*Data Analytics*

**DS-520 DATA ANALYSIS AND DECISION MAKING**

**Where are your eyes?**

The objectifying gaze, often referred to as “ogling” or “checking out,” can have many adverse consequences. A group of researchers used eye-tracking technology to better understand the nature and causes for this gaze. They asked 29 women and 36 men to look at images of college-aged women. Each woman had the same clothes and neutral expression but varied in body shape (ideal, average, and below average). Prior to looking at the images, each participant was told to focus on either the appearances or personalities of the women. Here is a summary of the amount of time (in milliseconds) the eyes focused on the chest of the women:

- Plot the means. Do you think there is an interaction? Explain your answer.
- Do you think the marginal means would be useful for understanding the results of this study? Explain why or why not.
- The researchers broke these results down further using body shape as a third factor. Describe why the inclusion of this factor complicates the analysis. In other words, why is this not a standard 2×2×3 experiment?

- A study reported the following results for data analyzed using a two-way ANOVA at the 5% significance level:
- What can you conclude from the information given?
- What additional information would you need to write a summary of the results for this study?

**3**. The effects of two stimulant drugs. An experimenter was interested in investigating the effects of two stimulant drugs (labeled A and B). She divided 25 rats equally into five groups (placebo, Drug A low, Drug A high, Drug B low, and Drug B high) and, 20 minutes after injection of the drug, recorded each rat’s activity level (higher score is more active). The following table summarizes the results:- Plot the means versus the type of treatment. Does there appear to be a difference in the activity level? Explain.
- Is it reasonable to assume that the variances are equal? Explain your answer and, if reasonable, compute sp.
- Give the degrees of freedom for the F statistic.
- The F statistic is 2.64. Find the associated P-value and state your conclusions.

- The National Survey of Student Engagement found that 87% of students report that their peers at least “sometimes” copy information from the Internet in their papers without reporting the source. Assume that the sample size is 430,000.
- Find the margin of error for 99% confidence.
- Here are some items from the report that summarizes the survey. More than 430,000 students from 730 four-year colleges and universities participated. The average response rate was 43% and ranged from 15% to 89%. Institutions pay a participation fee of between $3000 and $7500 based on the size of their undergraduate enrollment. Discuss these facts as possible sources of error in this study. How do you think these errors would compare with the error that you calculated in part (a)?

**5**. Food neophobia is a personality trait associated with avoiding unfamiliar foods. In one study of 564 children who were two to six years of age, the degree of food neophobia and the frequency of consumption of different types of food were measured. Here is a summary of the correlations:

Perform the significance test for each correlation and write a summary about food neophobia and the consumption of different types of food.

Laptops and other digital technologies with wireless access to the Internet are becoming more and more common in the classroom. While numerous studies have shown that these technologies can be used effectively as part of teaching, there is concern that these technologies can also distract learners if used for off-task behaviors.

In one study that looked at the effects of off-task multitasking with digital technologies in the classroom, a total of 145 undergraduates were randomly assigned to one of seven conditions.12 Each condition involved performing a task simultaneously during lecture. The study consisted of three 20-minute lectures, each followed by a 15-item quiz. The following table summarizes the conditions and quiz results (mean proportion correct):

- For this analysis, let’s consider the average of the three quizzes as the response. Compute this mean for each condition.
- The analysis of these average scores results in SSG=0.22178 and SSE=2.00238. Test the null hypothesis that the mean scores across all conditions are equal.
**Using the means from part (a) and the Bonferroni method, determine which pairs of means differ significantly at the 0.05 significance level.**(Hint: There are 21 pairwise comparisons, so the critical t-value is 3.095. Also, it is best to order the means from smallest to largest to help with pairwise comparisons.)- Summarize your results from parts (b) and (c) in a short report.