Chapter 22, Problem 1E

a.

To determine

Find the expected number of birth in each season if there is no “seasonal effect” on births.

Answer to Problem 1E

The expected number of birth in each season is 30 births per season.

Explanation of Solution

Given info:

The data represents the census of 120 students who are born in different season namely summer with 32, winter with 25, spring with 35 and fall with 28.

Calculation:

The general formula for expected count:

$\text{Expected count}=\text{observed frequency}\times \text{percentage of each season}$.

The expected count for each seasonal value can be obtained by the product of observed frequency with the percentage of each season:

Let observed value is the value of total students in the class which is 120 and percentage value is 0.25(25 percent) which does not change due to seasonal effect in the data.

$\begin{array}{c}\text{Expected count for summer}=\text{observed frequency}\times \text{percentage of summer}\\ =120\times 0.25\\ =30\end{array}$

Therefore, the remaining expected count for each season as follows:

Season	Expected count
Winter	30
Summer	30
Spring	30
Fall	30

Thus, the expected number of birth in each season is 30 births per season.

b.

To determine

Compute the ${\chi }^2$statistic value.

Answer to Problem 1E

The value of ${\chi }^2$ is 1.933.

Explanation of Solution

Calculation:

The test statistic can be obtained as follows:

$\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(\text{observed}-\text{Expected}\right)}^2}{\text{Expected}}}\\ =\frac{{\left(25-30\right)}^2}{30}+\frac{{\left(35-30\right)}^2}{30}+\frac{{\left(32-30\right)}^2}{30}+\frac{{\left(28-30\right)}^2}{30}\\ \simeq 1.933\end{array}$

Thus the value of test statistic is 1.933.

c.

To determine

Compute degrees of freedom for the ${\chi }^2$statistic value.

Answer to Problem 1E

The degree of freedom for the test statistic is 3.

Explanation of Solution

Calculation:

Degrees of freedom:

Here, there are four seasons.

The degrees of freedom is,

$\begin{array}{c}\text{Number of seasons}-1=4-1\\ =3\end{array}$

Thus, the degree of freedom for the test statistic is 3.