College

Run a regression analysis on the following bivariate set of data with [tex]$y$[/tex] as the response variable.



\[

\begin{array}{|r|r|}

\hline

x & y \\

\hline

92.7 & 133.8 \\

73.2 & 76.3 \\

107.8 & 141 \\

90.6 & 48.1 \\

53.5 & 5.6 \\

97.3 & 199.3 \\

89 & 116.7 \\

111.8 & 215.4 \\

75.5 & 16.4 \\

105.6 & 193.5 \\

73 & 75.1 \\

64.9 & -3.1 \\

\hline

\end{array}

\]



1. **Find the correlation coefficient** and report it accurate to three decimal places.

\[

r = 0.882

\]



2. **What proportion of the variation in [tex]$y$[/tex] can be explained by the variation in the values of [tex]$x$[/tex]?** Report answer as a percentage accurate to one decimal place.

\[

r^2 = 77.8\%

\]



3. **Based on the data, calculate the regression line** (each value to three decimal places).

\[

\hat{y} = 4.727x + 4

\]



4. **Predict what value (on average) for the response variable will be obtained from a value of 97.5 as the explanatory variable.** What is the predicted response value? (Report answer accurate to one decimal place.)

\[

\hat{y} = \boxed{122.6}

\]

Answer :

- Calculate the means of x and y values: $\bar{x} = 86.242$, $\bar{y} = 101.508$.
- Calculate the y-intercept: $b_0 = \bar{y} - b_1 \bar{x} = -306.156$.
- Form the regression line equation: $\hat{y} = -306.156 + 4.727x$.
- Predict the value of y when x = 97.5: $\hat{y} = 154.7$.
$\boxed{154.7}$

### Explanation
1. Problem Setup
We are given a dataset and asked to perform a regression analysis. We are given the correlation coefficient $r = 0.882$, the proportion of variation in $y$ explained by $x$ as $r^2 = 77.8\%$, and the slope of the regression line as $4.727$. We need to find the y-intercept of the regression line and then predict the value of $y$ when $x = 97.5$.

2. Calculating Means
First, we need to calculate the means of the $x$ and $y$ values. Using the given data, we find:
$$\bar{x} = \frac{\sum x_i}{n} = \frac{92.7 + 73.2 + 107.8 + 90.6 + 53.5 + 97.3 + 89 + 111.8 + 75.5 + 105.6 + 73 + 64.9}{12} = 86.2416666667$$
$$\bar{y} = \frac{\sum y_i}{n} = \frac{133.8 + 76.3 + 141 + 48.1 + 5.6 + 199.3 + 116.7 + 215.4 + 16.4 + 193.5 + 75.1 - 3.1}{12} = 101.508333333$$

3. Calculating the y-intercept
Next, we calculate the y-intercept ($b_0$) of the regression line using the formula:
$$b_0 = \bar{y} - b_1 \bar{x}$$
where $b_1$ is the slope of the regression line. We are given $b_1 = 4.727$. Substituting the values we have:
$$b_0 = 101.508333333 - 4.727 \times 86.2416666667 = 101.508333333 - 407.664358333 = -306.156025$$

4. Regression Line Equation
Now we have the regression line equation:
$$\hat{y} = -306.156 + 4.727x$$

5. Predicting the value of y
We can now predict the value of $y$ when $x = 97.5$:
$$\hat{y} = -306.156 + 4.727 \times 97.5 = -306.156 + 460.8825 = 154.726475$$
Rounding to one decimal place, we get $\hat{y} = 154.7$.

6. Final Answer
Therefore, the predicted response value is $154.7$.

### Examples
Regression analysis is a powerful tool used in many fields. For example, a real estate agent might use regression analysis to predict the price of a house based on its size, location, and other features. Similarly, a marketing analyst might use regression analysis to predict sales based on advertising spending, seasonality, and other factors. Understanding regression analysis helps in making informed decisions based on data.