Annual high temperatures in a certain location have been tracked for several years. Let [tex]$X$[/tex] represent the year and [tex]$Y$[/tex] the high temperature. Based on the data shown below, calculate the regression line (each value to two decimal places).
[tex]$y=$[/tex] x +
| x | y |
|---|---|
| 5 | 16.65 |
| 6 | 18.5 |
| 7 | 19.45 |
| 8 | 17.8 |
| 9 | 17.35 |
| 10 | 16.2 |
Answer (2)
Calculate the mean of x values: x ˉ = 7.5 .
Calculate the mean of y values: y ˉ = 17.658333... .
Calculate the slope: a = − 0.21 .
Calculate the y-intercept: b = 19.23 .
The regression line equation is: y = − 0.21 x + 19.23 .
Explanation
Understanding the Problem We are given a set of data points representing the annual high temperatures for several years. We want to find the regression line that best fits this data. The regression line has the form y = a x + b , where a is the slope and b is the y-intercept. Our goal is to calculate the values of a and b to two decimal places.
Calculating the Means First, we need to calculate the mean of the x values (years) and the mean of the y values (high temperatures). The x values are [5, 6, 7, 8, 9, 10] and the y values are [16.65, 18.5, 19.45, 17.8, 17.35, 16.2].
Calculating Mean Values The mean of the x values is calculated as: x ˉ = 6 5 + 6 + 7 + 8 + 9 + 10 = 6 45 = 7.5 The mean of the y values is calculated as: y ˉ = 6 16.65 + 18.5 + 19.45 + 17.8 + 17.35 + 16.2 = 6 105.95 = 17.658333...
Calculating the Slope Next, we calculate the slope a of the regression line using the formula: a = ∑ ( x i − x ˉ ) 2 ∑ ( x i − x ˉ ) ( y i − y ˉ ) We can calculate the numerator and denominator separately. Numerator: ( 5 − 7.5 ) ( 16.65 − 17.658333 ) + ( 6 − 7.5 ) ( 18.5 − 17.658333 ) + ( 7 − 7.5 ) ( 19.45 − 17.658333 ) + ( 8 − 7.5 ) ( 17.8 − 17.658333 ) + ( 9 − 7.5 ) ( 17.35 − 17.658333 ) + ( 10 − 7.5 ) ( 16.2 − 17.658333 ) = ( − 2.5 ) ( − 1.008333 ) + ( − 1.5 ) ( 0.841667 ) + ( − 0.5 ) ( 1.791667 ) + ( 0.5 ) ( 0.141667 ) + ( 1.5 ) ( − 0.308333 ) + ( 2.5 ) ( − 1.458333 ) = 2.520833 − 1.2625 + − 0.895833 + 0.070833 − 0.4625 − 3.645833 = − 3.675 Denominator: ( 5 − 7.5 ) 2 + ( 6 − 7.5 ) 2 + ( 7 − 7.5 ) 2 + ( 8 − 7.5 ) 2 + ( 9 − 7.5 ) 2 + ( 10 − 7.5 ) 2 = ( − 2.5 ) 2 + ( − 1.5 ) 2 + ( − 0.5 ) 2 + ( 0.5 ) 2 + ( 1.5 ) 2 + ( 2.5 ) 2 = 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5 So, a = 17.5 − 3.675 = − 0.21
Calculating the Y-Intercept Now, we calculate the y-intercept b using the formula: b = y ˉ − a x ˉ = 17.658333 − ( − 0.21 ) ( 7.5 ) = 17.658333 + 1.575 = 19.233333... Rounding to two decimal places, we get b = 19.23 .
Final Regression Line Equation Therefore, the regression line equation is: y = − 0.21 x + 19.23
Final Answer The regression line equation, rounded to two decimal places, is y = − 0.21 x + 19.23 .
Examples
Understanding the relationship between variables is crucial in many real-world scenarios. For instance, consider a farmer analyzing the relationship between rainfall and crop yield. By tracking rainfall (x) and measuring the resulting crop yield (y) over several seasons, the farmer can create a regression line. This line helps predict future yields based on expected rainfall, enabling better planning for irrigation and harvesting. Similarly, in business, regression analysis can help predict sales based on advertising expenditure, allowing for optimized marketing strategies. These models provide valuable insights for informed decision-making.
The regression line calculated from the data is y = − 0.21 x + 19.23 , using the means of x and y values to find the slope and y-intercept. The slope indicates a slight negative trend in temperatures over the years, while the y-intercept represents the estimated temperature at year 0. This model can be used for predictions based on the year.
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