Skip to main content

Section 24.3 Investigation 5.2: Teaching Morals

Exercises 24.3.1 The Study

Lee et al. (2014) examined whether classic stories about moral behavior influence whether children lie. They studied three moral stories (Pinocchio, The Boy Who Cried Wolf, and George Washington and the Cherry Tree) and one control story (The Tortoise and the Hare). Children ages 3-7 completed a temptation-resistance task and then were asked whether they had peeked. Suppose the results for children who peeked were:
Illustration for Investigation 5.2
Response \ Story Tortoise (control) George Washington Pinocchio Wolf Total
Confessed 20 22 13 16 71
Did not (lied) 44 22 31 30 127
Total 64 44 44 46 198

1. Observational Study or Experiment?

Was this an observational study or an experiment?
  • Experiment
  • Correct! The researchers determined which story each child heard, making this a randomized comparative experiment.
  • Observational study
  • Not quite. Ask yourself who determined which story each child heard: the children (observational) or the researchers (experiment)?
Solution.
This was a randomized comparative experiment, because the researchers determined which story each child heard.

2. Identify Variables.

Identify the response variable and the explanatory variable and classify them as quantitative or categorical.
Solution.
Explanatory variable: story type (categorical).
Response variable: whether the child confessed (categorical).

3. State Hypotheses.

State appropriate null and alternative hypotheses for this research question in symbols and/or in words. Be sure to define any symbols you use.
Solution.
Let \(\pi_j\) be the long-run probability of confession for story \(j\text{.}\)
\(H_0\!: \pi_{\text{tortoise}} = \pi_{\text{Washington}} = \pi_{\text{Pinocchio}} = \pi_{\text{wolf}}\)
\(H_a\!:\) not all of these probabilities are equal (at least one differs).
The following table displays both observed and expected counts:
Response \ Story Tortoise (control) George Washington Pinocchio Wolf Total
Confessed 20 (22.95) 22 (15.78) 13 (15.78) 16 (16.49) 71
Did not (lied) 44 (41.05) 22 (28.22) 31 (29.22) 30 (29.51) 127
Total 64 44 44 46 198
To model this randomized experiment under the null, we shuffle confession outcomes among story groups while preserving the group sizes.

4. Simulation with Applet.

Use the applet to generate a randomization distribution and compute an empirical p-value for this study.

Aside: Analyzing Two-way Tables Applet.

Hint. Applet Instructions
Enter the two-way table (without totals and using only one-word variable names) into the Sample Data box and press Use Table.
Analyzing Two-way Tables applet setup for Investigation 5.2
Make sure the story is being used as the explanatory variable or press the (explanatory, response) button. Check the Show Shuffle Options box, enter a large number of shuffles, and press Shuffle.
Use the pull-down menu to set the Statistic choice to the Chi-squared (\(\chi^2\)) statistic. Compute an empirical p-value based on the simulated chi-squared values.
Solution.
Answers will vary slightly by simulation run, but with many shuffles the empirical p-value should be around 0.16.

5. Overlay Chi-square Model.

Check the box to Overlay Chi-square distribution. Does the theoretical chi-squared distribution (with \(df=3\)) appear to be a reasonable model for the simulated null distribution? How are you deciding?
Solution.
Yes. The chi-squared distribution with \(df=3\) is a reasonable model for the simulated null distribution here, and the model-based p-value is close to the simulation p-value.
When data arise from a randomized experiment, the chi-squared model is generally appropriate if at least 80% of expected counts are at least 5 and all are at least 1.

6. Technology Output.

Use technology to carry out the chi-squared test as before. What are the values of the chi-squared statistic, degrees of freedom, and p-value?
Solution.
A representative output is \(\chi^2 \approx 5.20\) with \(df=3\) and \(p\approx 0.16\text{.}\)

7. Largest Cell Contributions.

Examine the chi-squared contributions for each cell ("residuals" are square roots of these values). Which cell(s) contribute the most to the overall chi-squared sum? Compare the observed counts to the expected counts for those cells. What do these comparisons reveal about which types of stories seem to make children more likely to confess?
Solution.
The largest contributions come from the George Washington column (both the confessed and did-not-confess cells).
For George Washington, the observed confessed count is larger than expected and the observed did-not-confess count is smaller than expected, suggesting this story may increase confession relative to the others.

Study Conclusions.

Examining the conditional proportions that confessed across the four stories (0.313, 0.500, 0.295, 0.348), we see the children were more likely to confess when read the George Washington story compared to the negative consequences stories or the neutral story. A chi-squared test (valid because all expected cell counts are larger than 5) however does not find these differences to be statistically significant (\(X^2 = 5.202\text{,}\) p-value = 0.158). We do not have convincing evidence that the type of story influences Canadian children’s likelihood of confessing their indiscretion in this situation. Still, the results are in the direction conjectured by the researchers, and larger sample sizes may find significant results if this study was repeated.

Subsection 24.3.2 Practice Problem 5.2

Suppose we compare only George Washington to the control (Tortoise and the Hare).

Checkpoint 24.3.1. Validity of Chi-squared Test.

Would a chi-squared test be valid for these data? How are you deciding?

Checkpoint 24.3.2. Degrees of Freedom.

Checkpoint 24.3.3. Chi-squared Test Output.

Use technology to calculate the chi-squared test statistic and p-value. What would you conclude?

Checkpoint 24.3.4. Two-sample \(z\)-test Comparison.

Use technology to calculate a two-sample \(z\)-test for these data. How does the (two-sided) p-value compare to what you found in Checkpoint 5.2.c?

Checkpoint 24.3.5. Fisher’s Exact Test Comparison.

Use Fisher’s Exact Test to calculate a p-value for these data. How does the p-value compare?
You have attempted of activities on this page.