StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

11.8

Expert-verifiedFound in: Page 639

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.

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.

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.

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.

use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places car manufacturers are interested in whether there is a relationship between the size of the car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent).To test this, suppose that $800$ car owners were randomly surveyed with the results in Table $11.44$. Conduct a test of independence.

Family Size | Sub & Compact | Mid-size | Full-size | Van & Truck |

$1$ | $20$ | $35$ | $40$ | $35$ |

$2$ | $20$ | $50$ | $70$ | $80$ |

$3-4$ | $20$ | $50$ | $100$ | $90$ |

$5+$ | $20$ | $30$ | $70$ | $70$ |

**Table 11.44**

94% of StudySmarter users get better grades.

Sign up for free