## Problem 1: Analyzing Ticket Costs by Country of Residence

**Problem Description:** In this SPSS assignment, we conducted a statistical analysis to determine if the cost of tickets for a particular event differs based on the country of residence. We used a one-way ANOVA test to explore this hypothesis. The results of the analysis indicated that there was a significant difference in the average cost of tickets among three groups: Americans, British, and individuals from other countries.

## Solution

**Key Findings:
**

- The average cost of tickets for Americans was $64.65, with a standard deviation of $70.65.
- British attendees had the lowest average ticket cost of $23.39, with a standard deviation of $21.11.
- Other countries had an average cost of $26.47, with a standard deviation of $48.60.
- The ANOVA test showed a significant difference in ticket costs (F(2,1305) = 65.195, p < .001).
- Post hoc tests revealed that American ticket costs were significantly different from both British and other countries, while there was no significant difference between British and other countries.

**Here are the descriptive statistics for the cost of tickets:**

Country of Residence |
N |
Mean |
Std. Deviation |
Std. Error |
95% Confidence Interval for Mean |
Minimum |
Maximum |
---|---|---|---|---|---|---|---|

American | 258 | 64.65 | 70.65 | 4.40 | (55.99, 73.31) | 0.00 | 512.33 |

British | 301 | 23.39 | 21.11 | 1.22 | (20.99, 25.79) | 0.00 | 153.46 |

Other | 749 | 26.47 | 48.60 | 1.78 | (22.99, 29.96) | 0.00 | 512.33 |

Total | 1308 | 33.30 | 51.76 | 1.43 | (30.49, 36.10) | 0.00 | 512.33 |

We also conducted a Kruskal-Wallis test as a non-parametric alternative to the ANOVA test to check for the normality assumption. The Kruskal-Wallis test confirmed that the assumption of normality was violated, and the cost of tickets was significantly different based on the country of residence.

## Problem 2: Analyzing the Relationship Between Age and Weight

**Problem Description: **The second part of the assignment involved analyzing the relationship between a person's age and their weight. We conducted a correlation analysis to explore this relationship. The results revealed a strong negative correlation between age and weight, suggesting that as age increases, weight tends to decrease.

## Solution

**Key Findings:
**

- The correlation coefficient (Pearson's r) between age and weight was -0.973, indicating a strong negative correlation (p < .001).

**The correlation analysis was presented as follows:
**

Variable |
Pearson Correlation |
Sig. (2-tailed) |
N |
---|---|---|---|

Age | 1 | 27 | |

Weight | -0.973** | 0.000 | 27 |

## Problem 3: Regression Analysis for Predicting Relapse

**Problem Description: **In the third part of the assignment, we conducted a multiple regression analysis to predict the number of relapses based on several independent variables. The analysis included model summary, ANOVA results, and coefficients. The goal was to identify significant predictors of relapse.

## Solution

**Key Findings:
**

- The model summary indicated that clinical judgment of depression, propensity for substance abuse, amount of social support, and perception of social support were significant predictors of relapse. However, the amount of daily stress was not a significant predictor.
- Perception of social support had the highest impact on relapse, with the lowest p-value and the largest coefficient estimate.
- The overall model was significant (F(5,114) = 20.667, p < .001), and it explained 45.3% of the variation in the dependent variable.

We also discussed the assumptions of linearity, normality of residuals, and multicollinearity, concluding that these assumptions were met, except for homoscedasticity.

## Problem 4: Assessing Reliability of a Scale

**Problem Description: **In the fourth part of the assignment, we conducted a reliability analysis to assess the internal consistency of the Brink Depression Scale (BDS). We used both split-half reliability and Cronbach's alpha to determine the reliability of the scale.

## Solution

**Key Findings:
**

- Split-half reliability: The scale's reliability estimate was 0.845.
- Cronbach's alpha: The scale exhibited high internal consistency, with a reliability of 0.856. Removing item 12 increased the alpha value to 0.863, suggesting improved reliability.

We also discussed the potential misleading aspect of the data presentation, which was corrected to provide percentages over the total, making the comparison more accurate.

## Problem 5: In-depth Reliability Analysis

**Problem Description:** In the fifth part of the assignment, we conducted an in-depth reliability analysis using Cronbach's alpha and examined the impact of removing an item (item 12) on the scale's reliability.

## Solution

**Key Findings:
**

- The scale showed high internal consistency with a Cronbach's alpha of 0.856.
- Removing item 12 improved the reliability to a Cronbach's alpha of 0.863.

## Problem 6: Data Visualization Evaluation

**Problem Description: **In the final part of the assignment, we assessed a data visualization, pointing out a potentially misleading aspect in the graph presentation. The misleading aspect was corrected to provide a more accurate representation of the data, comparing percentages over the total, rather than within groups.

## Solution

- Yes there is something potentially misleading. The first view of the graph makes it appears as if most of democrats agreed with the court. However, a second look at the graph revealed that the percentage presented here is percentage within group but not over total. For example assuming 200 people were sampled with 80 democrats, 70republicans and 50 independents. The percentage in the plot is to divide the number of agreements with 80 for democrats, 70 for republicans and 50 for independents.
- The changes I will add is to divide the agreement by the total such that it can become comparable