Chapter 14, Problem 36CP

Good friends and marital status Is the number of good friends associated with marital status? For GSS data with marital status measured with the categories (married, widowed, divorced, separated, never married), an ANOVA table reports F = 0.80 based on df₁ = 4, df₂ = 612.

1. a. Introduce notation and specify the null hypothesis and the alternative hypothesis for the ANOVA F test.
2. b. Based on what you know about the F distribution, would you guess that the test statistic value of 0.80 provides strong evidence against the null hypothesis? Explain.
3. c. Software reports a P-value of 0.53. Explain how to interpret it.

To determine

State the hypotheses and describe the notations.

Answer to Problem 36CP

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$

Alternative hypothesis:

$H_a:$ Not all means are equal.

Explanation of Solution

Given that $F=0.8,d.f_1=4\text{ and }d.f_2=612$.

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$

That is, all the means are equal.

Alternative hypothesis:

$H_a:$ Not all means are equal.

That is, at least one pair of means differ.

Here, ${\mu }_1$, ${\mu }_2$, ${\mu }_3$, ${\mu }_4$, and ${\mu }_5$ denote the mean number of good friends for married, widowed, divorced, separated, and never married.

To determine

Check whether test statistics value of 0.80 provide strong evidence against the null hypothesis.

Explanation of Solution

If F statistic is 1, then the null hypothesis will be accepted. If the F statistic is more than 1, then null hypothesis will be rejected.

Here, the value of F statistic is 0.8, which is close to 1. Thus, it does not provide strong evidence against the null hypothesis.

To determine

Interpret $\text{P}-\text{value}=\text{0}\text{.53}$.

Explanation of Solution

Rejection rule:

If $\text{P}-\text{value}\le \alpha$, reject $H_0$. Otherwise do not reject $H_0$.

Conclusion:

Here, the P-value is 0.53.

Here, $\text{P}-\text{value}>0.05$.

That is, P-value is greater than the 0.05.

Therefore, the null hypothesis is not rejected.

Thus, there is not enough evidence to conclude that at least one pair of population mean is not equal.