## Posts Tagged ‘critical-t’

### Forecast Friday Topic: Multiple Regression Analysis

June 17, 2010

(Ninth in a series)

Quite often, when we try to forecast sales, more than one variable is often involved. Sales depends on how much advertising we do, the price of our products, the price of competitors’ products, the time of the year (if our product is seasonal), and also demographics of the buyers. And there can be many more factors. Hence, we need to measure the impact of all relevant variables that we know drive our sales or other dependent variable. That brings us to the need for multiple regression analysis. Because of its complexity, we will be spending the next several weeks discussing multiple regression analysis in easily digestible parts. Multiple regression is a highly useful technique, but is quite easy to forget if not used often.

Another thing to note, regression analysis is often used for both time series and cross-sectional analysis. Time series is what we have focused on all along. Cross-sectional analysis involves using regression to analyze variables on static data (such as predicting how much money a person will spend on a car based on income, race, age, etc.). We will use examples of both in our discussions of multiple regression.

Determining Parameter Estimates for Multiple Regression

When it comes to deriving the parameter estimates in a multiple regression, the process gets both complicated and tedious, even if you have just two independent variables. We strongly advise you to use the regression features of MS-Excel, or some statistical analysis tool like SAS, SPSS, or MINITAB. In fact, we will not work out the derivation of the parameters with the data sets, but will provide you the results. You are free to run the data we provide on your own to replicate the results we display. I do, however, want to show you the equations for computing the parameter estimates for a three-variable (two independent variables and one dependent variable), and point out something very important.

Let’s assume that sales is your dependent variable, Y, and advertising expenditures and price are your independent variables, X1 and X2, respectively. Also, the coefficients – your parameter estimates will have similar subscripts to correspond to their respective independent variable. Hence, your model will take on the form:

Now, how do you go about computing α, β1 and β2? The process is similar to that of a two-variable model, but a little more involved. Take a look:

The subscript “i” represents the individual oberservation.  In time series, the subscript can also be represented with a “t“.

What do you notice about the formulas for computing β1 and β2? First, you notice that the independent variables, X1 and X2, are included in the calculation for each coefficient. Why is this? Because when two or more independent variables are used to estimate the dependent variable, the independent variables themselves are likely to be related linearly as well. In fact, they need to be in order to perform multiple regression analysis. If either β1 or β2 turned out to be zero, then simple regression would be appropriate. However, if we omit one or more independent variables from the model that are related to those variables in the model, we run into serious problems, namely:

Specification Bias (Regression Assumptions Revisited)

Recall from last week’s Forecast Friday discussion on regression assumptions that 1) our equation must correctly specify the true regression model, namely that all relevant variables and no irrelevant variables are included in the model and 2) the independent variables must not be correlated with the error term. If either of these assumptions is violated, the parameter estimates you get will be biased. Looking at the above equations for β1 and β2, we can see that if we excluded one of the independent variables, say X2, from the model, the value derived for β1 will be incorrect because X1 has some relationship with X2. Moreover, X2‘s values are likely to be accounted for in the error terms, and because of its relationship with X1, X1 will be correlated with the error term, violating the second assumption above. Hence, you will end up with incorrect, biased estimators for your regression coefficient, β1.

Omitted Variables are Bad, but Excessive Variables Aren’t Much Better

Since omitting relevant variables can lead to biased parameter estimates, many analysts have a tendency to include any variable that might have any chance of affecting the dependent variable, Y. This is also bad. Additional variables means that you need to estimate more parameters, and that reduces your model’s degrees of freedom and the efficiency (trustworthiness) of your parameter estimates. Generally, for each variable – both dependent and independent – you are considering, you should have at least five data points. So, for a model with three independent variables, your data set should have 20 observations.

Another Important Regression Assumption

One last thing about multiple regression analysis – another assumption, which I deliberately left out of last week’s discussion, since it applies exclusively to multiple regression:

No combination of independent variables should have an exact linear relationship with one another.

OK, so what does this mean? Let’s assume you’re doing a model to forecast the effect of temperature on the speed at which ice melts. You use two independent variables: Celsius temperature and Fahrenheit temperature. What’s the problem here? There is a perfect linear relationship between these two variables. Every time you use a particular value of Fahrenheit temperature, you will get the same value of Celsius temperature. In this case, you will end up with multicollinearity, an assumption violation that results in inefficient parameter estimates. A relationship between independent variables need not be perfectly linear for multicollinearity to exist. Highly correlated variables can do the same thing. For example, independent variables such as “Husband Age” and “Wife Age,” or “Home Value” and “Home Square Footage” are examples of independent variables that are highly correlated.

You want to be sure that you do not put variables in the model that need not be there, because doing so could lead to multicollinearity.

Now Can We Get Into Multiple Regression????

Wasn’t that an ordeal? Well, now the fun can begin! I’m going to use an example from one of my old graduate school textbooks, because it’s good for several lessons in multiple regression. This data set is 25 annual observations to predict the percentage profit margin (Y) for U.S. savings and loan associations, based on changes in net revenues per deposit dollar (X1) and number of offices (X2). The data are as follows:

 Year Percentage Profit Margin (Yt) Net Revenues Per Deposit Dollar (X1t) Number of Offices (X2t) 1 0.75 3.92 7,298 2 0.71 3.61 6,855 3 0.66 3.32 6,636 4 0.61 3.07 6,506 5 0.70 3.06 6,450 6 0.72 3.11 6,402 7 0.77 3.21 6,368 8 0.74 3.26 6,340 9 0.90 3.42 6,349 10 0.82 3.42 6,352 11 0.75 3.45 6,361 12 0.77 3.58 6,369 13 0.78 3.66 6,546 14 0.84 3.78 6,672 15 0.79 3.82 6,890 16 0.70 3.97 7,115 17 0.68 4.07 7,327 18 0.72 4.25 7,546 19 0.55 4.41 7,931 20 0.63 4.49 8,097 21 0.56 4.70 8,468 22 0.41 4.58 8,717 23 0.51 4.69 8,991 24 0.47 4.71 9,179 25 0.32 4.78 9,318

Data taken from Spellman, L.J., “Entry and profitability in a rate-free savings and loan market.” Quarterly Review of Economics and Business, 18, no. 2 (1978): 87-95, Reprinted in Newbold, P. and Bos, T., Introductory Business & Economic Forecasting, 2nd Edition, Cincinnati (1994): 136-137

What is the relationship between the S&Ls’ profit margin percentage and the number of S&L offices? How about between the margin percentage and the net revenues per deposit dollar? Is the relationship positive (that is, profit margin percentage moves in the same direction as its independent variable(s))? Or negative (the dependent and independent variables move in opposite directions)? Let’s look at each independent variable’s individual relationship with the dependent variable.

Net Revenue Per Deposit Dollar (X1) and Percentage Profit Margin (Y)

Generally, if revenue per deposit dollar goes up, would we not expect the percentage profit margin to also go up? After all, if the S & L is making more revenue on the same dollar, it suggests more efficiency. Hence, we expect a positive relationship. So, in the resulting regression equation, we would expect the coefficient, β1, for net revenue per deposit dollar to have a “+” sign.

Number of S&L Offices (X2) and Percentage Profit Margin (Y)

Generally, if there are more S&L offices, would that not suggest either higher overhead, increased competition, or some combination of the two? Those would cut into profit margins. Hence, we expect a negative relationship. So, in the resulting regression equation, we would expect the coefficient, β2, for number of S&L offices to have a “-” sign.

Are our Expectations Correct?

Do our relationship expectations hold up?  They certainly do. The estimated multiple regression model is:

Yt = 1.56450 + 0.23720X1t – 0.000249X2t

What do the Parameter Estimates Mean?

Essentially, the model says that if net revenues per deposit dollar (X1t) increase by one unit, then percentage profit margin (Yt) will – on average – increase by 0.23720 percentage points, when the number of S&L offices is fixed. If the number of offices (X2t) increases by one, then percentage profit margin (Yt) will decrease by an average of 0.000249 percentage points, when net revenues are fixed.

Do Changes in the Independent Variables Explain Changes in The Dependent Variable?

We compute the coefficient of determination, R2, and get 0.865, indicating that changes in the number of S&L offices and in the net revenue per deposit dollar explain 86.5% of the variation in S&L percentage profit margin.

Are the Parameter Estimates Statistically Significant?

We have 25 observations, and three parameters – two coefficients for the independent variables, and one intercept – hence we have 22 degrees of freedom (25-3). If we choose a 95% confidence interval, we are saying that if we resampled and replicated this analysis 100 times, the average of our parameter estimates will be contain the true parameter approximately 95 times. To do this, we need to look at the t-values for each parameter estimate. For a two-tailed 95% significance test with 22 degrees of freedom, our critical t-value is 2.074. That means that if the t-statistic for a parameter estimate is greater than 2.074, then there is a strong positive relationship between the independent variable and the dependent variable; if the t-statistic for the parameter estimate is less than -2.074, then there is a strong negative relationship. This is what we get:

 Parameter Value T-Statistic Significant? Intercept 1.5645000 19.70 Yes B1t 0.2372000 4.27 Yes B2t (0.0002490) (7.77) Yes

So, yes, all our parameter estimates are significant.

Next Forecast Friday: Building on What You Learned

I think you’ve had enough for this week! But we are still not finished. We’re going to stop here and continue with further analysis of this example next week. Next week, we will discuss computing the 95% confidence interval for the parameter estimates; determining whether the model is valid; and checking for autocorrelation. The following Forecast Friday (July 1) blog post will discuss specification bias in greater detail, demonstrating the impact of omitting a key independent variable from the model.

### Forecast Friday Topic: Simple Regression Analysis (Continued)

June 3, 2010

(Seventh in a series)

Last week I introduced the concept of simple linear regression and how it could be used in forecasting. I introduced the fictional businesswoman, Sue Stone, who runs her own CPA firm. Using the last 12 months of her firm’s sales, I walked you through the regression modeling process: determining the independent and dependent variables, estimating the parameter estimates, α and β, deriving the regression equation, calculating the residuals for each observation, and using those residuals to estimate the coefficient of determination – R2 – which indicates how much of the change in the dependent variable is explained by changes in the independent variable. Then I deliberately skipped a couple of steps to get straight to using the regression equation for forecasting. Today, I am going to fill in that gap, and then talk about a couple of other things so that we can move on to next week’s topic on multiple regression.

Revisiting Sue Stone

Last week, we helped Sue Stone develop a model using simple regression analysis, so that she could forecast sales. She had 12 months of sales data, which was her dependent variable, or Y, and each month (numbered from 1 to 12), was her independent variable, or X. Sue’s regression equation was as follows:

Where i is the period number corresponding to the month. So, in June 2009, i would be equal to 6; in January 2010, i would be equal to 13. Of course, since X is the month number, X=i in this example. Recall that Sue’s equation states that each passing month is associated with an average sales increase of \$479.02, suggesting her sales are on an upward trend. Also note that Sue’s R2=.917, which says 91.7% of the change in Sue’s monthly sales is explained by changes in the passing months.

Are these claims valid? We need to do some further work here.

Are the Parameter Estimates Statistically Significant?

Measuring an entire population is often impossible. Quite often, we must measure a sample of the population and generalize our findings to the population. When we take an average or standard deviation of a data set that is a subset of the population, our values are estimates of the actual parameters for the population’s true average and standard deviation. These are subject to sampling error. Likewise, when we perform regression analysis on a sample of the population, our coefficients (a and b) are also subject to sampling error. Whenever we estimate population parameters (the population’s true α and β), we are frequently concerned that they might actually have values of zero. Even though we have derived values a=\$9636.36 and b=\$479.02, we want to perform a statistical significance test to make sure their distance from zero is meaningful and not due to sampling error.

Recall from the May 25 blog post, Using Statistics to Evaluate a Promotion, that in order to do significance testing, we must set up a hypothesis test. In this case, our null hypothesis is that the true population coefficient for month – β – is equal to zero. Our alternative hypothesis is that β is not equal to zero:

H0: β = 0

HA: β≠ 0

Our first step here is to compute the standard error of the estimate, that is, how spread out each value of the dependent variable (sales) is from the average value of sales. Since we are sampling from a population, we are looking for the estimator for the standard error of the estimate. That equation is:

Where ESS is the error sum of squares – or \$2,937,062.94 – from Sue’s equation; n is the sample size, or 12; k is the number of independent variables in the model, in this case, just 1. When we plug those numbers into the above equation, we’re dividing the ESS by 10 and then taking the square root, so Sue’s estimator is:

sε = \$541.95

Now that we know the estimator for the standard error of the estimate, we need to use that to find the estimator for the standard deviation of the regression slope (b). That equation is given by:

Remember from last week’s blog post that the sum of all the (x-xbar) squared values was 143. Since we have the estimator for the standard error of the estimate, we divide \$541.95 by the square root of 143 to get an Sb = 45.32. Next we need to compute the t-statistic. If Sue’s t-statistic is greater than her critical t-value, then she’ll know the parameter estimate of \$479.02 is significant. In Sue’s regression, she has 12 observations, and thus 10 degrees of freedom: (n-k-1) = (12-1-1) = 10. Assuming a 95% confidence interval, her critical t is 2.228. Since parameter estimates can be positive or negative, if her t value is less than -2.228 or greater than 2.228, Sue can reject her null hypothesis and conclude that her parameter estimates is meaningfully different from zero.

To compute the t-statistic, all Sue needs to do is divide her b1 coefficient (\$479.02) by her sb (\$45.32). She ends up with a t-statistic of 10.57, which is significant.

Next Sue must do the same for her intercept value, a. To do this, Sue, must compute the estimator of the standard deviation of the intercept (a). The equation for this estimate is:

All she needs to do is plug in her numbers from earlier: her sε = \$541.95; n=12; she just takes her average x-bar of 6.5 and squares it, bringing it to 42.25; and the denominator is the same 143. Working that all in, Sue gets a standard error of 333.545. She divides her intercept value of \$9636.36 by 333.545 and gets a t-statistic of 28.891, which exceeds the 2.228 critical t, so her intercept is also significant.

Prediction Intervals in Forecasting

Whew! Aren’t you glad those t-statistics calculations are over? If you run regressions in Excel, these values will be calculated for you automatically, but it’s very important that you understand how they were derived and the theory behind them. Now, we move back to forecasting. In last week’s post, we predicted just a single point with the regression equation. For January 2010, we substituted the number 13 for X, and got a point forecast for sales in that month: \$15,863.64. But Sue needs a range, because she knows forecasts are not precise. Sue wants to develop a prediction interval. A prediction interval is simply the point forecast plus or minus the critical t value (2.228) for a desired level of confidence (95%, in this example) times the estimator of the standard error of the estimate (\$541.95). So, Sue’s prediction interval is:

\$15,863.64 ± 2.228(\$541.95)

= \$15,863.64 ± \$1,207.46

\$14,656.18_____\$17,071.10

So, since Sue had chosen a 95% level of confidence, she can be 95% confident that January 2010 sales will fall somewhere between \$14,656.18 and \$17,071.10

Recap and Plan for Next Week’s Post

Today, you learned how to test the parameter estimates for significance to determine the validity of your regression model. You also learned how to compute the estimates of the standard error of the estimates, as well as the estimators of the standard deviations of the slope and intercept. You then learned how to derive the t-statistics you need to determine whether those parameter estimates were indeed significant. And finally, you learned how to derive a prediction interval. Next week, we begin our discussion of multiple regression. We will begin by talking about the assumptions behind a regression model; then we will talk about adding a second independent variable into the model. From there, we will test the model for validity, assess the model against those assumptions, and generate projections.