Questions: Question 21 of 25 Click to download the data in your preferred format. CSV Excel JMP Mac-Text Minitab PC-Text R SPSS TI CrunchIt! Make a scatterplot using year as the explanatory variable and temperature as the response variable. The least-squares regression line is temperature = -7.8 + (0.0111 × year) What would you predict for the annual average temperature for 2014 based on this line? Report your answer in degrees to two decimal places. temperature = 14.56 - Celsius The predicted temperature from the least-squares regression equation for 2014 is within 0.05 degrees of the actual temperature. The correlation between these variables is r=0.675. What percentage of the variation in temperature can be explained by the straight-line dependence on year? Round your answer to the nearest whole percent.

Question 21 of 25 Click to download the data in your preferred format. CSV Excel JMP Mac-Text Minitab PC-Text R SPSS TI CrunchIt! Make a scatterplot using year as the explanatory variable and temperature as the response variable. The least-squares regression line is temperature = -7.8 + (0.0111 × year) What would you predict for the annual average temperature for 2014 based on this line? Report your answer in degrees to two decimal places. temperature = 14.56 - Celsius The predicted temperature from the least-squares regression equation for 2014 is within 0.05 degrees of the actual temperature. The correlation between these variables is r=0.675. What percentage of the variation in temperature can be explained by the straight-line dependence on year? Round your answer to the nearest whole percent.
Transcript text: Question 21 of 25 Click to download the data in your preferred format. CSV Excel JMP Mac-Text Minitab PC-Text R SPSS TI Crunchlt! Make a scatterplot using year as the explanatory variable and temperature as the response variable. The least-squares regression line is temperature $=-7.8+(0.0111 \times$ year $)$ What would you predict for the annual average temperature for 2014 based on this line? Report your answer in degrees to two decimal places. temperature $=$ 14.56 $\square$ - Celsius The predicted temperature from the least-squares regression equation for 2014 is within 0.05 degrees of the actual temperature. The correlation between these variables is $r=0.675$. What percentage of the variation in temperature can be explained by the straight-line dependence on year? Round your answer to the nearest whole percent.
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Solution

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Solution Steps

Step 1: Predict the annual average temperature for 2014

The least-squares regression line is given by: \[ \text{temperature} = -7.8 + (0.0111 \times \text{year}) \] Substitute 2014 for the year: \[ \text{temperature} = -7.8 + (0.0111 \times 2014) \] Calculate the temperature: \[ \text{temperature} = -7.8 + 22.3254 = 14.5254 \] Rounded to two decimal places, the predicted temperature for 2014 is 14.53°C.

Step 2: Calculate the percentage of variation explained

The correlation coefficient is given as \( r = 0.675 \). The percentage of variation explained by the regression line is given by \( r^2 \times 100\% \). Calculate \( r^2 \): \[ r^2 = 0.675^2 = 0.455625 \] Convert to percentage: \[ 0.455625 \times 100\% = 45.5625\% \] Rounded to the nearest whole percent, the percentage of variation explained is 46%.

Final Answer

  1. The predicted temperature for 2014 is 14.53°C.
  2. 46% of the variation in temperature can be explained by the straight-line dependence on year.

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