# Conducting and Analyzing Statistical Tests

## Conducting and Analyzing Statistical Tests

• Submitted By: scrowe7
• Date Submitted: 06/23/2013 7:06 PM
• Category: Psychology
• Words: 594
• Page: 3
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Conducting and Analyzing Statistical Tests
Sherry Liezl Crowe
PSY210: Psychological Statistics
Argosy University

06/16/2013

A study wants to examine the relationship between student anxiety for an exam and the number of hours studied. The data is as follows:
Student Anxiety Scores | Study Hours |
5 | 1 |
10 | 6 |
5 | 2 |
11 | 8 |
12 | 5 |
4 | 1 |
3 | 4 |
2 | 6 |
6 | 5 |
1 | 2 |
| |
1. Why is a correlation the most appropriate statistic?
2. What is the null and alternate hypothesis?
3. What is the correlation between student anxiety scores and number of study hours? Select alpha and interpret your findings. Make sure to note whether it is significant or not and what the effect size is.
4. How would you interpret this?
5. What is the probability of a type I error? What does this mean?
6. How would you use this same information but set it up in a way that allows you to conduct a t-test? An ANOVA?

Conducting and Analyzing Statistical Tests
Correlation gives the relationship or the degree of association between two variables. Since we are interested in examining the relationship between the variables student anxiety for an exam and the number of hours studied, therefore correlation would be the most appropriate statistic for this study.
Null hypothesis: Student anxiety for an exam does not have a significant correlation with the number of hours studied. H0: = 0 (The correlation coefficient is zero)
Alternate Hypothesis: Student anxiety for an exam has a significant correlation with the number of hours studied. H1: ≠ 0 (The correlation coefficient is non zero)
The correlation coefficient between student anxiety scores and number of study hours is 0.5654.

We choose significance level, = 0.05
Sample Size, n = 10, therefore, degrees of freedom, df = (n-2) = (10 - 2) = 8
Critical Value for significance level, = 0.05 and df = 8 is given as ± 2.3060 (from t-distribution...