Chapter 14, Problem 37SE

The article “Birth Order and Political Success” (Psych. Reports, 1971: 1239–1242) reports that among 31 randomly selected candidates for political office who came from families with four children, 12 were firstborn, 11 were middle born, and 8 were last born. Use this data to test the null hypothesis that a political candidate from such a family is equally likely to be in any one of the four ordinal positions.

To determine

Test the null hypothesis that the political candidate from a family with four children is equally likely within one of the four positions.

Answer to Problem 37SE

There is sufficient evidence to conclude that political candidate from a family with four children is equally likely within one of the four positions.

Explanation of Solution

Given info:

A study conducted with 31 political candidates states the 12 candidates were firstborn, 11 candidates were middle born and 8 candidates were last born.

Calculation:

The claim is to test whether the political candidate from a family with four children is equally likely within one of the four positions. If the claim is rejected, then the political candidate from a family with four children is not equally likely within one of the four positions.

A family with four children has first, second, third and fourth child. For an equally likely event the probability would be $\frac{1}{4}=0.25$. But, there are three categories given firstborn, middle born and last born. Hence, the corresponding probabilities would be 0.25, $0.5\left(=0.25+0.25\right)$ and 0.25.

Testing the hypothesis:

Null hypothesis:

$H_0:p_1=0.25,p_2=0.50,p_3=0.25$

That is, the political candidate in a four children family is equally likely to occur.

Alternative hypothesis:

$H_a:$ At least one of the observed proportions differs from the expected proportion.

That is, the observed proportion for at least one category is not equal to the expected proportion.

Expected frequency:

The expected frequency for each group is calculated as follows,

$\text{Expected frequency}=n{p}_i$

Where,

n is the total number of observed frequency.

$p_i$ is the proportion corresponding to a particular group.

The expected frequency for the first category:

$\begin{array}{c}\text{Expected frequency}=\left(31\right)\left(0.25\right)\\ =7.75\end{array}$

The expected frequency for the second category:

$\begin{array}{c}\text{Expected frequency}=\left(31\right)\left(0.50\right)\\ =15.5\end{array}$

The expected frequency for the third category:

$\begin{array}{c}\text{Expected frequency}=\left(31\right)\left(0.25\right)\\ =7.75\end{array}$

Test statistic:

Software procedure:

Step-by-step procedure to obtain the test statistic using MINITAB is given below:

• Choose Stat > Tables > Chi-Square Goodness-of-Fit Test (One Variable).
• In Observed counts, enter the column of Number of political candidates.
• In Category names, enter the column of Birth order.
• Under Test, select the column of Expected frequency in Proportions specified by historical counts.
• Click OK.

Output obtained from MINITAB is given below:

Decision rule:

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}<\alpha$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

The P-value is 0.162 and the level of significance is 0.10.

The P-value is greater than the level of significance.

That is, $0.162\left(=P\text{-value}\right)>0.10\left(=\alpha \right)$

Hence, the null hypothesis is not rejected.

Thus, there is sufficient evidence to conclude that the political candidate from a family with four children is equally likely within one of the four positions.