The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12.

$\begin{array}{|ll|}\hline \text{Number of Pets}& \text{Percent}\\ 0& 18\\ 1& 25\\ 2& 30\\ 3& 18\\ 4+& 9\\ \hline\end{array}$

A random sample of $1,000$students from the Eastern United States resulted in the data in Table 11.13.

$\begin{array}{|ll|}\hline \text{Number of Pets}& \text{Frequency}\\ 0& 210\\ 1& 240\\ 2& 320\\ 3& 140\\ 4+& 90\\ \hline\end{array}$

At the $1\%$ significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p-value?

The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12.

$\begin{array}{|ll|}\hline \text{Number of Pets}& \text{Percent}\\ 0& 18\\ 1& 25\\ 2& 30\\ 3& 18\\ 4+& 9\\ \hline\end{array}$

A random sample of $1,000$students from the Eastern United States resulted in the data in Table 11.13.

$\begin{array}{|ll|}\hline \text{Number of Pets}& \text{Frequency}\\ 0& 210\\ 1& 240\\ 2& 320\\ 3& 140\\ 4+& 90\\ \hline\end{array}$

The total frequencies have now been determined to be $1000$. Now, multiply each expected frequency by $1000$to get the following table;

$\begin{array}{|lll|}\hline \text{Number of Pets}& \text{Percent}& \text{Expected Frequency}\\ 0& 18& 18\times 1000=18000\\ 1& 25& 25\times 1000=25000\\ 2& 30& 30\times 1000=30000\\ 3& 18& \begin{array}{l}18\times 1000=18000\\ 18\times 1000=18000\end{array}\\ 4+& 9& 9\times 1000=9000\\ \hline\end{array}$

The null hypothesis is expressed as follows:

$H_0$: Students in the Eastern United States had the same "number of pets" distribution as the student population in the United States as a whole.

And the alternative hypothesis is expressed as follows:

$H_a$: The "number of pets" distribution among students in the Eastern United States differs from the national distribution.

Let's do the calculation in Excel, and the p-value may be found by using the CHIDIST () formula as shown below:

$\therefore {\chi }^2\text{test statistic}=98010.156\approx 98010.16$

$p-\text{value}=0$

The null hypothesis is rejected, and it can be inferred that there is enough data to conclude that the distribution of "number of pets" among students in the Eastern United States differs from that of the entire United States student population.