91Ó°ÊÓ

In Montreal, Canada, an experiment was done with parents of children who were thought to have a high risk of committing crimes when they became teenagers (Tremblay et al., 1996 ). Some of the families were randomly assigned to receive parental training, and the others were not. Out of 43 children whose parents were randomly assigned to the parental training group, 6 had been arrested by the age of \(15 .\) Out of 123 children whose parents were not in the parental training group, 37 had been arrested by age 15 . a. Find and compare the percentages of children arrested by age \(15 .\) Is this what researchers might have hoped? b. Create a two-way table from the data, and test whether the treatment program is associated with arrests. Use a significance level of \(0.05\). c. Do a two-proportion \(z\) -test, testing whether the parental training lowers the rate of bad results. Use a significance level of \(0.05\). d. Explain the difference in the results of the chi-square test and the twoproportion \(z\) -test. e. Can you conclude that the treatment causes the better result? Why or why not?

Step by step solution

01

Calculation of percentages

To calculate the percentages of children arrested by 15, divide the number of arrests per group by the total number in each group and multiply by 100. For the parental training group, \(\frac{6}{43} * 100 ≈ 13.95% \) and for the non-training group, \(\frac{37}{123} * 100 ≈ 30.08% \). This means that less children whose parents were trained were arrested, indicating a positive outcome.

02

Creating a two-way table

Create a two-way table with 'Parental Training' and 'No Parental Training' as columns. In the rows, group the results as 'Arrested' and 'Not Arrested'. Fill in the counts from the question prompt.

03

Chi-square test

The Chi-square test is used to reveal if the observed distribution of cases falls within the expected one. Use a significance level of 0.05. Input the counts from the two-way table into a Chi-square test calculator, along with the sample sizes of the two groups (43 and 123). Record the results.

04

Two-proportion z-test

The two-proportion z-test reveals if the proportions from two groups are significantly different. Use a significance level of 0.05, and input the counts and sample sizes per group. Perform the test and record the results.

05

Explanation

The results from the Chi-square test show if there is a significant association between the two groups, that is, if incidence of arrest is connected with parental training. The z-test indicates if the proportions of the two groups are distinct. Compare and contrast these results.

06

Conclusion

Indicate if the parental training causes a decrease in arrest incidence, based on the statistical evidence available. Consideration should be given to the statistical results, as well as to external confounding factors not examined in this study, such as socio-economic status, parenting style beyond the training, etc.

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