MESA8450 Spring 2024 Multilevel Regression Models Assignment 4

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MESA8450 Spring 2024

Multilevel Regression Models

Homework Assignment 4

.For this assignment, you will need to use the Assign4_Data_L1L2.sav (or Assign4_Data_L1.sav
and Assign4_Data_L2.sav) file provided on Canvas. Please do not use SPSS for this assignment
since it provides different results that the ones reported for the data below. For each question
below, present the output relevant to the question asked

Descriptive Statistics

1. Use descriptive analyses, summarize the data in the SPSS files. For example, you could create a table with the mean, standard error and standard deviation for at each time point, and by treatment v. control group. Write a short description of the descriptive findings.

According to the data presented in the table, it is evident that the mean score for both the control and treatment groups is increasing over each time period. Notably, the treatment group exhibits a higher mean score than the control group at each time point. Specifically, the control group's mean scores for the four time points are 213.9, 230.9, 233.4, and 243.3, while the standard deviations are 20.9, 20.9, 23.2, and 24.3. However, the standard deviations suggest that the variability in the outcome scores is increasing. In contrast, for the treatment group, the mean scores for the four-time points are 242.0, 251.6, 270.0, and 288.2, whereas the standard deviations are 27.1, 22.9, 26.0, and 19.9.

The Unconditional Linear Random Coefficients Growth Model:

2. Using Outcome as the dependent variable, estimate the Linear Random Coefficients

Growth Model by adding time (uncentered) into the level-1 model and allowing the slope
to vary randomly. Present the statistical model as separate models and as a mixed model
(i.e., all in one regression equation)

Level 1 Model:

Level 2 Model:

Mixed Model:  

The maximum number of level-1 units = 100
The maximum number of level-2 units = 25
The maximum number of iterations = 100
Method of estimation: full maximum likelihood

Final Results - Iteration 2

Iterations stopped due to small change in likelihood function
σ² = 207.74770
Standard error of σ² = 41.54954
τ

Standard errors of τ

Approximate confidence intervals of tau variances
INTRCPT1 : (193.116,855.149)
TIME : (3.347,141.525)
τ (as correlations)

Confidence intervals of τ correlations

The value of the log-likelihood function at iteration 2 = -4.434876E+02

Final estimation of fixed effects:

Final estimation of fixed effects

(with robust standard errors)

Final estimation of variance components

3. Using the Graph Data option in HLM (or some other package), plot a line graph to show the growth trajectory for the groups (x-axis: time, y-axis: outcome). What does the graph indicate about the variability in initial status and the variability in growth trajectories?

Based on the graph below, it can be observed that there is a considerable variation in the initial status of the groups as well as their growth trajectories. Some students experience a significant increase in their scores, while some show a decline in their performance over time. Additionally, there are students with both high and low growth rates.

4. Fixed Effects:

a. Interpret the meaning of the fixed effects (intercept and slope), and stating the statistical hypothesis, indicate whether the coefficient associated with time is significantly different from zero.

Based on the data analysis, we can conclude that the intercept value is statistically significant and different from zero at a significance level of 0.001. This means that the average outcome value at initial status (time=0) across all individuals is 229.44 points.

 , with ,

 Test the hypothesis:

Since the p-value is less than 0.001, then we can reject the null hypothesis and conclude that the intercept is statistically significantly different from zero.

Based on the analysis of the data, then we can conclude that the slope value is statistically significant and different from zero at a significance level of 0.001. This meaning that the average growth rate (slope) over each period is 12.78 points.

, with ,

Test the hypothesis:

Since the p-value is less than 0.001, we can reject the null hypothesis and conclude that the slope is statistically significantly different from zero.

5. Random Effects:

a. What are the values of the random effects, σ², τ₀₀, and τ₁₁ 

b. Is there significant variability in the intercept, τ00 (e.g., initial status)?

 Test the hypothesis

Since the  and p-value is less than 0.001, then we can reject the null hypothesis and conclude that there is significant variability in the intercept.

           c. Is there significant variability among the individual trajectories from the average
growth trajectory, τ11 (time slope)?

Test the hypothesis.

Since the  and the p-value is 0.034, which is less than 0.05. We can reject the null hypothesis and conclude that there is significant variability among the individual trajectories from the average growth trajectoryτ11 (time slope).

6. Interpret the correlation between initial status and individual growth.

The following shows correlation between initial status and individual growth

It seems that there is a strong positive correlation between initial status and individual growth, indicating that those who begin with a higher initial status tend to experience more significant growth over time compared to those who start with a lower initial status.