Questions: An engineer wants to determine how the weight of a gas-powered car, x, affects gas mileage, y. The accompanying data represent the weights of various domestic cars and ti gallon in the city for the most recent model year. Complete parts (a) through (d) below. (c) A certain gas-powered car weighs 3640 pounds and gets 17 miles per gallon. Is the miles per gallon of this car above average or below average for cars of this weight? The estimated average miles per gallon for cars of this weight is miles per gallon. The miles per gallon of this car is average for cars of this weight. (Round to three decimal places as needed.) (d) Would it be reasonable to use the least-squares regression line to predict the miles per gallon of a hybrid gas and electric car? Why or why not? A. No, because the hybrid is a different type of car. B. Yes, because the hybrid is partially powered by gas. C. Yes, because the absolute value of the correlation coefficient is greater than the critical value for a sample size of n=10. D. No, because the absolute value of the correlation coefficient is less than the critical value for a sample size of n=10.

An engineer wants to determine how the weight of a gas-powered car, x, affects gas mileage, y. The accompanying data represent the weights of various domestic cars and ti gallon in the city for the most recent model year. Complete parts (a) through (d) below. (c) A certain gas-powered car weighs 3640 pounds and gets 17 miles per gallon. Is the miles per gallon of this car above average or below average for cars of this weight? The estimated average miles per gallon for cars of this weight is miles per gallon. The miles per gallon of this car is average for cars of this weight. (Round to three decimal places as needed.) (d) Would it be reasonable to use the least-squares regression line to predict the miles per gallon of a hybrid gas and electric car? Why or why not? A. No, because the hybrid is a different type of car. B. Yes, because the hybrid is partially powered by gas. C. Yes, because the absolute value of the correlation coefficient is greater than the critical value for a sample size of n=10. D. No, because the absolute value of the correlation coefficient is less than the critical value for a sample size of n=10.
Transcript text: An engineer wants to determine how the weight of a gas-powered car, x , affects gas mileage, y . The accompanying data represent the weights of various domestic cars and ti gallon in the city for the most recent model year. Complete parts (a) through (d) below. (c) A certain gas-powered car weighs 3640 pounds and gets 17 miles per gallon. Is the miles per gallon of this car above average or below average for cars of this weight? The estimated average miles per gallon for cars of this weight is $\square$ miles per gallon. The miles per gallon of this car is $\square$ average for cars of this weight. (Round to three decimal places as needed.) (d) Would it be reasonable to use the least-squares regression line to predict the miles per gallon of a hybrid gas and electric car? Why or why not? A. No, because the hybrid is a different type of car. B. Yes, because the hybrid is partially powered by gas. C. Yes, because the absolute value of the correlation coefficient is greater than the critical value for a sample size of $\mathrm{n}=10$. D. No, because the absolute value of the correlation coefficient is less than the critical value for a sample size of $\mathrm{n}=10$.
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Solution

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Solution Steps

Step 1: Calculate Means

The means of the weights (\(x\)) and gas mileage (\(y\)) are calculated as follows:

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 3450.0 \]

\[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 18.5 \]

Step 2: Calculate Correlation Coefficient

The correlation coefficient (\(r\)) is computed to assess the strength and direction of the linear relationship between weight and gas mileage:

\[ r = -0.9636 \]

Step 3: Calculate Slope (\(\beta\)) and Intercept (\(\alpha\))

The slope (\(\beta\)) and intercept (\(\alpha\)) of the regression line are determined using the following formulas:

Numerator for \(\beta\):

\[ \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 630300 - 10 \cdot 3450.0 \cdot 18.5 = -7950.0 \]

Denominator for \(\beta\):

\[ \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 119850000 - 10 \cdot 3450.0^2 = 825000.0 \]

Thus, the slope is:

\[ \beta = \frac{-7950.0}{825000.0} = -0.0096 \]

The intercept is calculated as:

\[ \alpha = \bar{y} - \beta \bar{x} = 18.5 - (-0.0096) \cdot 3450.0 = 51.7455 \]

Step 4: Equation of the Line of Best Fit

The equation of the line of best fit is given by:

\[ y = 51.7455 - 0.0096x \]

Step 5: Interpretation of Slope and Intercept

For every pound added to the weight of the car, gas mileage in the city will decrease by \( -0.0096 \) mile(s) per gallon, on average. A weightless car will get \( 51.7455 \) miles per gallon.

Step 6: Estimate Average Mileage for a Specific Weight

For a car weighing \( 3640 \) pounds, the estimated average miles per gallon is calculated as follows:

\[ \text{Estimated mileage} = \beta \cdot 3640 + \alpha = -0.0096 \cdot 3640 + 51.7455 = 16.8015 \]

Step 7: Comparison with Actual Mileage

The actual mileage of the car is \( 17 \) miles per gallon. Comparing this with the estimated mileage:

\[ \text{Actual mileage} = 17 \quad \text{and} \quad \text{Estimated mileage} = 16.8015 \]

Since \( 17 > 16.8015 \), the miles per gallon of this car is above average for cars of this weight.

Final Answer

  • For every pound added to the weight of the car, gas mileage in the city will decrease by \( -0.0096 \) mile(s) per gallon, on average.
  • A weightless car will get \( 51.7455 \) miles per gallon.
  • The estimated average miles per gallon for a car weighing \( 3640 \) pounds is \( 16.8015 \).
  • The miles per gallon of this car is above average for cars of this weight.

\[ \boxed{\text{Slope: } -0.0096, \text{ Intercept: } 51.7455, \text{ Estimated MPG: } 16.8015, \text{ Comparison: Above average}} \]

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