SOLUTION: Run a regression analysis on the following bivariate set of data with y as the response variable. x y 48.7 22.9 47.8 19.7 45.8 37.1 48.8 25.3 43.6 43.8 45.7 37.4 50.1 17.8

Question 1167202: Run a regression analysis on the following bivariate set of data with y as the response variable.
x y
48.7 22.9
47.8 19.7
45.8 37.1
48.8 25.3
43.6 43.8
45.7 37.4
50.1 17.8
45.5 31.5
46.1 36
44.5 41.7
44.8 30.8
43.1 40.3
Find the correlation coefficient and report it accurate to three decimal places.
r = -.899

What proportion of the variation in y can be explained by the variation in the values of x? Report answer as a percentage accurate to one decimal place. (If the answer is 0.84471, then it would be 84.5%...you would enter 84.5 without the percent symbol.)
r² = 80.8%
Based on the data, calculate the regression line (each value to three decimal places)
y = -3.615 x + 199.047

Predict what value (on average) for the response variable will be obtained from a value of 49.3 as the explanatory variable. Use a significance level of α = 0.05 to assess the strength of the linear correlation.
What is the predicted response value? (Report answer accurate to one decimal place.)
y =

Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import numpy as np
from scipy.stats import linregress
# Data
x = np.array([48.7, 47.8, 45.8, 48.8, 43.6, 45.7, 50.1, 45.5, 46.1, 44.5, 44.8, 43.1])
y = np.array([22.9, 19.7, 37.1, 25.3, 43.8, 37.4, 17.8, 31.5, 36.0, 41.7, 30.8, 40.3])
# Run linear regression
slope, intercept, r_value, p_value, std_err = linregress(x, y)
# 1. Correlation coefficient (r)
r = r_value
# 2. Coefficient of determination (r^2)
r_squared = r_value**2
r_squared_percent = r_squared * 100
# 3. Regression line: y = intercept + slope * x
# The slope is 'b' and the intercept is 'a' in the standard form y = a + bx
# The user's form is y = b x + a, so:
b = slope
a = intercept
# 4. Predicted response value (y_hat) for x = 49.3
x_predict = 49.3
y_predicted = intercept + slope * x_predict
# Print results with required formatting
print(f"Correlation coefficient (r): {r:.3f}")
print(f"Coefficient of determination (r^2, percent): {r_squared_percent:.1f}")
print(f"Regression slope (b): {b:.3f}")
print(f"Regression intercept (a): {a:.3f}")
print(f"Predicted y for x=49.3: {y_predicted:.1f}")
print(f"P-value for correlation significance (used for assessing strength): {p_value:.4f}")
print(f"Significance Level (alpha): 0.05")
```
```text?code_stdout&code_event_index=2
Correlation coefficient (r): -0.899
Coefficient of determination (r^2, percent): 80.8
Regression slope (b): -3.615
Regression intercept (a): 199.047
Predicted y for x=49.3: 20.9
P-value for correlation significance (used for assessing strength): 0.0001
Significance Level (alpha): 0.05
```
That's an excellent idea to double-check the calculations\! Running the regression analysis confirms your intermediate results and provides the final prediction.
Here are the confirmed results based on the data:
* **Correlation coefficient ($r$):**
$$r = -0.899$$
* **Proportion of variation in $y$ explained by $x$ ($r^2$):**
$$r^2 = (-0.899)^2 \approx 0.8082$$
As a percentage, this is **$80.8\%$**.
* **Regression line:**
$$\hat{y} = 199.047 - 3.615x$$
$$\hat{y} = -3.615x + 199.047$$
-----
## Predicted Response Value and Correlation Strength
The final step is to predict the response value and assess the correlation strength.
### Predicted Response Value
To predict the response variable ($\hat{y}$) for an explanatory variable value of $x = 49.3$, we substitute $49.3$ into the regression equation:
$$\hat{y} = -3.615(49.3) + 199.047$$
$$\hat{y} \approx -178.29495 + 199.047$$
$$\hat{y} \approx 20.75205$$
Rounding to one decimal place:
$$\hat{y} = \mathbf{20.8}$$
### Assessing Correlation Strength ($\alpha=0.05$)
To assess the strength of the linear correlation, we compare the **p-value** from the regression analysis to the significance level ($\alpha$).
* **P-value** $\approx 0.0001$
* **Significance Level** ($\alpha$) $= 0.05$
Since the p-value ($0.0001$) is much smaller than $\alpha$ ($0.05$), we reject the null hypothesis (that there is no linear correlation). Therefore, the linear correlation between $x$ and $y$ is **statistically significant** and is confirmed to be **strong**.
What is the predicted response value? (Report answer accurate to one decimal place.)
y = **20.8**
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