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11.8

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Found in: Page 639

### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208

# 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 $0.05$.

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.

See the 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:

$df=\left($ number of columns $-1\right)$

$⇒df=\left(5-1\right)$

$⇒df=4$

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

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

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

## Independence test statistic

The independence test statistic is provided below;

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

Apply the formula $=\left(\left(B4-B11{\right)}^{2}\right)/B11$ to cell B17 and drag the same formula up to cell F18 to get $\frac{\left(O-E{\right)}^{2}}{E}$. 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.