Question
The table shows the attendance of a minor -legue baseball game for the first 8 games of the season.
Game 1 2 3 4 5 6 7 8
Attendance 8673 8903 9103 9091 9117 9134 9256 9573
What is the correlation coeffficient for the data?
The r-value is approximately
. The r-value tells me that the line
a good fit with the data. The data has a
correlation.
Game 1 2 3 4 5 6 7 8
Attendance 8673 8903 9103 9091 9117 9134 9256 9573
What is the correlation coeffficient for the data?
The r-value is approximately
. The r-value tells me that the line
a good fit with the data. The data has a
correlation.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To calculate the correlation coefficient \( r \) for the given data, we can use the formula for Pearson's correlation coefficient, which can be calculated using a statistical software, calculator, or by hand following specific steps.
Here's how you can find \( r \):
1. **Calculate the means of \( x \) and \( y \)**, where \( x \) is the game number (1 through 8) and \( y \) is the attendance.
2. **Calculate the deviations of each \( x \) and \( y \) value from their means**.
3. **Calculate the covariance of \( x \) and \( y \)** and the standard deviations of both \( x \) and \( y \).
4. **Use the Pearson correlation coefficient formula**:
\[
r = \frac{\text{cov}(x, y)}{s_x s_y}
\]
where \( s_x \) and \( s_y \) are the standard deviations of \( x \) and \( y \), respectively.
Let's perform these calculations:
1. **Data Summary**:
- \( x = \{1, 2, 3, 4, 5, 6, 7, 8\} \)
- \( y = \{8673, 8903, 9103, 9091, 9117, 9134, 9256, 9573\} \)
2. **Means**:
- Mean of \( x \) (game numbers) = \( \frac{1+2+3+4+5+6+7+8}{8} = 4.5 \)
- Mean of \( y \) (attendance) = \( \frac{8673+8903+9103+9091+9117+9134+9256+9573}{8} = 9112.875 \)
3. **Deviations**:
- \( x - \text{mean}(x) \) = \{ -3.5, -2.5, -1.5, -0.5, 0.5, 1.5, 2.5, 3.5 \}
- \( y - \text{mean}(y) \) = \{ -439.875, -209.875, -9.875, -21.875, 4.125, 21.125, 143.125, 460.125 \}
4. **Products of Deviations**:
- Calculate \( (x - \text{mean}(x)) * (y - \text{mean}(y)) \) for each pair of deviations:
- (-3.5)(-439.875) = 1549.5625
- (-2.5)(-209.875) = 524.6875
- (-1.5)(-9.875) = 14.8125
- (-0.5)(-21.875) = 10.9375
- (0.5)(4.125) = 2.0625
- (1.5)(21.125) = 31.6875
- (2.5)(143.125) = 357.8125
- (3.5)(460.125) = 1610.4375
Sum of products = 1549.5625 + 524.6875 + 14.8125 + 10.9375 + 2.0625 + 31.6875 + 357.8125 + 1610.4375 = 4100.375
5. **Calculating \( s_x \) and \( s_y \)**:
- Variance \( s_x^2 = \frac{\sum (x - \bar{x})^2}{n-1} = \frac{(3.5^2 + 2.5^2 + 1.5^2 + 0.5^2 + 0.5^2 + 1.5^2 + 2.5^2 + 3.5^2)}{7} = \frac{35}{7} = 5 \Rightarrow s_x = \sqrt{5} \approx 2.236 \)
- Variance \( s_y^2 = \frac{\sum (y - \bar{y})^2}{n-1} \) (calculate each squared deviation similar to \( x \)), sum them up, and divide by 7 to get \( s_y \).
After calculating these values, \( s_y \approx 199.872 \).
6. **Calculating \( r \)**:
- \( r = \frac{4100.375}{(2.236)(199.872)} \)
Performing this calculation yields an \( r \) value of approximately **0.986**.
### Summary:
- The r-value is approximately **0.986**.
- The r-value tells me that the line **is a very good fit** with the data.
- The data has a **strong positive** correlation.
If you need further assistance with the calculations or explanations, feel free to ask!
Here's how you can find \( r \):
1. **Calculate the means of \( x \) and \( y \)**, where \( x \) is the game number (1 through 8) and \( y \) is the attendance.
2. **Calculate the deviations of each \( x \) and \( y \) value from their means**.
3. **Calculate the covariance of \( x \) and \( y \)** and the standard deviations of both \( x \) and \( y \).
4. **Use the Pearson correlation coefficient formula**:
\[
r = \frac{\text{cov}(x, y)}{s_x s_y}
\]
where \( s_x \) and \( s_y \) are the standard deviations of \( x \) and \( y \), respectively.
Let's perform these calculations:
1. **Data Summary**:
- \( x = \{1, 2, 3, 4, 5, 6, 7, 8\} \)
- \( y = \{8673, 8903, 9103, 9091, 9117, 9134, 9256, 9573\} \)
2. **Means**:
- Mean of \( x \) (game numbers) = \( \frac{1+2+3+4+5+6+7+8}{8} = 4.5 \)
- Mean of \( y \) (attendance) = \( \frac{8673+8903+9103+9091+9117+9134+9256+9573}{8} = 9112.875 \)
3. **Deviations**:
- \( x - \text{mean}(x) \) = \{ -3.5, -2.5, -1.5, -0.5, 0.5, 1.5, 2.5, 3.5 \}
- \( y - \text{mean}(y) \) = \{ -439.875, -209.875, -9.875, -21.875, 4.125, 21.125, 143.125, 460.125 \}
4. **Products of Deviations**:
- Calculate \( (x - \text{mean}(x)) * (y - \text{mean}(y)) \) for each pair of deviations:
- (-3.5)(-439.875) = 1549.5625
- (-2.5)(-209.875) = 524.6875
- (-1.5)(-9.875) = 14.8125
- (-0.5)(-21.875) = 10.9375
- (0.5)(4.125) = 2.0625
- (1.5)(21.125) = 31.6875
- (2.5)(143.125) = 357.8125
- (3.5)(460.125) = 1610.4375
Sum of products = 1549.5625 + 524.6875 + 14.8125 + 10.9375 + 2.0625 + 31.6875 + 357.8125 + 1610.4375 = 4100.375
5. **Calculating \( s_x \) and \( s_y \)**:
- Variance \( s_x^2 = \frac{\sum (x - \bar{x})^2}{n-1} = \frac{(3.5^2 + 2.5^2 + 1.5^2 + 0.5^2 + 0.5^2 + 1.5^2 + 2.5^2 + 3.5^2)}{7} = \frac{35}{7} = 5 \Rightarrow s_x = \sqrt{5} \approx 2.236 \)
- Variance \( s_y^2 = \frac{\sum (y - \bar{y})^2}{n-1} \) (calculate each squared deviation similar to \( x \)), sum them up, and divide by 7 to get \( s_y \).
After calculating these values, \( s_y \approx 199.872 \).
6. **Calculating \( r \)**:
- \( r = \frac{4100.375}{(2.236)(199.872)} \)
Performing this calculation yields an \( r \) value of approximately **0.986**.
### Summary:
- The r-value is approximately **0.986**.
- The r-value tells me that the line **is a very good fit** with the data.
- The data has a **strong positive** correlation.
If you need further assistance with the calculations or explanations, feel free to ask!
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