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Q.1

Expert-verifiedFound in: Page 652

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation?

When the number of degrees of freedom for a chi-square distribution is $25,$ the population mean is $25$ and the standard deviation is $7.07.$

Given in the question that, We need to find the population mean and standard deviation if the number of degrees of freedom for a chi-square distribution is$25$.

The population mean in a chi-square distribution is given as

${\mu}_{{x}^{2}}=df$

and the standard deviation is calculated as

${\sigma}_{{x}^{2}}=\sqrt{2\times df}$

The degree of freedom is now set to$25$ in the question, and the population mean is determined as follows:

${\mu}_{{x}^{2}}=df$

$=25$

$\Rightarrow {\mu}_{{x}^{2}}=25$

Also known as standard deviation, it is determined as follows:

${\sigma}_{{x}^{2}}=\sqrt{2\times df}$

$=\sqrt{2\times 25}\phantom{\rule{0ex}{0ex}}=\sqrt{50}\phantom{\rule{0ex}{0ex}}=7.071\phantom{\rule{0ex}{0ex}}\approx 7.07$

$\Rightarrow {\sigma}_{{x}^{2}}=7.07$

Some travel agents claim that honeymoon hot spots vary according to the age of the bride. Suppose that $280$ recent brides were interviewed as to where they spent their honeymoons. The information is given in Table $11.46$. Conduct a test of independence

Location | $20-29$ | $30-39$ | $40-49$ | $50$ and over |

Niagara Falls | $15$ | $25$ | $25$ | $20$ |

Poconos | $15$ | $25$ | $25$ | $10$ |

Europe | $10$ | $25$ | $15$ | $5$ |

Virgin Islands | $20$ | $25$ | $15$ | $5$ |

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**

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