AP Statistics |
Unit Test #3 |

_____ / 45

Category |
Whole |
Universal |
ART |
Other |

Policies |
320 |
280 |
240 |
160 |

1. |
Over the past 5 years, an insurance company has had a mix of 40% whole life policies, 20% universal life policies, 25% annual renewable-term (ART) policies, and 15% other types of policies. A change in this mix over the long haul could require a change in the commission structure, reserves, and possibly investments. A sample of 1,000 policies issued over the last few months gave the results shown here. Use these data to test the hypothesis that there has been a change in the distribution of policy types at this insurance company. Use $\alpha =0.05$. |
(15) |

${H}_{0}:$ the distribution of policy types is 40% whole life, 20% universal life, 25% ART and 15% other (as they have been in the past)

${H}_{a}:$ the distribution of policy types has changed from past values in some way.

This requires that the sample was obtained randomly, that the sample is smaller than 10% of the population, and that all expected counts are at least five. I’ll have to assume that the sample was obtained randomly. Furthermore, I’ll have to assume that this company has issued at least 10,000 policies. The expected counts are as follows…

Category |
Whole |
Universal |
ART |
Other |

Policies |
$\left(1000\right)\left(0.4\right)=400$ |
$\left(1000\right)\left(0.2\right)=200$ |
$\left(1000\right)\left(0.25\right)=250$ |
$\left(1000\right)\left(0.15\right)=150$ |

All expected counts are at least five. The test statistic is ${\chi}^{2}=\frac{{\left(320-400\right)}^{2}}{400}+\frac{{\left(280-200\right)}^{2}}{200}+\frac{{\left(240-250\right)}^{2}}{250}+\frac{{\left(160-150\right)}^{2}}{150}\approx 49.0667$. With three degrees of freedom, $P\left({\chi}^{2}>49.0667\right)\approx 0$.

If the distribution of policy types is 40% whole life, 20% universal life, 25% ART and 15% other, then I can expect a sample of 1000 policies to be this far (or farther) away from the expected proportions in almost no trials. Since $0<0.05$, this is significant at the 5% level. I reject the null hypothesis. The data provide evidence that there has been a change in the distribution of policy types.

Opinion |
Favor |
Indifferent |
Oppose |

Already Union Member |
140 |
42 |
18 |

Non-Member |
70 |
198 |
132 |

2. |
A speaker who advises managers on how to avoid being unionized claims that worker opinions about unions are independent of actual union membership. A random sample of 600 industrial workers yielded the data shown above. Test the speaker’s claim at the 10% level of significance. |
(15) |

${H}_{0}:$ Union membership and opinion are independent

${H}_{a}:$ Union membership and opinion are not independent

This requires that the sample was obtained randomly, that the sample is smaller than 10% of the population, and that all expected counts are at least five. I’m told that the sample was obtained randomly. I’m certain that there are at least 6000 industrial workers that could be interviewed. The expected counts are as follows…

Opinion |
Favor |
Indifferent |
Oppose |

Already Union Member |
$\frac{\left(200\right)\left(210\right)}{600}=70$ |
$\frac{\left(200\right)\left(240\right)}{600}=80$ |
$\frac{\left(200\right)\left(150\right)}{600}=50$ |

Non-Member |
$\frac{\left(400\right)\left(210\right)}{600}=140$ |
$\frac{\left(400\right)\left(240\right)}{600}=160$ |
$\frac{\left(400\right)\left(150\right)}{600}=100$ |

All expected counts are at least five. The test statistic is ${\chi}^{2}=\frac{{\left(140-70\right)}^{2}}{70}+\frac{{\left(42-80\right)}^{2}}{80}+\frac{{\left(18-50\right)}^{2}}{50}+\frac{{\left(70-140\right)}^{2}}{140}+\frac{{\left(198-160\right)}^{2}}{160}+\frac{{\left(132-100\right)}^{2}}{100}\approx 162.795$. With two degrees of freedom, $P\left({\chi}^{2}>162.795\right)\approx 0$.

If Union membership and opinion are independent, then I can expect to find a sample of 1000 workers with a distribution that is this far (or farther) from expected in almost no trials. Since $0<0.10$, this is significant at the 10% level. I reject the null hypothesis. The data provide evidence that Union membership and opinion are not independent.

Incidence |
Control |
Low Dose |
High Dose |

Tumors Present |
10 |
14 |
19 |

No Tumors |
90 |
86 |
81 |

3. |
A carcinogenicity study was conducted to examine the tumor potential of a drug product scheduled for initial testing in humans. A total of 300 rats were studied for a 6-month period. At the beginning of the study, 100 rats were randomly assigned to the control group, 100 to the low-dose group, and the remaining 100 to the high-dose group. On each day of the study, the rats in the control group received an injection of an inert solution, whereas those in the drug groups received an injection of the solution plus drug. The sample data are shown in the accompanying table. Do these data provide evidence at the 1% level of significance of an association between the amount of the drug and the presence of tumors in the rats? |
(15) |

${H}_{0}:$ the proportion of rats with tumors present is the same in all three dosage groups

${H}_{a}:$ the proportion of rats with tumors present is not the same in all three dosage groups

This requires that the rats were randomly assigned to the three dosage groups and that all expected counts are at least five. I’m told that the rats were randomly assigned to one of the three groups. The expected counts are as follows…

Incidence |
Control |
Low Dose |
High Dose |

Tumors Present |
$\frac{\left(43\right)\left(100\right)}{300}\approx 14.33$ |
$\frac{\left(43\right)\left(100\right)}{300}\approx 14.33$ |
$\frac{\left(43\right)\left(100\right)}{300}\approx 14.33$ |

No Tumors |
$\frac{\left(257\right)\left(100\right)}{300}\approx 85.67$ |
$\frac{\left(257\right)\left(100\right)}{300}\approx 85.67$ |
$\frac{\left(257\right)\left(100\right)}{300}\approx 85.67$ |

All expected counts are at least five. The test statistic is ${\chi}^{2}=\frac{{\left(10-14.33\right)}^{2}}{14.33}+\frac{{\left(14-14.33\right)}^{2}}{14.33}+\frac{{\left(19-14.33\right)}^{2}}{14.33}+\frac{{\left(90-85.67\right)}^{2}}{85.67}+\frac{{\left(86-85.67\right)}^{2}}{85.67}+\frac{{\left(81-85.67\right)}^{2}}{85.67}\approx 3.3119$. With two degrees of freedom, $P\left({\chi}^{2}>3.3119\right)\approx 0.1909$.

If the proportion of rats with tumors present is the same in all three dosage groups, then I can expect to find proportions that are this different (or more different) in about 19.09% of trials. Since $0.1909>0.01$, this is not significant at the 1% level of significance. I fail to reject the null hypothesis. The data do not provide evidence that the proportion of rats with tumors present is not the same in all three dosage groups.

Page last updated 14:20 2021-03-29