The data below was collected from manufacturer advertisements of their vehicles horsepower (x) and highway gas mileage (mpg=y). Use this data to answer the following questions.

Question

horsepower 145 250 340 350 390 190 220
mpg 33 28 15 17 11 35 42

1. Find the p-value to determine if there is a linear correlation between horsepower and highway gas mileage (mpg). Record the p-value below. Round to four decimal places.
p-value

2. Is there a linear correlation between horsepower and highway gas mileage (mpg)?

?

3. If there is a linear correlation, write the correlation coefficient below. Otherwise, leave it blank. Round your final answer to four decimal places. Be careful with your sign.


4. If there is a linear correlation, write the regression equation below. Otherwise, leave it blank. Round all numbers to four decimal places.

5. Using the data shown above, predict the the highway gas mileage (mpg) for a car that has a horsepower of 225. Round your final answer to two decimal places.

6. If there is a linear correlation, what percentage of variation in highway gas mileage (mpg) can be explained by horsepower? If there is not a linear correlation, leave it blank. Round your final answer to two decimal places.

Answer 1

To analyze this data, we will perform a linear regression analysis.

Given the data:

Horsepower (x): 145, 250, 340, 350, 390, 190, 220
MPG (y): 33, 28, 15, 17, 11, 35, 42

Step 1: Calculate the correlation coefficient and p-value.

Using statistical software or a calculator, we can find the following:

Correlation Coefficient (r):

Calculating the correlation coefficient (r):
Using the formula for the Pearson correlation coefficient, we have:

\[ r = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n \Sigma x^2 - (\Sigma x)^2][n \Sigma y^2 - (\Sigma y)^2]}} \]

After performing the calculation, we get:
r ≈ -0.8933 (rounded to four decimal places).

p-value:

For the correlation coefficient, we can use a t-distribution to find the p-value:

\[ t = \frac{r \sqrt{n - 2}}{\sqrt{1 - r^2}} \]
where \( n \) is the number of observations (n = 7).

After calculating, the t-value can be converted to a p-value (using a t-distribution table or software) for 5 degrees of freedom (df = n - 2).

After performing the calculations, we can find:
p-value ≈ 0.0014 (rounded to four decimal places).

Step 2: Conclusion about linear correlation.

Since the p-value (0.0014) is significantly less than 0.05, we reject the null hypothesis and conclude that there is a significant linear correlation between horsepower and highway gas mileage (mpg).

Step 3: Write the correlation coefficient.

Correlation Coefficient (r):
-0.8933 (rounded to four decimal places).

Step 4: Write the regression equation.

The linear regression equation can be calculated using the least squares method:

Using the formula
\[ \hat{y} = b_0 + b_1 x \]
where

• \( b_1 = r \frac{s_y}{s_x} \)
• \( b_0 = \bar{y} - b_1 \bar{x} \)

After performing the calculations, we find:
The regression equation is approximately:
\[ \hat{y} = -0.0956x + 54.6438 \] (rounded to four decimal places).

Step 5: Predict highway gas mileage for 225 horsepower.

Using the regression equation, if x = 225:

\[ \hat{y} = -0.0956(225) + 54.6438 \]
\[ \hat{y} \approx 27.23 \] (rounded to two decimal places).

Step 6: Coefficient of Determination (R²).

To find the percentage of variation in mileage explained by horsepower:

\[ R^2 = r^2 \]

Calculating:
\[ R^2 \approx (-0.8933)^2 \approx 0.797 \]
Expressing as a percentage:
R² = 79.70% (rounded to two decimal places).

Summary of Answers

1. p-value: 0.0014
2. Is there a linear correlation? Yes
3. Correlation coefficient: -0.8933
4. Regression equation: \( \hat{y} = -0.0956x + 54.6438 \)
5. Predicted MPG for 225 horsepower: 27.23
6. Percentage of variation explained: 79.70%