Posts Tagged ‘degree of freedom’

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.

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Using Statistics to Evaluate a Promotion

May 25, 2010

Marketing – as much as cashflow – is the lifeblood of any business. No matter how good your product or service may be, it’s worthless if you can’t get it in front of your customers and get them to buy it. So all businesses, large and small, must engage in marketing. And we see countless types of marketing promotions or tactics being tried: radio and TV commercials, magazine and newspaper advertisements, public relations, coupons, email blasts, and so forth. But are our promotions working? The merchant John Wannamaker, often dubbed the father of modern advertising is said to have remarked, “Half the money I spend on advertising is wasted; the trouble is I don’t know which half.”

Some basic statistics can help you evaluate the effectiveness of your marketing and take away much of the mystique Wannamaker complained about. When deciding whether to do a promotion, managers and business owners have no way of knowing whether it will succeed; in fact, in today’s economy, budgets are still tight. The cost to roll out a full promotion can wipe out an entire marketing budget if it proves to be a fiasco. This is why many businesses do a test before doing a complete rollout. The testing helps to reduce the amount of uncertainty involved in an all-out campaign.

Quite often, large companies need to choose between two or more competing campaigns for rollout. But how do they know which will be effective? Consider the example of Jenny Kaplan, owner of K-Jen, a New Orleans-style restaurant. K-Jen serves up a tasty jambalaya entrée, which is priced at $10.00. Jenny believes that the jambalaya is a draw to the restaurant and believes that by offering a discount, she can increase the average amount of the table check. Jenny decides to issue coupons via email to patrons who have opted-in to receive such promotions. She wants to knock a dollar off the price of the jambalaya as the offer, but doesn’t know whether customers would respond better to an offer worded as “$1.00 off” or as “10% off.” So, Jenny decides to test the two concepts.

Jenny goes to her database of nearly 1,000 patrons and randomly selects 200 patrons. She decides to send half of those a coupon for $1.00 off for jambalaya, and the other half a coupon for 10% off. When the coupon offer expires 10 days later, Jenny finds that 10 coupons were redeemed for each offer – a redemption rate of 10% each. Jenny observes that either wording will get the same number of people to respond. But she wonders which offer generated the largest table check. So she looks at the guest checks to which the coupons were stapled. She notices the following:

Guest Check Amounts

 

Offer

 
 

$1.00 off

10% Off

 
 

$38.85

$50.16

 
 

$36.97

$54.44

 
 

$35.94

$32.20

 
 

$54.17

$32.69

 
 

$68.18

$51.09

 
 

$49.47

$46.18

 
 

$51.39

$57.72

 
 

$32.72

$44.30

 
 

$22.59

$59.29

 
 

$24.13

$22.94

 

 

Jenny quickly computes the average for each offer. The “$1.00 off” coupon generated an average table check of $41.44; the “10% off” coupon generated an average of $45.10. At first glance, it appears that the 10% off promotion generated a higher guest check. But is that difference meaningful, or is it due to chance? Jenny needs to do further analysis.

Hypothesis Testing

How does Jenny determine if the 10% off coupon really did better than the $1.00 off coupon? She can use statistical hypothesis testing, which is a structured analytical method for comparing the difference between two groups – in this case, two promotions. Jenny starts her analysis by formulating two hypotheses: a null hypothesis, which states that there is no difference in the average check amount for either offer; and an alternative hypothesis, which states that there is, in fact, a difference in the average check amount between the two offers. The null hypothesis is often denoted as H0, and the alternative hypothesis is denoted as HA. Jenny also refers to the $1.00 off offer as Offer #1, and the 10% off offer as Offer #2. She wants to compare the means of the two offers, the means of which are denoted as μ1 and μ2, respectively. Jenny writes down her two hypotheses:

H0: The average guest check amount for the two offers is equal.

HA: The average guest check amount for the two offers is not equal.

Or, more succinctly:

H0: μ12

HA: μ1≠μ2

 

Now, Jenny is ready to go to work. Note that the symbol μ denotes the population she wants to measure. Because Jenny did her test on a portion – a sample – of her database, the averages she computed were the sample average, which is denoted as . As we stated earlier, the average table checks for the “$1.00 off” and “10% off” offers were 1=$41.44 and 2=$45.10, respectively. Jenny needs to approximate μ using . She must also compute the sample standard deviation, or s for each offer.

Computing the Sample Standard Deviation

To compute the sample standard deviation, Jenny must subtract the mean of a particular offer from each of its check amounts in the sample; square the difference; sum them up; divide by the total observations minus 1(9) and then take the square root:

$1.00 Off

Actual Table Check

Average Table Check

Difference

Difference Squared

$38.85

$41.44

-$2.59

$6.71

$36.97

$41.44

-$4.47

$19.99

$35.94

$41.44

-$5.50

$30.26

$54.17

$41.44

$12.73

$162.03

$68.18

$41.44

$26.74

$714.97

$49.47

$41.44

$8.03

$64.46

$51.39

$41.44

$9.95

$98.98

$32.72

$41.44

-$8.72

$76.06

$22.59

$41.44

-$18.85

$355.36

$24.13

$41.44

-$17.31

$299.67

   

Total

$1,828.50

   

S21=

$203.17

   

S1=

$14.25

 

10% Off

Actual Table Check

Average Table Check

Difference

Difference Squared

$50.16

$45.10

$5.06

$25.59

$54.44

$45.10

$9.34

$87.22

$32.20

$45.10

-$12.90

$166.44

$32.69

$45.10

-$12.41

$154.03

$51.09

$45.10

$5.99

$35.87

$46.18

$45.10

$1.08

$1.16

$57.72

$45.10

$12.62

$159.24

$44.30

$45.10

-$0.80

$0.64

$59.29

$45.10

$14.19

$201.33

$22.94

$45.10

-$22.16

$491.11

   

Total

$1,322.63

   

S22=

$146.96

   

S2=

$12.12

 

Notice the denotation of S2. That is known as the variance. The variance and the standard deviation are used to measure the average distance between each data point and the mean. When data are normally distributed, about 95% of all observations fall within two standard deviations from the mean (actually 1.96 standard deviations). Hence, approximately 95% of the guest checks for the $1.00 off offer should fall between $41.44 ± 1.96*($14.25) or between $13.51 and $69.37. All ten fall within this range. For the 10% off offer, about 95% will fall between $45.10 ± 1.96*($12.12), or between $21.34 and $68.86. All 10 observations also fall within this range.

Degrees of Freedom and Pooled Standard Deviation

Jenny noticed two things immediately: first, that the 10% off coupon has the higher sample average, and second each individual table check is closer to it mean than it is for the $1.00 off coupon. Also notice that when we were computing the sample standard deviation for each offer, Jenny divided by 9, and not 10. Why? Because she was making estimates of the population standard deviation. Since samples are subject to error, we must account for that. Each observation gives us information into the population’s actual values. However, Jenny had to make an estimate based on that sample, so she gives up one observation to account for the sampling error – that is, she lost a degree of freedom. In this example, Jenny has 20 total observations; since she estimated the population standard deviation for both offers, she lost two degrees of freedom, leaving her with 18 (10 + 10 – 2).

Knowing the remaining degrees of freedom, Jenny must pool the standard deviations, weighting them by their degrees of freedom. This would be especially evident if the sample sizes of the two offers were not equal. The pooled standard deviation is given by:

FYI – n is simply the sample size. Jenny then computes the pooled standard deviation:

S2p = ((9 * $203.17) + (9 * $146.96))) / (10 + 10 – 2)

= ($1,828.53 + $1,322.64)/18

= $3,151.17/18

= $175.07

Now take the square root: $13.23

Hence, the pooled standard deviation is $13.23

Computing the t-Test Statistic

Now the fun begins. Jenny knows the sample mean of the two offers; she knows the hypothesized difference between the two population means (which we would expect to be zero, if the null hypothesis said they were equal); she knows the pooled standard deviation; she knows the sample size; and she knows the degrees of freedom. Jenny must now calculate the t-Test statistic. The t-Test Statistic, or the t-value, represents the number of estimated standard errors the sample average is from that of the population. The t-value is computed as follows:

 

So Jenny sets to work computing her t-Test Statistic:

t = (($41.44 – $45.10) – (0)) / ($13.23) * SQRT(1/10 + 1/10)

= -$3.66 / ($13.23 * SQRT(1/5))

=-$3.66 / ($13.23 * .45)

=-$3.66/$5.92

= -0.62

This t-statistic gives Jenny a basis for testing her hypothesis. Jenny’s t-statistic indicates that the difference in sample table checks between the two offers is 0.62 standard errors below the hypothesized difference of zero. We now need to determine the critical t – the value that we get from a t-distribution table that is available in most statistics textbooks and online. Since we are estimating with a 95% confidence interval, and since we must account for a small sample, our critical t-value is adjusted slightly from the 1.96 standard deviations from the mean. For 18 degrees of freedom, our critical t is 2.10. The larger the sample size, the closer to 1.96 the critical t would be.

So, does Jenny Accept or Reject her Null Hypothesis (Translation: Is the “10% Off” Offer Better than the “$1.00 Off” Offer)?

Jenny now has all the information she needs to determine whether one offer worked better than the other. What does the critical t of 2.10 mean? If Jenny’s t-statistic is greater than 2.10, or (since one offer can be lower than the other), less than -2.10, then she would reject her null hypothesis, as there is sufficient evidence to suggest that the two means are not equal. Is that the case?

Jenny’s t-statistic is -0.62, which is between -2.10 and 2.10. Hence, it is within the parameters. Jenny should not reject H0, since there is not enough evidence to suggest that one offer was better than the other at generating higher table checks. In fact, there’s nothing to say that the difference between the two offers is due to anything other than chance.

What Does Jenny Do Now?

Basically, Jenny can conclude that there’s not enough evidence that the “$1.00 off” coupon was worse/better than the “10% off” coupon in generating higher table check amounts, and vice-versa. This does not mean that our hypotheses were true or false, just that there was not enough statistical evidence to say so. In this case, we did not accept the null hypothesis, but rather, failed to reject it. Jenny can do a few things:

  1. She can run another test, and see if the same phenomenon holds.
  2. Jenny can accept the fact that both offers work equally well, and compare their overall average table checks to those of who ordered jambalaya without the coupons during the time the offer ran; if the coupons generated average table checks that were higher (using the hypothesis testing procedures outlined above) than those who paid full price, then she may choose to rollout a complete promotion using either or both of the offers described above.
  3. Jenny may decide that neither coupon offer raised average check amounts and choose not to do a full rollout after all.

So Why am I Telling You This?

The purpose of this blog post was to take you step-by-step into how you can use a simple concept like t-tests to judge the performance of two promotion concepts. Although a spreadsheet like Excel can run this test in seconds, I wanted to walk you through the theory in laymen’s terms, so that you can grasp the theory, and then apply it to your business. Analysights is in the business of helping companies – large and small – succeed at marketing, and this blog post is one ingredient in the recipe for your marketing success. If you would like some assistance in setting up a promotion test or in evaluating the effectiveness of a campaign, feel free to contact us at www.analysights.com.