T-Tests And ANOVA Assignment Discussion Paper
201WK5Assign: t-Tests and ANOVA
A one-way ANOVA was performed to determine if three different overall satisfaction with material well-being have different means. A sample of 935 participants was recruited. 367 of the participants (M = 12.71, SD = 2.353, 95% CI = [12.47, 12.95]) had no housing problem, 264 of them (M = 11.97, SD = 2.588, 95% CI = [11.66, 12.28]) had one housing problem, and 304 of them (M = 10.57, SD = 2.594, 95% CI = [10.28, 10.86]) had two or more housing problems (see Table 1) T-Tests And ANOVA Assignment Discussion Paper.
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Table 1. Descriptive statistics
A Levene’s test was conducted to determine whether or not the three groups (overall satisfaction levels with material well-being) have equal variances. The Levene’s test based on the mean reveals that we are unable to reject the null hypothesis as there is no sufficient evidence to say that the variance between the three groups is significantly different (F (2, 932) = 2.109, p = 0.122). The p-value is not less than 0.05; as such, we fail to reject the null hypothesis. In other words, the three groups have equal variance thereby meeting the assumption for performing an ANOVA test (see Table 2)T-Tests And ANOVA Assignment Discussion Paper.
Table 2. Test of homogeneity of variances
A one-way ANOVA test was conducted to test the null hypothesis that all the means in the three groups are equal, and the alternative hypothesis that at least one population mean is different from the rest. The test results reveal that the null hypothesis should be rejected, and alternative hypothesis accepted (F (2, 932) = 61.674, p = 0.000). The p-value is less than 0.05, and so we reject the null hypothesis and accept the alternative hypothesis. This means that there is sufficient evidence to say that there is a statistically significant difference between the means of the three groups with the conclusion that at least one of the group means is different from the rest (see Table 3).
Table 3. ANOVA
Turkey’s test for multiple comparisons found that the mean responses were significantly different between all the three groups as all the p-values are less than 0.05 (see Table 4) T-Tests And ANOVA Assignment Discussion Paper
You are a DNP-Prepared nurse tasked with evaluating patient care at your practice compared to patient care at affiliated practices. You have noticed that a key complaint from your patients concerns the wait times associated with each patient visit. Based on these complaints, you have decided to compare the wait times at your practice to the wait times at affiliated practices. After recording the wait times at each practice, for 50 individual patients at each practice, you are now prepared to analyze your data. What approach will you use to analyze the data?In the scenario provided, you might decide to use, the Analysis of Variance (ANOVA) approach. “ANOVA is a statistical procedure that compares data between two or more groups or conditions to investigate the presence of differences between those groups on some continuous dependent variable” (Gray & Grove, 2020). ANOVA is often a recommended statistical technique, as it has low chance of error for determining differences between three or more groups. For this Assignment, analyze the ANOVA statistics provided in the ANOVA Exercises SPSS Output document. Examine the results to determine the differences and reflect on how you would interpret these results. Summarize your interpretation of the ANOVA statistics provided in the Week 5 ANOVA Exercises SPSS Output document. Note: Interpretation of the ANOVA output should include identification of the p-value to determine whether the differences between the group means are statistically significant. Be sure to accurately evaluate each of the results presented (descriptives, ANOVA results, and multiple comparisons using post-hoc analysis)