Question

Do families and singles have the same distribution of cars? Use a level of significance of $0.05$. Suppose that $100$randomly selected families and $200$randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table $11.20.$Do families and singles have the same distribution of cars? Test at a level of significance of $0.05$.

Short answer

The alpha value has been set at $0.05$. Because $p-value<\alpha$ , the null hypothesis, $H_0$, will be rejected. As a result, the null hypothesis is rejected, whereas the alternative hypothesis is accepted. As a result, there is sufficient data to establish that the distribution of cars among families and singles is not equal.

Step-by-step solution

Given information

Given in the question that, use a level of significance of $0.05$. Suppose that $100$randomly selected families and $200$randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table 11.20. We need to test at a$0.05$ significance level that whether the families and singles have the same distribution of cars.

Explanation

The null hypothesis is as follows:

$H_0$: Car distribution is the same for families and singles.

The alternate hypothesis is as follows:

$H_a$: Car ownership is not evenly distributed among families and singles.

The following formula can be used to calculate degrees of freedom:

$df=($number of columns $-1)$

$\Rightarrow df=(5-1)$

$\Rightarrow df=4$

The table of observed values has already been provided. Let's now use the formula below to determine the predicted frequencies:

$E=\frac{\text{(row total}\left)\right(\text{column total})}{\text{overall total}}$

Let's use Excel to determine the expected (E) values, as shown below.

Independence test statistic 

The independence test statistic is provided below;

$=\sum _{i\times j}=\frac{(O-E{)}^2}{E}$ is a test statistic.

Apply the formula $=\left((B4-B11{)}^2\right)/B11$to cell B17 and drag the same formula up to cell F18 to get $\frac{(O-E{)}^2}{E}$. After that, add up the totals of the columns and rows. The following is a table of test statistics:

As a result, the test statistic is$62.912$

In Excel, the p-value can be determined using the CHIDIST () formula, as illustrated below:

As a result, the $p$value is zero.