The source of this training is the authoritative website of statistical solutions . In this post, in fact, with a simple and tangible example, you will learn how to perform linear regression analysis in principle in SPSS software .
Introductory regression training video in SPSS
Regression, But the main article of regression training…. This statistic is based on the 2006 FBI (Federal Police) crime statistics. In this statistical sample, we want to know the relationship between how and the number of crimes in the city. First we need to see if there is a linear relationship between the data or not. In this way we check the data scatter curve.
Dispersion diagrams show a linear relationship between data. Which allows us to perform a regression analysis. We can also examine the correlation of the Pearson bivariate functions and find that these two variables are highly correlated (r = 0.959 and p <0.001). Second, we need to look at multivariate normality.
There may not be multivariate normality in this statistical sample. Calmograph-Smirnov test confirms this assumption (p = 0.002 and p = 0.006). A ln on these two variables proves this problem and the multivariate normality as (p = 0.543 and p = 0.
Now we can do linear regression analysis. Linear regression path as: Analyze / Regression / Linear
In this simple statistical example, we only need to add variables such as (log-pop) and (log-murder) to our model as independent and dependent variables. We can run other statistical indicators that are needed to confirm the validity of linear regression analysis.
Performing these statistical processes that explain alignment and the Durbin-Watson test are useful for self-correlation detection. To test the variance of the residuals, we consider a special chart in the plots menu.
SPSS program command in linear regression
The results of the first table summarize the model and show the general statistical process. The expected in our model is 0.756, which with the result of the test concludes that linear regression can describe 76.1% of the variance of the data.
In the camera-Watson test it is d = 2.323 and between 2.5> d> 1.5, so it can be assumed that there is no first-order autocorrelation between the data (in other words). With F = 156.2 and a degree of freedom of 50, the importance of this test increases and it can be assumed that in our model there is a linear relationship between the variables.
The following table shows the regression coefficients and the width of the origin and their size. Linear regression analysis says that the linear regression function should be Y = -13.067 + 1.222X. This does not mean that the crime rate will increase 1.2 times per 1,000 population because we take ln from the variables.
If we do the linear regression analysis again with the main variables, we get the equation y = 11.85 + 6.7 * 10-5, which shows that for every 10,000 people, the crime rate increases 6.7 times.
Linear regression analysis examines false assumptions in which the coefficient is zero. The t test says that both the width of the origin and the variable have significant values (p <0.001). And therefore non-zero. This table also contains beta weights (which express the value of the independent variables) and contains linear statistics.
Since we have only one independent variable, we do not address its value. The next issue to consider is the homogeneity and normality of the residuals.
The histogram shows that the residues have an approximately normal distribution. The Q – Q curve with critical values of z * indicates that there is less error in the linear regression of our model.
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