Question

The China Health and Nutrition Survey aims to examine the effects of the health, nutrition, and family planning policies and programs implemented by national and local governments. It, for example, 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.

Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
---|---|---|---|---|---|

education | 4 | 4142.09 | 1035.52 | 1.26 | 0.2846 |

Residuals | 794 | 653047.83 | 822.48 |

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

- Ho: μ
_{1}= μ_{2}= μ_{3}= μ_{4}

Ha: At least one of the means is different - Ho: μ
_{1}= μ_{2}= μ_{3}= μ_{4}= μ_{5}

Ha: At least one of the means is different - Ho: μ
_{1}= μ_{2}= μ_{3}= μ_{4}= μ_{5}

Ha: μ_{1}≠ μ_{2}≠ μ_{3}≠ μ_{4}≠ μ_{5} - Ho: μ
_{1}= μ_{2}= μ_{3}= μ_{4}= μ_{5}

Ha: At least one pair of means is the same

(b) Carry out the hypothesis test:

The value of the test statistic is: *(please round to
two decimal places)*

The p-value for this test is: *(please round to four
decimal places)*

Explain the conclusion of the test in the context of the study:

- Since p<α we fail to reject the null hypothesis
- Since p ≥ α we do not have enough evidence to reject the null hypothesis
- Since p ≥ α we accept the null hypothesis that all groups spend the same average number of hours per week taking care of their young children
- Since p<α we reject the null hypothesis and accept the alternative that at least one group does not spend the same average number of hours per week taking care of their young children
- Since p ≥ α we reject the null hypothesis and accept the alternative that not all groups spend the same average number of hours per week taking care of their young children

Answer #1

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

Ho will be all groups have same mean.

Ha will be that at least one group does not spend the same average number of hours per week taking care of their young children

5 levels of education.

- Ho: μ1 = μ2 = μ3 = μ4 = μ5

Ha: At least one of the means is different

2)

it is an one way anova test.

The test statistic = F

from the table, F = 1.26

**1.26**

The p-value for this test is:

it is denoted by "Pr(>F)" in table

P-value = 0.2846

**0.2846**

reject Ho if P-value < a

**Since p ≥ α we do not
have enough evidence to reject the null
hypothesis.**

we do not have sufficeint evidence to conclude that at least one group does not spend the same average number of hours per week taking care of their young children.

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