The basic approach is to use the following regression model, employing the notation from Definition 3 of Method of Least Squares for Multiple Regression:

where the odds function is as given in the following definition.

**Definition 1**: *Odds*(*E*) is the **odds** that event *E* occurs, namely

Where *p* has a value 0 ≤ *p* ≤ 1 (i.e. *p* is a probability value), we can define the odds function as

**Observation**: For our purposes, the odds function has the advantage of transforming the probability function, which has values from 0 to 1, into an equivalent function with values between 0 and ∞. When we take the natural log of the odds function, we get a range of values from -∞ to ∞*.*

**Definition 2**: The **logit** function is the log of the odds function, namely logit(*E*) = ln *Odds*(*E*), or

**Definition 3**: Based on the logistic model as described above, we have

where *π* = *P*(E). It now follows that (see Exponentials and Logs):

and so

Here we switch to the model based on the observed sample (and so the *π* parameter is replaced by its sample estimate *p*, the *β _{j}* coefficients are replaced by the sample estimates

*b*and the error term

_{j}*ε*is dropped). For our purposes we take the event

*E*to be that the dependent variable y has value 1. If y takes only the values 0 or 1, we can think of

*E*as success and the complement

*E′*of

*E*as failure. This is as for the trials in a binomial distribution.

Just as for the regression model studied in Regression and Multiple Regression, a sample consists of *n* data elements of the form (y* _{i}*,

*x*

_{i1}, x*,…, x*), but for logistic regression each y

_{ik}*only takes the value 0 or 1. Now let*

_{i}*E*= the event that y

_{i}*= 1 and*

_{i}*p*=

_{i}*P*(

*E*). Just as the regression line studied previously provides a way to predict the value of the dependent variable y from the values of the independent variables

_{i}*x*

_{1}

*, …, x*in for logistic regression we have

_{k}Note too that since the y* _{i}* have a proportion distribution, by Property 2 of Proportion Distribution,

*var*(y

*) =*

_{i}*p*(1 –

_{i}*p*).

_{i}**Observation**: In the case where *k* = 1, we have

Such a curve has sigmoid shape:

**Figure 1 – Sigmoid curve for p**

The values of *b*_{0}* *and *b*_{1} determine the location direction and spread of the curve. The curve is symmetric about the point where *x = -b*_{0 }*/ b*_{1}*.* In fact, the value of *p* is 0.5 for this value of *x*.

**Observation**: Logistic regression is used instead of ordinary multiple regression because the assumptions required for ordinary regression are not met. In particular

- The assumption of the linear regression model that the values of y
- The assumption of the linear regression model that the variance of y is constant across values of
*x*(homogeneity of variances) also cannot be met with a binary variable. Since the variance is*p*(1*–p*)*p*= .10 or .90, the variance is (.1)(.9) = .09, and so as*p*approaches 1 or 0, the variance approaches 0. - Using the linear regression model, the predicted values will become greater than one and less than zero if you move far enough on the x-axis. Such values are theoretically inadmissible for probabilities.

For the logistics model, the least squares approach to calculating the values of the coefficients *b _{i}* cannot be used; instead the maximum likelihood techniques, as described below, are employed to find these values.

**Definition 4**: The **odds ratio** between two data elements in the sample is defined as follows:

Using the notation *p _{x}* =

*P*(

*x*), the

**log odds ratio**of the estimates is defined as

**Observation**: In the case where

Furthermore, for any value of *d*

Note too that when *x* is a dichotomous variable,

E.g. when *x* = 0 for male and *x* = 1 for female, then represents the odds ratio between males and females. If for example *b*_{1} = 2, and we are measuring the probability of getting cancer under certain conditions, then = 7.4, which would mean that the odds of females getting cancer would be 7.4 times greater than males under the same conditions.

**Observation**: The model we will use is based on the binomial distribution, namely the probability that the sample data occurs as it does is given by

Taking the natural log of both sides and simplifying we get the following definition.

**Definition 5**: The **log-likelihood** statistic is defined as follows:

where the y* _{i}* are the observed values while the

*p*are the corresponding theoretical values.

_{i}**Observation**: Our objective is to find the maximum value of *LL* assuming that the *p _{i}* are as in Definition 3. This will enable us to find the values of the

*b*coordinates. It might be helpful to review Maximum Likelihood Function to better understand the rest of this topic.

_{i}**Example 1**: A sample of 760 people who received doses of radiation between 0 and 1000 rems was made following a recent nuclear accident. Of these 302 died as shown in the table in Figure 2. Actually each row in the table represents the midpoint of an interval of 100 rems (i.e. 0-100, 100-200, etc.).

**Figure 2 – Data for Example 1 plus probability and odds**

Let *E _{i}* = the event that a person in the

*i*th interval survived. The table also shows the probability

*P*(

*E*) and odds

_{i}*Odds*(

*E*) of survival for a person in each interval. Note that

_{i}*P*(

*E*

_{i}) = the percentage of people in interval

*i*who survived and

In Figure 3 we plot the values of *P*(*E _{i}*) vs.

*i*and ln

*Odds*(

*E*) vs.

_{i}*i*. We see that the second of these plots is reasonably linear.

**Figure 3 – Plot of probability and ln odds**

Given that there is only one independent variable (namely *x* = # of rems), we can use the following model

Here we use coefficients *a* and *b* instead of *b*_{0} and *b*_{1} just to keep the notation simple.

We show two different methods for finding the values of the coefficients *a* and *b*. The first uses Excel’s Solver tool and the second uses Newton’s method. Before proceeding it might be worthwhile to click on Goal Seeking and Solver to review how to use Excel’s Solver tool and Newton’s Method to review how to apply Newton’s Method. We will use both methods to maximize the value of the log-likelihood statistic as defined in Definition 5.

**Sample Size**: The recommended minimum sample size for logistic regression is given by 10*k*/*q* where *k* = the number of independent variables and *q* = the smaller of the percentage of cases with y = 0 or y = 1, with a minimum of 100.

For Example 1, *k* = 1 and *q* = 302/760 = .397, and so 10*k*/*q* = 25.17. Thus a minimum sample of size 100 is recommended.

Charles,

Thanks for the amazing tool! I’ve tried your logit function on one of my data sets and found perfect agreement with other methods of calculating logit models. Very much appreciated!

I have a question on the applicability for another data set, in which the I have thousands of repeats on non-independent data. For example, we tested n participants on a test with a binary outcome with a number of independent variables (including repeats) in a fully repeated measures design. I’ve converted the binary outcome results into probabilities and done a multiple regression on the probabilities; however, I’m not satisfied with the resultant curve as for certain levels of the independent variables the regression model probability exceeds 1. Is it correct/justifiable to apply a logistic regression approach?

Andrew,

Sorry, but I don’t have a clear picture of your design.

Charles

Sorry that I wasn’t clear. I have data from a laboratory test with human subjects with several independent variables using a fully repeated measures design and a binary dependent variable. I’m unsure if it would be correct to use a logistic regression in this situation. Thanks

Andrew,

Sorry, but I don’t know what you mean by a fully repeated design.

Charles

Repeated measures in that all participants are tested under all combinations of independent variables, such that all manipulations are within subjects factors.

Andrew,

Based on what I understand about your situation, it seems like binary logistic regression could be a good choice.

In a previous comment, you said that the “levels of the independent variables the regression model probability exceeds 1”. This should not happen in logistic regression.

Charles

I was curious if you could provide some advice. Long forgotten are my ‘out-of-context’ statistics courses and having stumbled about your blog I hope make a comeback!

I’m looking to analyze the outcome of lots of simulations. At a high level I am comparing success vs failure of the simulation based on several different parameters. I want to get a report on the best parameter set to most likely have successful runs in the future.

In other words: y = f(x1,x2,x3,…,xn) where y is 0 or 1 and the x values are categorical (i.e., they only take on specific values though not the same number of options).

It seems logistic regression you described could be what I need but I could really use your 2cents on if I’m barking up the wrong tree. Thank you in advance!

Scott,

Although you haven’t provided me with enough information to say for sure, it does sound like logistic regression could be a suitable approach.

Charles

I do appreciate your feedback.

Thanks for the extraordinarily useful page. I’ve downloaded the examples-part 2 and am looking at the multinomial logit example 1. I apologize in advance for asking such a basic question, but if I look at the Odds calculated for cured for Dosage 20, Gender 0, we see 0 observed cured.

It isn’t clear to me how the Odds for cured could be anything other than zero.

As I say, it is a very basic question, thanks for an excellent site!

Richard, when you say “if I look at the Odds calculated for cured for Dosage 20, Gender 0, we see 0 observed cured”, are you referring to some worksheet on the website or to some other that you are working on? In general if the number of successes is 0, the the odds will be 0.

Charles

Hi Charles,

Say we are doing a binary logistic regression of y with respect to each of the separate variables x_1,x_2, x_3, i.e., we are logistically regressing x_1|y , x_2|y and x_3|y ; x_i is the independent variable in each case. Assume all pairs (b_i, b_j) have the same sign combination, i.e., b_i positive, b_j negative, etc.

Question: Is there a meaningful way of defining a new variable x from x_1,x_2, x_3 , say x=x_1+x_2+x_3 , so that we can regress x|y ? Specifically, here the x_i are measures of efficiency and y is a measure of control, so I want to measure the overall effect of control on efficiency.

Thanks in Advance.

Does the new variable have to be a linear combination of the independent variables?

Charles

I would prefer that it was, but it is not necessary. Could you please tell me for both cases, or please give me refs?

Thanks Again.

You can define a new variable p which is calculated as the formula for p on the following webpage

http://www.real-statistics.com/logistic-regression/basic-concepts-logistic-regression/

There is no linear version. The logit function shown on that webpage is linear.

Charles

Thanks a lot, Charles, very helpful.

I see, so please let me know if I am on the right track:

If we have Y regressed against, say, 1/3(X_1+X_2+X_3) , and we have our cutoff , say X_1>A, X_2>A, X_3>A (same for all ), then we want to model

P(E)/(1-P(E))

For E= Event that P(X_1=1 and X_2=1 and X_3=1 )? And here E is the event where each of X_1, X_2, X_3 are above their respective cutoff values, say x_1, x_2, x_3 respectively? Do we then need to know the joint distribution ? I think in this case it is safe to assume independence of the X_i.

Thanks Again.

Never mind, sorry, I just realized my mistake. Please feel free to delete my previous post.

When we deal with a time series data for x(i)s, among which collinearity is highly likely present, what model can I use to draw inference about the probability between x(i) and a binary dependent variable y(i) ?

Mich,

Sorry, but I don’t understand your question.

Charles

My appology. I used wrong term. The subject of my analysis is a pair of time series [ x(t), y(t) ], where y(i) is a binominal variable taking either value of 0 or 1. Does the presence of “serial correlation” and/or even “heteroskedascity” among x(t) breach any assumption of the model? If that is the case, do we have any resolution to carry out the logistic regression? Hope that this question makes sense. Sorry again.

Mich,

See page 13 and 14 of

http://people.stern.nyu.edu/jsimonof/classes/2301/pdf/logistic.pdf

Charles

Dear Charles

Thank you very much for the info. I really appreciate it.

Michio

Hi, Charles

Similar to the discussion on multiple linear regression, is there a discussion addressing outliers and influential observations in undertaking multiple logistic regression? Do you have a recommended suite of Real Statistics features that one should use?

Regards,

Rich

Rich,

I have not yet included this on the website or in the software, but I really should. I will look into this for a future release. In the meantime, here is an article that may be helpful:

http://scialert.net/fulltext/?doi=jas.2011.26.35&org=11

Charles

Hi, Charles

If you incorporate a discussion/tool for outlier detection in logistic regressions, perhaps you could include something on power as well?

Regards,

Rich

Rich,

I plan to add information about power and sample size for logistic regression. Just haven’t gotten to it yet.

Charles

Pingback: Examples of logistic regression and multinomial logistic regression - MathHub

Dear Charles,

I have followed your method using my data set, which has a sample size of ~500 and ~150 unique x-values, however only ~30 of the population died. This results in many infinity and negative infinity values for the ln Odds(E) graph. Is it still possible to create this graph and obtain the coefficients for the line of best fit?

My end goal is to create an ROC curve to find out how good the x-values are at predicting the outcome (lived/died).

Thanks a lot,

Ally

Ally,

I suggest that you create an example that adheres to the situation that you describe and try logistic regression to see what happens. My guess is that there is a good chance that it will still be possible to create the ROC curve that you are looking for.

Charles

Mr Charles,

The information you have provided here is truly helpful for someone like me who is studying statistics. Would you mind elaborating little bit how do we get the likelihood statistic as a binomial distribution?Thank you

Sorry, but I don’t understand your question. Are you looking for the likelihood statistic for the binomial distribution?

Charles

Apologies for not being clear with my question.I’m asking about the likelihood function which is used for evaluating the model.The formula which you have used to compare between the predicted probability and the actual outcome and which is converted to log likelihood statistic which is being maximized.The formula mentioned above definition 5 on this page.

Chirag,

I suggest that you look at the following webpage

http://www.real-statistics.com/general-properties-of-distributions/maximum-likelihood-function/

The idea is that the probability that the sample will occur as it does is a product of probabilities. This is because the sample elements are independent of each other; i.e. P(A and B) = P(A) x P(B) when A and B are independent events. Thus the likelihood function L is a product of probability function values (that are dependent on certain parameters). For logistic regression, the probability function is the pdf for the binary distribution.

Since the sample that was observed actually did occur, the approach we use is to find the values of the parameters that maximize L(i.e. that make the sample events most likely)

Charles

Dear Charles,

Greetings. As a novice, I must say I’ve found your website immensely helpful.

I have the following data which I need to build a regression model for. Kindly suggest how I might go about it…

Dependent Variable: Binary Categorical

Independent Variables: Many are categorical and many are interval

I need to see if it is possible to build a model and test it. Kindly advise.

Rahul,

Logistic regression is a possible choice. See the webpages on Logistic Regression.

Charles

I am so sorry for the last respond, it might be confusing because I forgot to put ‘0’

below is the data I am working on:

age leave stay total

18.6 0 2 2

18.8 2 0 2

18.9 1 0 1

19 1 0 1

19.1 0 1 1

19.2 2 1 3

19.3 0 1 1

19.4 0 1 1

19.5 1 0 1

19.6 0 2 2

19.8 4 2 6

19.9 1 1 2

20 2 2 4

20.1 3 4 7

20.2 1 3 4

20.3 4 1 5

20.4 5 9 14

20.5 3 4 7

20.6 3 7 10

20.7 4 6 10

20.8 4 8 12

20.9 5 16 21

21 7 8 15

21.1 5 10 15

21.2 5 13 18

21.3 7 16 23

21.4 6 17 23

21.5 14 13 27

21.6 12 13 25

21.7 12 17 29

21.8 17 9 26

21.9 7 23 30

22 20 32 52

22.1 19 32 51

22.2 21 49 70

22.3 23 55 78

22.4 25 36 61

22.5 27 46 73

22.6 22 54 76

22.7 32 31 63

22.8 32 50 82

22.9 50 39 89

23 39 65 104

23.1 40 81 121

23.2 42 89 131

23.3 44 80 124

23.4 71 66 137

23.5 50 84 134

23.6 64 94 158

23.7 57 95 152

23.8 59 108 167

23.9 77 88 165

24 61 105 166

24.1 66 118 184

24.2 70 131 201

24.3 73 114 187

24.4 66 108 174

24.5 64 137 201

24.6 62 159 221

24.7 66 135 201

24.8 86 130 216

24.9 96 126 222

25 82 152 234

25.1 75 135 210

25.2 88 143 231

25.3 72 149 221

25.4 92 156 248

25.5 89 171 260

25.6 95 173 268

25.7 80 172 252

25.8 103 150 253

25.9 94 183 277

26 92 156 248

26.1 95 185 280

26.2 97 196 293

26.3 86 179 265

26.4 95 208 303

26.5 72 187 259

26.6 94 169 263

26.7 89 195 284

26.8 84 167 251

26.9 86 157 243

27 83 181 264

27.1 90 177 267

27.2 75 187 262

27.3 76 191 267

27.4 75 206 281

27.5 72 197 269

27.6 66 207 273

27.7 84 150 234

27.8 77 174 251

27.9 59 152 211

28 67 166 233

28.1 64 164 228

28.2 64 174 238

28.3 65 165 230

28.4 82 200 282

28.5 80 195 275

28.6 78 179 257

28.7 83 155 238

28.8 89 148 237

28.9 74 150 224

29 68 153 221

29.1 69 125 194

29.2 71 180 251

29.3 67 156 223

29.4 55 187 242

29.5 75 164 239

29.6 63 151 214

29.7 58 164 222

29.8 58 143 201

29.9 64 146 210

30 62 111 173

30.1 55 159 214

30.2 66 160 226

30.3 54 147 201

30.4 49 149 198

30.5 58 141 199

30.6 50 163 213

30.7 50 179 229

30.8 51 143 194

30.9 52 129 181

31 56 136 192

31.1 46 138 184

31.2 40 134 174

31.3 65 147 212

31.4 49 145 194

31.5 58 125 183

31.6 34 184 218

31.7 51 136 187

31.8 43 125 168

31.9 47 116 163

32 43 129 172

32.1 43 115 158

32.2 52 142 194

32.3 41 114 155

32.4 42 132 174

32.5 61 137 198

32.6 45 157 202

32.7 50 127 177

32.8 43 117 160

32.9 50 114 164

33 49 120 169

33.1 32 129 161

33.2 35 117 152

33.3 51 134 185

33.4 41 131 172

33.5 49 131 180

33.6 40 154 194

33.7 39 101 140

33.8 36 135 171

33.9 45 108 153

34 42 92 134

34.1 41 115 156

34.2 48 114 162

34.3 40 117 157

34.4 38 132 170

34.5 40 114 154

34.6 25 134 159

34.7 27 110 137

34.8 52 96 148

34.9 35 106 141

35 38 89 127

35.1 23 103 126

35.2 38 90 128

35.3 34 101 135

35.4 34 106 140

35.5 38 108 146

35.6 45 112 157

35.7 25 103 128

35.8 37 91 128

35.9 22 84 106

36 31 93 124

36.1 37 81 118

36.2 18 90 108

36.3 39 91 130

36.4 26 96 122

36.5 30 98 128

36.6 33 102 135

36.7 28 71 99

36.8 27 74 101

36.9 28 64 92

37 33 74 107

37.1 19 78 97

37.2 25 81 106

37.3 31 72 103

37.4 27 88 115

37.5 31 89 120

37.6 25 83 108

37.7 25 69 94

37.8 16 88 104

37.9 28 64 92

38 19 43 62

38.1 23 54 77

38.2 25 73 98

38.3 21 71 92

38.4 19 76 95

38.5 21 64 85

38.6 17 70 87

38.7 27 68 95

38.8 21 63 84

38.9 18 66 84

39 16 54 70

39.1 22 59 81

39.2 20 72 92

39.3 14 45 59

39.4 17 69 86

39.5 19 72 91

39.6 16 65 81

39.7 12 64 76

39.8 16 51 67

39.9 12 42 54

40 15 59 74

40.1 19 41 60

40.2 10 59 69

40.3 10 57 67

40.4 18 60 78

40.5 17 77 94

40.6 17 51 68

40.7 13 60 73

40.8 13 51 64

40.9 19 52 71

41 10 66 76

41.1 22 61 83

41.2 8 63 71

41.3 19 37 56

41.4 16 51 67

41.5 13 56 69

41.6 15 75 90

41.7 9 54 63

41.8 19 56 75

41.9 14 50 64

42 16 51 67

42.1 15 52 67

42.2 13 59 72

42.3 12 56 68

42.4 22 62 84

42.5 16 67 83

42.6 13 84 97

42.7 9 82 91

42.8 15 67 82

42.9 19 63 82

43 19 70 89

43.1 21 74 95

43.2 17 83 100

43.3 12 80 92

43.4 17 73 90

43.5 13 73 86

43.6 10 75 85

43.7 15 72 87

43.8 13 69 82

43.9 9 59 68

44 12 69 81

44.1 12 79 91

44.2 14 71 85

44.3 12 87 99

44.4 9 72 81

44.5 11 82 93

44.6 14 83 97

44.7 14 68 82

44.8 14 55 69

44.9 12 77 89

45 11 72 83

45.1 7 63 70

45.2 13 61 74

45.3 9 65 74

45.4 12 64 76

45.5 12 75 87

45.6 15 71 86

45.7 7 68 75

45.8 9 47 56

45.9 13 57 70

46 10 41 51

46.1 12 46 58

46.2 12 43 55

46.3 16 44 60

46.4 9 55 64

46.5 6 45 51

46.6 12 41 53

46.7 10 47 57

46.8 3 36 39

46.9 8 44 52

47 8 34 42

47.1 9 35 44

47.2 13 41 54

47.3 4 51 55

47.4 11 41 52

47.5 7 31 38

47.6 3 55 58

47.7 3 43 46

47.8 8 37 45

47.9 8 30 38

48 8 36 44

48.1 7 33 40

48.2 6 31 37

48.3 5 27 32

48.4 6 35 41

48.5 6 38 44

48.6 9 38 47

48.7 4 27 31

48.8 5 37 42

48.9 9 29 38

49 6 25 31

49.1 10 24 34

49.2 8 35 43

49.3 6 31 37

49.4 5 33 38

49.5 5 27 32

49.6 6 31 37

49.7 4 30 34

49.8 6 30 36

49.9 7 27 34

50 2 20 22

50.1 13 24 37

50.2 4 32 36

50.3 3 28 31

50.4 3 22 25

50.5 1 32 33

50.6 1 24 25

50.7 5 24 29

50.8 4 20 24

50.9 4 22 26

51 1 14 15

51.1 2 23 25

51.2 3 21 24

51.3 4 16 20

51.4 3 16 19

51.5 3 22 25

51.6 5 20 25

51.7 3 14 17

51.8 2 20 22

51.9 4 11 15

52 3 18 21

52.1 4 12 16

52.2 2 11 13

52.3 2 13 15

52.4 2 14 16

52.5 7 11 18

52.6 4 13 17

52.7 1 25 26

52.8 5 6 11

52.9 2 11 13

53 5 11 16

53.1 1 7 8

53.2 2 11 13

53.3 1 14 15

53.4 2 28 30

53.5 6 9 15

53.6 3 10 13

53.7 2 13 15

53.8 5 8 13

53.9 1 8 9

54 0 8 8

54.1 1 12 13

54.2 0 3 3

54.3 0 12 12

54.4 0 7 7

54.5 2 6 8

54.6 1 7 8

54.7 0 12 12

54.8 0 9 9

54.9 0 5 5

55 8 3 11

55.1 49 1 50

55.6 1 1 2

55.7 0 1 1

55.8 1 1 2

55.9 1 0 1

56.1 1 0 1

56.4 0 1 1

56.5 0 2 2

56.6 0 1 1

56.7 1 0 1

56.9 0 1 1

57 2 0 2

57.1 1 0 1

57.3 0 1 1

57.8 0 1 1

58.5 0 1 1

58.7 1 0 1

60.6 1 0 1

61.2 1 0 1

70.7 0 1 1

Grand Total 10476 27764 38240

Thank you for your kindness, Sir.

Regard,

Please send me an Excel file with your data by email along with the results that you obtained that you say are different from mine. Please make sure that when you use the Real Statistics software that you don’t include the totals for each item in the input data range.

Charles

dear Mr. Charles,

I have sent an email to your address. Please kindly check.

regards,

Ikhlas

I wish to used this model to predict the attrition probability, but I wonder when seeing my P(E) and Odds(E) plot because it seems different from yours, do I need to check the data plot to decide whether my data could be modeling as binary logistic regression? if so, how to check it?

When you say that your P(E) and Odds(E) are different from mine, do you mean that you got a different answer for one of the examples on the website?

Charles

I am not using the example data, but I use mine, in the picture your P(E) and Odds(E) probability linearly decreasing by the x- axis getting bigger, but my data doesn’t, so my point is, is there any requirement when choosing the binary logistic regression as a model?

The assumptions are: the dependent variable has exactly two outcomes; observations are independent; no collinearity, outliers or high leverage/influencer values; the model fits (i.e. there is a linear relationship between the independent variables and the logit transformation of the dependent variable).

If you send me an Excel file with your data, I will try to figure out what is happening.

Charles

Dear Charles,

I’ve sent an email to you few days ago, please kindly review my data.

Best regard,

Ikhlas

Dear Ikhlas,

I haven’t forgotten your email, and will get to it shortly.

Charles

How do we get the P(E) values? Thank you.

The referenced webpage describes how to get the P(E) values. The webpage sometimes uses the notation p for P(E). The other webpages on the Logistic Regression topic gets into more detail. You can also download the spreadsheets with all the examples shown on the website to get even more information.

Charles

Thank you ????

I hope that my response was helpful. If not, please provide some more detail about what you are looking for.

Charles

I tried the logistic regression tool with a data-set of Rows-142829 and Columns- 47 and my system hangs each time. Please can you let me know what is the capability of the addin. Also recommend me some way i can apply logistic regression on the same.

Assuming that the data is in raw format (and not summary format), there are two possible approaches:

1. Try using the

Logistic Regressiondata analysis tool, making sure thatRaw dataoption is selected, but theShow summary in outputoption is not selected. To use the summary output option the number of rows must be less than 65,000.2. If the above approach doesn’t work, use the LogitMatches function as described on the webpage http://www.real-statistics.com/logistic-regression/real-statistics-functions-logistic-regression/

With this amount of data, the analysis will be quite slow.

Charles

Dear Charles,

I was wondering how the sigmoid curve is plotted. It seems that the probabilty p given by p=1/(1+EXP(-B0-B1*x)) is plotted on the y-axis, what did you use on the x-axis? Seems to be from -3 to 3

Thanks in advance

Jip,

Yes, from -3 to 3.

Charles

Hi Charles, I’m trying to predict how suitable a prospective client would be for me; based on historical of my already existing clients. This will be based on 5 independent categorical variables.What do you suggest a suitable regression techniques would be?

You have several choices. First you need to decide whether your dependent variable is categorical or not.

If it is categorical then some version of logistic regression could be used. If there are two outcomes then the model is a binary logistic regression model (e.g. client is suitable or client is unsuitable). If there are multiple categories then you could use multinomial logistic regression or more likely ordinal logistic regression (e.g. client is a top prospect, client is a good prospect, client is a fair prospect or client is not a suitable prospect).

With only categorical variables you could also use log-linear models.

With a quantitative dependent variable (e.g. client is assigned a numeric rating from 0 to 100, with 100 representing clients that are most suitable), you could multiple regression employing dummy variables to handle the categorical independent variables.

All of these approaches are described on the Real Statistics website.

Charles

Dear Sir,

Thank You for your great explanations. Logic regression is used for item response theory also. Do you have any explanations about that?

Venura,

I don’t have any further information about this subject on the website. The site is evolving all the time and new information is constantly being added. What sort of explanations on this subject would be most helpful to you?

Charles

Sir,

we can use logistic regression in IRT to evaluate further students’ results,that’s what I asked.

Sorry Venura, I should have read your question more carefully. I now see that you are referencing item response theory. This is a topic that I need to add to the website.

Charles

In example 1, there’s a plot of the Odds(E). The plot shown looks to me like it is not the Odds(E), but rather the logit = ln Odds(E). Is this correct?

Alan,

Yes, you are correct. I have now changed the plot to ln Odds(E). Thanks for catching this.

Charles

Sir

I am at little confused about the value of yi in definition 5. In definition 5 you said yi is observed value. But I think if it is a binomial distribution y can only take the value of either 1 or 0, how could it be observed value of p?

Colin

Colin,

yi only takes a value of 0 or 1. pi is the predicted value, which can take any value between 0 and 1. Of course we will use the pi as a way of estimating the yi (even for unobserved yi). If pi > .5 (or some other preassigned value) then pi predicts a value for yi of 1 and pi < .5 predicts a value for yi of 0. Charles

Sir

I am sorry. I still don’t understand. In the “Finding Logistic Regression Coefficients using Excel’s Solver” you said yi “is the observed probability of survival in the ith of r intervals” and the value of yi in Figure 1 of “Finding Logistic Regression Coefficients using Excel’s Solver” does not take the value of either 0 or 1, which makes me confused.

Colin

Colin,

You are correct. It is a little confusing. I should have said that yi = the fraction of subjects in the ith interval that survived. I have now made this revision on the webpage.

Charles