College

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



\[

\begin{tabular}{|r|r|}

\hline \multicolumn{1}{|c|}{$x$} & \multicolumn{1}{c|}{$y$} \\

\hline 92.7 & 133.8 \\

\hline 73.2 & 76.3 \\

\hline 107.8 & 141 \\

\hline 90.6 & 48.1 \\

\hline 53.5 & 5.6 \\

\hline 97.3 & 199.3 \\

\hline 89 & 116.7 \\

\hline 111.8 & 215.4 \\

\hline 75.5 & 16.4 \\

\hline 105.6 & 193.5 \\

\hline 73 & 75.1 \\

\hline 64.9 & -3.1 \\

\hline

\end{tabular}

\]



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

- [tex]r = \square[/tex]



2. **What proportion of the variation in [tex]$y$[/tex] can be explained by the variation in the values of [tex]$x$[/tex]?** Report the 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.)

- [tex]r^2 = \square \%[/tex]



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

- [tex]\hat{y} = \square x + \square[/tex]



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 the answer accurate to one decimal place.)

- [tex]\hat{y} = \square[/tex]

Answer :

- Calculate the correlation coefficient: $r = 0.878$.
- Calculate the coefficient of determination: $r^2 = 77.1\%$.
- Determine the regression line: $\hat{y} = 3.706x - 218.114$.
- Predict the value of y for x = 97.5: $\hat{y} = 143.2$.

### Explanation
1. Understanding the Problem
We are given a bivariate data set and asked to perform a regression analysis. This involves finding the correlation coefficient, the proportion of variation in $y$ explained by $x$ (coefficient of determination), the regression line equation, and a prediction for $y$ given a specific value of $x$.

2. Calculating Correlation Coefficient
First, we need to calculate the correlation coefficient $r$. The formula for $r$ is $$r = \frac{\sum (x_i - x_{bar})(y_i - y_{bar})}{(n-1)s_x s_y}$$, where $x_{bar}$ and $y_{bar}$ are the means of $x$ and $y$ respectively, $s_x$ and $s_y$ are the standard deviations of $x$ and $y$ respectively, and $n$ is the number of data points. After performing the calculations, the correlation coefficient is found to be $r = 0.878$.

3. Calculating Coefficient of Determination
Next, we calculate the coefficient of determination $r^2$, which represents the proportion of the variance in the dependent variable that is predictable from the independent variable. $$r^2 = (0.878)^2 = 0.771$$. Converting this to a percentage, we get $77.1\%$.

4. Determining the Regression Line
Now, we determine the regression line equation, which has the form $\hat{y} = bx + a$, where $b$ is the slope and $a$ is the y-intercept. The slope $b$ is calculated as $$b = r \frac{s_y}{s_x}$$. The y-intercept $a$ is calculated as $$a = y_{bar} - b x_{bar}$$. After performing the calculations, we find that $b = 3.706$ and $a = -218.114$. Therefore, the regression line equation is $\hat{y} = 3.706x - 218.114$.

5. Predicting the Value of y
Finally, we predict the value of $y$ for $x = 97.5$ using the regression line equation: $$\hat{y} = 3.706(97.5) - 218.114 = 143.2$$.

6. Final Answer
The correlation coefficient is $r = 0.878$, the proportion of variation in $y$ explained by $x$ is $77.1\%$, the regression line equation is $\hat{y} = 3.706x - 218.114$, and the predicted value of $y$ for $x = 97.5$ is $143.2$.

### Examples
Regression analysis is used in various fields, such as finance, to predict stock prices based on historical data. For example, if you have data on a company's revenue and stock price over several years, you can use regression analysis to create a model that predicts the stock price based on the revenue. This can help investors make informed decisions about buying or selling stocks. Similarly, in marketing, regression analysis can be used to predict sales based on advertising expenditure.