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Section 25.3 Investigation 5.2: Teaching Morals

Exercises 25.3.1 The Study

Lee et al. (2014) examined whether some classic stories about moral behavior actually influence whether or not kids lie. They examined the stories of β€œPinocchio”, β€œThe Boy Who Cried Wolf”, and β€œGeorge Washington and the Cherry Tree”, as stories commonly used by teachers and parents to promote honesty, though in different ways (negative consequences of lying for the first two vs. positive consequences of truth telling in the third).
Illustration for Investigation 5.2
Two hundred and sixty-eight Canadian children aged 3–7 years were recruited for the study (children begin to tell lies around 2–3 years of age). Children participated in a β€œtemptation-resistance task” that has been used widely to study whether children choose to lie to hide a transgression (essentially peeking at the answer when left alone in the room for one minute) and then were read one of the three stories or a control story β€” β€œThe Tortoise and the Hare”. (The reader did not know whether or not they had peeked.) After the story, the child was asked whether or not he or she had peeked. Suppose the results for the children who peeked turned out like this:
Response \ Story Tortoise and Hare (control) George Washington Pinocchio Boy Who Cried 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 the observed and the expected counts.
Response \ Story Tortoise and Hare (control) George Washington Pinocchio Boy Who Cried 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 (28.22) 30 (29.51) 127
Total 64 44 44 46 198
We see that the observed counts are not always equal to the expected counts, but perhaps the random assignment created groups that were slightly different prior to the start of the study, and these differences merely reflect those random variations. Of course we can investigate this by simulating binomial random samples as in Investigation 5.1, but this time we want to model the random assignment process rather than the random sampling process. In other words, we will assume whether or not a child confesses is not influenced by which story they are read, and we will shuffle and redistribute these outcomes among the explanatory variable groups.
Use the applet to generate a randomization distribution and compute an empirical p-value for this study.
  • 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 Sample Data window with the story two-way table entered
  • 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.

4. Simulation with Applet.

Report your simulation-based p-value:
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.

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 the two-way table arises from a randomized experiment, we can apply the chi-squared distribution to predict the randomization distribution of the chi-squared statistic as long as at least 80% of the expected cell counts are at least 5 and all of the individual expected cell counts are at least one.

6. Technology Output.

What are the values of the chi-squared statistic, degrees of freedom, and p-value?
Chi-squared statistic:
Degrees of freedom:
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 25.3.2 Practice Problem 5.2

Suppose we had decided in advance to only compare the George Washington group to the control group (Tortoise and the Hare) to see whether there is a difference in the probability of confessing after hearing these two stories.

Checkpoint 25.3.1. Validity of Chi-squared Test.

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

Checkpoint 25.3.2. Degrees of Freedom.

What are the degrees of freedom for this chi-squared statistic?

Checkpoint 25.3.3. Chi-squared Test Output.

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

Checkpoint 25.3.4. 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 Question 3?

Checkpoint 25.3.5. Fisher’s Exact Test Comparison.

Use Fisher’s Exact Test to calculate a p-value for these data. How does the p-value compare?
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