You have found the regression line for a set of data points to be: ŷ = 30.23x + 173.52. Use the line to predict the value of y when x = 48.
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I don't remember finding that particular regression line... – Jun 18 '15 at 03:38
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The answer posted by @Quasar below is extremely misleading. – Michael Hardy Jun 18 '15 at 05:16
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Really looks like a homework problem I wouldn't want to contribute to – Cheeku Jun 18 '15 at 07:15
4 Answers
Plug $x=48$ into the original equation: $$\hat y = 30.23(48)+173.52$$ $$\hat y = 1624.56$$
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The "fitted regression line" - the one that you provided - is the one that, when given an x value, will provide (if it's a good regression!) the best estimate of y.
Thus, for any possible value of x, you can simply plug it into the regression equation and find an estimate (that's what the hat means, usually!) for y. There should be hats over 30.23 and 173.52 when they are $\alpha$ and $\beta$, or other stand-ins for your regression intercept and coefficient, too, because they're estimated parameters obtained from your ordinary least squares regression.
Edited for clarity/correction thanks to comment from @michael hardy.
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If you say $\hat y=\hat\alpha x+\hat\beta$, and find that $\hat\alpha = 30.23$, then one should not write $\widehat{30.23}$. The number $30.23$ is a based on observed data; it is a statistic. The parameter $\alpha$, on the other hand, is unobservable; it is not a statistic. ${}\qquad{}$ – Michael Hardy Jun 18 '15 at 03:52
We may predict that, for $x=48$, then $$ ŷ = 30.23 \times \color{blue}{48} + 173.52 =\color{blue}{1624.56.} $$
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Here's a quick overview of what you mean, by regressing $y$ on $x$.
Linear regression essentially tries to establish a straight-line relationship between a variable of interest $y$ and a factor $x$ (sometimes multiple factors $x_{1},x_{2},\cdots,x_{k}$) that is(are) believed to explain(cause) $y$.
Linear regression assumes that $y$ is a linear function of $x$. Any $y_{i}$ can be computed, if $x_{i}$ is known.
$y_{i}=a+bx_{i}$
As an example, you believe that the average returns on Google's stock trading on the stock exchange has a significant relationship to the average returns on the S&P 500 stock index.
Let's denote,
The expected annual returns on Google's stock by $E(R_{google})$.
The expected annual returns on the S&P500 index by $E(R_{market})$.
You would then study the historical data of Google's stock returns, say over the last 25 years and likewise the S&P 500 index returns for the past 25 years. In reality, you are collecting a sample.
Since we are performing linear regression, we assume that $E(R_{google})$ is a linear function of $E(R_{market})$.
$E(R_{google})=a+\beta\cdot{E(R_{market})}$
Using the sample data, mathematicians then use a method like Ordinary Least Squares(OLS) to estimate the coefficients $a$ and $\beta$. You would then arrive at an equation like (I just mocked it up),
$E(R_{google})=4+(1.2){E(R_{market})}$
or like the one you gave us,
$\hat{y}=30.23x+173.52$
Graphically, this is equivalent of plotting the sample data - the set of $(x_{i},y_{i})$'s with $E(R_{market})$ on the X-axis, and $E(R_{google})$ on Y-axis and passing a straight line through these points which is the best fit.
Once you've built a regression model, you could predict the dependent variable $y$ for any value of the independent variable $x$. You forecast $y$ by simply plugging in the x-value in the regression equation.
Regression is used in various disciplines to study different phenomena e.g. man-power planning a company, predicting inflation in an economy, predicting the long-run exchange rates, predicting the volatility of option prices, predicting the energy demands given the temperature and day of the year etc.
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This answer is quite incorrect. Linear regression is not about fitting straight lines; that's only one special case. The assumption that the reason it is called "linear" is that one fits a line is false. See this answer explaining why fitting polynomials by least squares is considered a form of linear regression. ${}\qquad{}$ – Michael Hardy Jun 18 '15 at 05:14
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