Question

4.42 Child care hours, Part II. Exercise 4.22 introduces the China Health and Nutrition Survey which, among other things, collects information on number of hours Chinese parents spend taking care of their children under age 6. The side by side box plots below show the distribution of this variable by educational attainment of the parent. Also provided below is the ANOVA output for comparing average hours across educational attainment categories.

150

100

50

0

college

lower middle school primary school

technical or vocational

F value Pr(>F) 1.26 0.2846

upper middle school

Df Sum Sq education 4 4142.09 Residuals 794 653047.83

Mean Sq 1035.52 822.48

1. (a) Write the hypotheses for testing for a difference between the average number of hours spent on child care across educational attainment levels.

2. (b) What is the conclusion of the hypothesis test?

Answer 1

Objective: To test whether there is a significant difference between the average number of hours spent on child care across educational attainment levels. Let denote the mean number of hours spend on child care with parent's educational attainment of 4 levels in ascending order respectively.

(a)

To test: At 5% level of significance, say,

Vs    Not all means are equal

The test statistic F ratio is obtained from the given table:  

Entering the given data in the ANOVA table:

We get F ratio = 1.26, with p-value = 0.2846

Comparing the p-value of the test with the significance level, since, p-value = 0.2846 > 0.05, we fail to reject H0. We do not have sufficient evidence to reject the null hypothesis as the probability of obtaining the result as obtained now, when the null hypothesis is true (p-value), is large, which implies that the result obtained might just be due to chance, there is no effect as such.

We may conclude that there is no significant difference between the average number of hours spent on child care across educational attainment levels based on the given data.