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To find the value of the linear correlation coefficient [tex]\( r \)[/tex] between the scores of Test A and Test B, we need to follow these steps:
1. List the data points:
- Test A scores: 48, 52, 58, 44, 43, 43, 40, 51, 59
- Test B scores: 73, 67, 73, 59, 58, 56, 58, 64, 74
2. Calculate the means of Test A and Test B scores:
- Mean of Test A ([tex]\( \bar{A} \)[/tex]):
[tex]\[ \bar{A} = \frac{48 + 52 + 58 + 44 + 43 + 43 + 40 + 51 + 59}{9} = \frac{438}{9} \approx 48.67 \][/tex]
- Mean of Test B ([tex]\( \bar{B} \)[/tex]):
[tex]\[ \bar{B} = \frac{73 + 67 + 73 + 59 + 58 + 56 + 58 + 64 + 74}{9} = \frac{582}{9} \approx 64.67 \][/tex]
3. Calculate the deviations from the mean for each test:
- For Test A: [tex]\( (A_i - \bar{A}) \)[/tex]
- For Test B: [tex]\( (B_i - \bar{B}) \)[/tex]
4. Find the product of the deviations for each pair of scores:
- [tex]\( (A_i - \bar{A})(B_i - \bar{B}) \)[/tex]
5. Sum these products to get the covariance:
[tex]\[ \text{Cov}(A, B) = \frac{\sum_{i=1}^n (A_i - \bar{A})(B_i - \bar{B})}{n-1} \][/tex]
6. Calculate the standard deviations of both A and B:
[tex]\[ \sigma_A = \sqrt{\frac{\sum_{i=1}^n (A_i - \bar{A})^2}{n-1}} \][/tex]
[tex]\[ \sigma_B = \sqrt{\frac{\sum_{i=1}^n (B_i - \bar{B})^2}{n-1}} \][/tex]
7. Finally, compute the correlation coefficient [tex]\( r \)[/tex] using the formula:
[tex]\[ r = \frac{\text{Cov}(A, B)}{\sigma_A \sigma_B} \][/tex]
After performing all the calculations, we would find that the correlation coefficient [tex]\( r \)[/tex] measures the strength and direction of the linear relationship between the two tests' scores. For the given data, the correlation coefficient [tex]\( r \)[/tex] is found to be approximately [tex]\( 0.867 \)[/tex].
Hence, the value of the linear correlation coefficient [tex]\( r \)[/tex] is [tex]\( \boxed{0.867} \)[/tex].
1. List the data points:
- Test A scores: 48, 52, 58, 44, 43, 43, 40, 51, 59
- Test B scores: 73, 67, 73, 59, 58, 56, 58, 64, 74
2. Calculate the means of Test A and Test B scores:
- Mean of Test A ([tex]\( \bar{A} \)[/tex]):
[tex]\[ \bar{A} = \frac{48 + 52 + 58 + 44 + 43 + 43 + 40 + 51 + 59}{9} = \frac{438}{9} \approx 48.67 \][/tex]
- Mean of Test B ([tex]\( \bar{B} \)[/tex]):
[tex]\[ \bar{B} = \frac{73 + 67 + 73 + 59 + 58 + 56 + 58 + 64 + 74}{9} = \frac{582}{9} \approx 64.67 \][/tex]
3. Calculate the deviations from the mean for each test:
- For Test A: [tex]\( (A_i - \bar{A}) \)[/tex]
- For Test B: [tex]\( (B_i - \bar{B}) \)[/tex]
4. Find the product of the deviations for each pair of scores:
- [tex]\( (A_i - \bar{A})(B_i - \bar{B}) \)[/tex]
5. Sum these products to get the covariance:
[tex]\[ \text{Cov}(A, B) = \frac{\sum_{i=1}^n (A_i - \bar{A})(B_i - \bar{B})}{n-1} \][/tex]
6. Calculate the standard deviations of both A and B:
[tex]\[ \sigma_A = \sqrt{\frac{\sum_{i=1}^n (A_i - \bar{A})^2}{n-1}} \][/tex]
[tex]\[ \sigma_B = \sqrt{\frac{\sum_{i=1}^n (B_i - \bar{B})^2}{n-1}} \][/tex]
7. Finally, compute the correlation coefficient [tex]\( r \)[/tex] using the formula:
[tex]\[ r = \frac{\text{Cov}(A, B)}{\sigma_A \sigma_B} \][/tex]
After performing all the calculations, we would find that the correlation coefficient [tex]\( r \)[/tex] measures the strength and direction of the linear relationship between the two tests' scores. For the given data, the correlation coefficient [tex]\( r \)[/tex] is found to be approximately [tex]\( 0.867 \)[/tex].
Hence, the value of the linear correlation coefficient [tex]\( r \)[/tex] is [tex]\( \boxed{0.867} \)[/tex].
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