4. A research team collected data on participants in a longitudinal study. They

Question

4. A research team collected data on participants in a longitudinal study. They observed that the average extent of rebellious symptoms observed at age 12 was 10.54, with an observed standard deviation of 3.42 (the divisor in the underlying variance calculation was ). They observed that the average measure of aggression at age 8 was 3.54, with an observed standard deviation of 0.64 (the divisor in the underlying variance calculation was ). The Pearson product moment correlation coefficient between the two variables was 0.32. The research team seeks to estimate the regression of extent of rebellious symptoms at age 12 on aggression at age 8.

a. Find the estimated regression equation of extent of rebellious symptoms on the aggression measure. Find the 95% confidence interval for the slope in this equation. (15 points).

b. Complete the analysis of variance table for this regression and test the null hypothesis that the slope is zero at levels of significance 0.10, 0.05, and 0.01. (15 points)

c. Use the least-squares prediction equation to predict the extent of rebellious symptoms at age 12 that will be shown by a participant who has a measure of aggression at age 8 to 5.0. Give a 95% prediction interval for the extent of rebellious symptoms at age 12 for this participant. (20 points)

These are final answer as following, but I wanna detail and please do all questions!!!!

Answer: a. The dependent variable is extent of rebellious symptoms, and the independent variable is academic achievement. The OLS estimate of the slope is 1.71, and its estimated variance is 0.05695. The 95% confidence interval is 1.242 to 2.18. Note that the confidence interval excludes 0 by a wide margin, documenting that the association is significant at the 0.05 level. b. Analysis of Variance Table for Problem 1: Source Degrees of freedom Sum of Squares Mean Square F statistic Regression on academic achievement 1 540.168 540.168 51.35 Error 450 4734.909 10.52 Total 451 5275.0764 Reject the null hypothesis that the slope is zero at the 0.01 level (and the 0.05 and 0.10 levels as well). The F-statistic is 51.35 which is larger than 6.74 (the 0.01 critical value for (1, 240) degrees of freedom) and 6.63 (the 0.01 critical value for (1, ) degrees of freedom). Excel reports that the 0.01 critical value is 6.6616, which also is smaller than the F-statistic. c. The fitted value of rebellious symptoms at age 12 for a participant with academic achievement 6.0 at age 8 is 13.04. The 95% prediction interval for rebellious symptoms at age 12 for a participant with academic achievement 6.0 at age 8 is .

Explanation / Answer

Solution:

a) The dependent variable is extent of rebellious symptoms, and the independent variable is academic achievement. The OLS estimate of the slope is 1.71, and its estimated variance is 0.05695. The 95% confidence interval is 1.242 to 2.18. Note that the confidence interval excludes 0 by a wide margin, documenting that the association is significant at the 0.05 level.
b) Analysis of Variance Table for Problem 1:

Reject the null hypothesis that the slope is zero at the 0.01 level (and the 0.05 and 0.10 levels as well). The F-statistic is 51,35 which is larger than 6.74(the 0.01 critical value for (1,240) degrees of freedom) and 6.63 (the 0.01critical value for (1, ) degrees of freedom). Excel reports that the 0.01 critical value is 6.6616, which also is smaller than the F-statistic.

c) The fitted value of rebellious symptoms at age 12 for a participant with academic achievement 6.0 at age 8 is 13.04. The 95% prediction interval for rebellious symptoms at age 12 for a participant with academic 6,0 at age 8 is 13.04 ± 6.40.

Source Degrees of freedom Sum of Squares Mean Square F statistic Regression on academic achievement 1 540.168 540.168 51.35 Error 450 4734.909 10.52 Total 451 5375.0764

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