Dear all, as an example the output of a logistic regression based on wald test is provided. For instance, for scenario(1), (Beta)^2/(Standard Error)^2 =
(0.551)^2/(0.362)^2=2.3167=2.32 &
p-value=1-pchisq(2.3167,1)=0.1279 (R command)
But I can not understand how wald statistic and its P-value are calculated for the scenario it self (7.291 and 0.121).
The Scenario (0)estimate is associated with the intercept - that is the estimate of the left out category for the 'mean' value of the Scenario category dummy set when all predictors are at the zero value. You are testing via the Wald test whether this logit estimate is zero - that is you are testing on a probability scale whether this estimate is significantly different from a probability of 50% - and it is.
I suggest you follow a course on logistic modeling - this one is free
http://www.bristol.ac.uk/cmm/learning/online-course/index.html
and these are the modules
Modules
Using quantitative data in research (watch video introduction)
Introduction to quantitative data analysis (watch video introduction)
Multiple regression
Multilevel structures and classifications (watch video introduction)
Introduction to multilevel modelling
Regression models for binary responses THIS ONE
Multilevel models for binary responses
Multilevel modelling in practice: Research questions, data preparation and analysis
Single-level and multilevel models for ordinal responses
Single-level and multilevel models for nominal responses
Three-level multilevel models
Cross-classified multilevel models
Multiple membership multilevel models
Missing Data
Multilevel Modelling of Repeated Measures Data
K
Hi,
The Wald test is a parametric statistical test named after the Hungarian statistician Abraham Wald. Whenever a relationship within or between data items can be expressed as a statistical model with parameters to be estimated from a sample, the Wald test can be used to test the true value of the parameter based on the sample estimate.
Use Wald's tests to examine the effect of single variables.
For sample sample t - test wasn't a suitable one. Instead of t - test, we use Wald test.
Could you write the formula by which 7.291 is obtained?
The Wald statistic is a type of chi-square statistic which can be thought of as a sum of independent squared z statistics. The z test for each parameter is B/SE. Thus the value 7.291 should be the sum of (B/SE)^2 for all of the four scenarios.
The univariate Wald test compares the difference between a maximum likelihood point estimate of a single parameter and a hypothesised value to its standard error:
Test statistic = (Estimate – hypothesised value)/ Standard error of estimate
When a null hypothesis of zero is used this reduces to the familiar ratio of the estimate divided by its standard error and the resultant test statistic is compared to a z distribution. If the value is greater than ±1.96 we commonly say that the results are significant at the 0.05 level; in repeated samples of the data we are unlikely to have got a value by chance as big as the estimate if the population value is really zero.
Another version of this test is to square all the values
Test statistic = (Estimate – hypothesised value)**2/ Variance of estimate
When this version is used, the test statistic needs to be compared not against a z distribution but against a chi-square distribution with one degree of freedom, with exactly the same results. The basis of both these univariate procedures is that the maximum likelihood estimates are assumed to be asymptotically normal.
The advantage of the Wald test over the standard z test is that it can be readily extended to a multivariate Wald test so that you can do multiple, complex hypothesis testing of a set of parameters simultaneously.
The principle behind this procedure was introduced by Wald, Abraham (1943) Tests of statistical hypotheses concerning several parameters when the number of observations is large Transactions of the American Mathematical Society, 54, 426-482 although many long-established procedures can be interpreted as Wald tests.
Kelvin took the time to give a much more comprehensive answer - I would just comment (in case it isn't obvious) that relying on the asymptotic properties of the Wald test is generally a bad idea unless you have at least fairly large samples and you are also confident that the procedures behaves well in your context. For a non-multilevel logistic regression you'd probably want means for the different scenarios to be in the range 0.2-0.8 and sample sizes of say 30+ per scenario for the inferences to be reasonably accurate.
Given you have access to R I'd probably run the regression in R via glm() and then get likelihood ratio tests and profile likelihood CIs instead. ISTR you can get the likelihood ratio tests from SPSS fairly easily also by comparing models with and without the relevant parameters.
The 7.291 Wald test score with 4 degrees of freedom is perhaps the one computed by the formula:
W = (b^)T (Var(b^))-1 b^ = (b^)T I(b^) b^ = (b^)T (XT V X) b^
where
b^: the estimated parameter (coefficient) vector with elements the individual coefficients for the levels of your polychotomous variable
Var(b^): the covariance matrix (https://en.wikipedia.org/wiki/Covariance_matrix) for the above coefficients
I(b^): the observed information matrix (https://en.wikipedia.org/wiki/Observed_information)
X: the design matrix (https://en.wikipedia.org/wiki/Design_matrix)
V = diag(P^i(1 - P^i)
P^i: The estimated probabilities.
However, should you consider to make use of it, it should be again underlined that the Wald statistic may lead to Type II error.
Reference: Hosmer DW, Lemeshow S: Applied Logistic Regression, 2nd ed, 2000, [Wiley]. Chapter 2, pp. 31-46.
Еssential logistic regression notes from my statistical notebook.
I am not satisfied when I see during manuscript revision that the authors represent only the odds ratio (OR) and p-value results from logistic regression model.
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