# **How to Use SPSS to run a linear regression and a One-Way ANOVA.**

**CASE STUDY: REGRESSION AND ANOVA**

: This variable corresponds to a continuous variable, with the ratio level of measurement. We obtain the following descriptive statistics from SPSS__ADDSCC__

The distribution of the variable is shown below in the form of a histogram:

: This variable corresponds to a continuous variable, with the ratio level of measurement. We obtain the following descriptive statistics from SPSS__GPA__

We have the histogram:

This variable corresponds to a discrete variable, with the nominal level of measurement. The descriptive statistics are shown below:__DROPOUT__

The distribution looks like:

: This is a discrete variable with the nominal level of measurement. The descriptive statistics we obtain with SPPS are__ENGL__

The distribution is given by

: This is a discrete variable with the nominal level of measurement. The descriptive statistics are__GENDER__

The distribution is shown below:

: This is a discrete variable with the nominal level of measurement. The basic descriptive statistics are__REPEAT__

The histogram looks like:

: This variable corresponds to a continuous variable with the ratio level of measurement. The descriptive statistics we find with SPSS are__IQ__

The histogram is given by:

** ENGG: **This variable corresponds to a discrete (ordinal level of measurement) variable. The basic statistics are

The distribution looks like:

This variable corresponds to a discrete variable, with the ordinal level of measurement. The descriptive statistics are:__SOCPROB__

The distribution looks like:

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**3A. INFERENTIAL PROCEDURES**

We analyze the degree of linear correlation between GPA and ADDSC using SPSS:

The correlation coefficient is equal to

\[\rho =-0.557\]which shows a significant level of linear association between GPA and ADDSC, based on the p-values shown in the table. In order to analyze the normality of these two variables, we proceed in the following way:

* ADDSC*: We are going to test

The results of the Kolmogorov-Smirnov test are shown below:

The p-value is equal to \(p = 0.200\), which shows that we ** cannot reject the null hypothesis of normality. **This implies that there is not enough evidence to reject the normality assumption

* GPA*: We are going to test

The results of the Kolmogorov-Smirnov test are shown below:

The p-value is equal to \( p = 0.000\), which means that we ** reject the null hypothesis of normality**. There is enough evidence to reject normality of GPA

__Now we perform a Linear Regression analysis__:

The adjusted

R Square coefficient is equal to \[\text{Adj }{{R}^{2}}=\text{0}\text{.308}\]which means that* only 30.8% of the variance of the dependent variable (GPA) is explained by the linear regression*.

__ANOVA analysis__:

The table shows that **the model is significant overall** (\(p=0.000\)).

The regression coefficients are

The p-value shows that ADDSC is a **significant predictor** (\(p=0.000\)), which means that the Regression Equation is

**3B. DIFFERENCE BETWEEN GROUPS**

We run a t-test to determine if there is any difference between the mean GPA of dropouts and non-dropouts. In other words, we are going to test the following hypotheses

\[\begin{align} & {{H}_{0}}:{{\mu }_{1}}={{\mu }_{2}} \\ & {{H}_{A}}:\,{{\mu }_{1}}\ne {{\mu }_{2}} \\ \end{align}\]where \({{\mu }_{1}}\) represents the mean GPA of dropouts, and \({{\mu }_{2}}\) represents the mean GPA of non-dropouts.

The following are the results using SPSS.

- The table shows that Levene’s Test is not significant,
**which means we can assume equal variances.**

- The p-value for the t-test is equal to $p=0.000$. That means that
*we have enough evidence*to claim that non-dropout students have higher GPA than dropout students.

**3C. DIFFERENCE AMONG GROUPS**

We need to test the following hypotheses:

\[\begin{align} & {{H}_{0}}\text{: }{{\mu }_{{{G}_{1}}}}={{\mu }_{{{G}_{2}}}}\text{=}{{\mu }_{{{G}_{3}}}}\text{ } \\ & {{H}_{A}}\text{: the means are not equal}\, \\ \end{align}\]We use Analysis of Variance, ANOVA, to determine whether there is or not a difference between the group means.

Levene's Test shows that ** we can consider the variances to be all the same, **since p = 0.603, which is not less than the significance level.

The ANOVA analysis shows evidence that ** there is enough evidence to claim that not all the means are the same**.

- Now that we know that there is a difference between the means, we need to perform a Post-Hoc analysis, which is shown below:

Using the LSD test, we conclude that

- The mean of the group 1 is greater than the other group means.

- The mean of group 2 is higher than the mean of group 3.

**3D. INDEPENDENCE OF CATEGORICAL VARIABLES**

We need to test the following hypotheses

\[\begin{align} & {{H}_{0}}\text{: ''Gender''}\text{ and ''Repeat'' are independent} \\ & {{H}_{A}}:\,\text{ ''Gender''}\text{ and ''Repeat'' are NOT independent} \\ \end{align}\]We have the following crosstabulation:

Using the Chi-Square Test for independence (the variables are categorical)

- The Chi-Square is significant, with a p-value of \(p = 0.025\), which implies that we have GENDER and REPEAT
at the \(\alpha =0.05\) significance level.*are dependent,*

**3E. EXTRA RESEARCH QUESTION**

Is the linear association between IQ and GPA statistically significant?

Using SPSS:

The correlation coefficient is equal to

\[r=0.443\]which indicates a significant level of linear association between GPA and IQ. This association is rather weak though, given that the R Square coefficient is \({{R}^{2}}=0.197\), which means that *only 19.7% of the variation in the GPA is explained by the IQ variable.*

The ANOVA analysis shows that model is significant overall, with a p-value equal to \(p=0.000\). The coefficients are:

The table shows that IQ is a significant predictor of GPA (\(p=0.000\)). The Regression Equation is equal to

\[\hat{y}=-7.22+0.03067x\]
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