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To determine the linear correlation coefficient of the relationship between the rainfall (in inches) and the yield of wheat (bushels per acre), we can follow these steps:
1. Understand the Data:
Let's consider the data given:
- Rainfall (in inches), [tex]\( x \)[/tex]: [9.4, 7.7, 123, 11.4, 17.7, 9.2, 5.9, 14.5, 149]
- Yield (bushels per acre), [tex]\( y \)[/tex]: [46.5, 422, 548, 55, 78.4, 45.2, 27.9, 72, 748]
2. Correlation Coefficient Formula:
The Pearson linear correlation coefficient [tex]\( r \)[/tex] is calculated using the formula:
[tex]\[ r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} \][/tex]
where:
- [tex]\( x_i \)[/tex] and [tex]\( y_i \)[/tex] are the individual sample points.
- [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex] are the means (average values) of the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] variables, respectively.
3. Compute the Means:
Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \bar{x} = \frac{\sum x_i}{n} \][/tex]
[tex]\[ \bar{y} = \frac{\sum y_i}{n} \][/tex]
where [tex]\( n \)[/tex] is the number of data points, which is 9 in our case.
4. Compute the Sums:
Evaluate the terms for sum of products of deviations, sum of squared deviations for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) \][/tex]
[tex]\[ \sum (x_i - \bar{x})^2 \][/tex]
[tex]\[ \sum (y_i - \bar{y})^2 \][/tex]
5. Compute [tex]\( r \)[/tex]:
Use these sums to find the correlation coefficient [tex]\( r \)[/tex].
Following these steps analytically, we should end up calculating the linear correlation coefficient [tex]\( r \)[/tex] based on the given data.
Upon calculating using these steps, the result is:
[tex]\[ r = 0.883704508386272 \][/tex]
Therefore, the linear correlation coefficient for the given rainfall and wheat yield data is approximately [tex]\( 0.884 \)[/tex].
From the given options, none of them directly match [tex]\( 0.884 \)[/tex], which we have precisely calculated. The closest to our result from the given options would have been [tex]\( 0.899 \)[/tex], but [tex]\( 0.884 \)[/tex] is the accurate linear correlation coefficient.
1. Understand the Data:
Let's consider the data given:
- Rainfall (in inches), [tex]\( x \)[/tex]: [9.4, 7.7, 123, 11.4, 17.7, 9.2, 5.9, 14.5, 149]
- Yield (bushels per acre), [tex]\( y \)[/tex]: [46.5, 422, 548, 55, 78.4, 45.2, 27.9, 72, 748]
2. Correlation Coefficient Formula:
The Pearson linear correlation coefficient [tex]\( r \)[/tex] is calculated using the formula:
[tex]\[ r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}} \][/tex]
where:
- [tex]\( x_i \)[/tex] and [tex]\( y_i \)[/tex] are the individual sample points.
- [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex] are the means (average values) of the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] variables, respectively.
3. Compute the Means:
Calculate the mean of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \bar{x} = \frac{\sum x_i}{n} \][/tex]
[tex]\[ \bar{y} = \frac{\sum y_i}{n} \][/tex]
where [tex]\( n \)[/tex] is the number of data points, which is 9 in our case.
4. Compute the Sums:
Evaluate the terms for sum of products of deviations, sum of squared deviations for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) \][/tex]
[tex]\[ \sum (x_i - \bar{x})^2 \][/tex]
[tex]\[ \sum (y_i - \bar{y})^2 \][/tex]
5. Compute [tex]\( r \)[/tex]:
Use these sums to find the correlation coefficient [tex]\( r \)[/tex].
Following these steps analytically, we should end up calculating the linear correlation coefficient [tex]\( r \)[/tex] based on the given data.
Upon calculating using these steps, the result is:
[tex]\[ r = 0.883704508386272 \][/tex]
Therefore, the linear correlation coefficient for the given rainfall and wheat yield data is approximately [tex]\( 0.884 \)[/tex].
From the given options, none of them directly match [tex]\( 0.884 \)[/tex], which we have precisely calculated. The closest to our result from the given options would have been [tex]\( 0.899 \)[/tex], but [tex]\( 0.884 \)[/tex] is the accurate linear correlation coefficient.
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