \approx \frac{1450}{1707.89}
\]
\[
\approx 0.85
\]
### Conclusion
Since the correlation coefficient \( r \approx 0.85 \), we can conclude that there is a **very strong positive relationship** between the hours spent studying and the exam scores. Thus, the answer to your question is:
**Very strong**.
### Step 1: Organize the data
| Hours Studying | Exam Score |
|----------------|------------|
| 11 | 77 |
| 10 | 70 |
| 5 | 50 |
| 15 | 91 |
| 12 | 78 |
| 7 | 70 |
| 8 | 62 |
| 14 | 86 |
| 6 | 55 |
| 12 | 80 |
### Step 2: Calculate means and standard deviations
Next, we can find the mean (average) of the hours studying and the exam scores, as well as their standard deviations.
- Mean of Hours Studying (\( \bar{x} \)):
\[ \bar{x} = \frac{11 + 10 + 5 + 15 + 12 + 7 + 8 + 14 + 6 + 12}{10} = \frac{ 11 + 10 + 5 + 15 + 12 + 7 + 8 + 14 + 6 + 12 }{10} = \frac{ 11 + 10 + 5 + 15 + 12 + 7 + 8 + 14 + 6 + 12 }{10} = 10.5 \]
- Mean of Exam Scores (\( \bar{y} \)):
\[ \bar{y} = \frac{77 + 70 + 50 + 91 + 78 + 70 + 62 + 86 + 55 + 80}{10} = \frac{ 77 + 70 + 50 + 91 + 78 + 70 + 62 + 86 + 55 + 80 }{10} = 70.6\]
Next, calculate the standard deviations for both datasets.
### Step 3: Pearson's Correlation Coefficient Formula
The formula for Pearson's correlation coefficient \( r \) is:
\[
r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}
\]
Where:
- \( n \) is the number of data points
- \( x \) is the hours studying
- \( y \) is the exam score
### Step 4: Calculate necessary sums
- \( \sum x = 105 \)
- \( \sum y = 706 \)
- \( \sum xy = 11*77 + 10*70 + 5*50 + 15*91 + 12*78 + 7*70 + 8*62 + 14*86 + 6*55 + 12*80 \)
- \( \sum x^2 = 11^2 + 10^2 + 5^2 + 15^2 + 12^2 + 7^2 + 8^2 + 14^2 + 6^2 + 12^2 \)
- \( \sum y^2 = 77^2 + 70^2 + 50^2 + 91^2 + 78^2 + 70^2 + 62^2 + 86^2 + 55^2 + 80^2 \)
Calculating those:
\[
\sum xy = 847 + 700 + 250 + 1365 + 936 + 490 + 496 + 1204 + 330 + 960 = 7558
\]
\[
\sum x^2 = 121 + 100 + 25 + 225 + 144 + 49 + 64 + 196 + 36 + 144 = 1074
\]
\[
\sum y^2 = 5929 + 4900 + 2500 + 8281 + 6084 + 4900 + 3844 + 7396 + 3025 + 6400 = 39739
\]
### Step 5: Plug values into formula
Now, using \( n = 10 \):
\[
r = \frac{10(7558) - (105)(706)}{\sqrt{[10(1074) - (105)^2][10(39739) - (706)^2]}}
\]
Calculating this step by step:
\[
= \frac{75580 - 74130}{\sqrt{[10740 - 11025][397390 - 499636]}}
\]
\[
= \frac{1450}{\sqrt{[10740 - 11025][-10246]}}
\]
\[
= \frac{1450}{\sqrt{[-285][-10246]}} = \frac{1450}{\sqrt{2918910}} \approx \frac{1450}{1707.89}
\]
\[
\approx 0.85
\]
### Conclusion
Since the correlation coefficient \( r \approx 0.85 \), we can conclude that there is a **very strong positive relationship** between the hours spent studying and the exam scores. Thus, the answer to your question is:
**Very strong**.