AP Statistics |
Unit Test #9 |

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A dairy has developed a line of lactose-free products that are more tolerable to those who are lactose-intolerant. To assess the potential market for these products, the dairy commissioned a market research study of individuals in its sales area. A random sample of 250 individuals showed that 86 of them suffer from milk intolerance.

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1. |
State and check the conditions for constructing a 90% confidence interval for the population proportion that suffer from milk intolerance. |
(7) |

The sample must have been obtained randomly, the sample must be smaller than 10% of the population, and both of $n\widehat{p}$ and $n\left(1-\widehat{p}\right)$ must be at least 10. I’m told that the sample is a random sample. I’ll have to assume that there are over 2500 individuals in this market area. $n\widehat{p}=86>10$ and $n\left(1-\widehat{p}\right)=164>10$.

2. |
Regardless of your answer to [1], construct and interpret a 90% confidence interval for the population proportion that suffer from milk intolerance. |
(7) |

With 90% confidence, ${z}^{*}\approx 1.6449$. The interval is $0.344\pm 1.6449\sqrt{\frac{\left(0.344\right)\left(0.656\right)}{250}}\approx \left(0.2946,0.3934\right)$. I am 90% confident that the population proportion that suffer from lactose intolerance is between 29.46% and 39.34%.

3. |
What do we mean when we say that |
(5) |

If we took many samples and constructed an interval from each sample, then about 90% of those intervals ought to contain the population proportion that suffer from lactose intolerance.

4. |
A follow-up study is planned, and the researchers want to reduce the margin of error to 2%. How large of a sample will be required for this? |
(5) |

$0.02=1.6449\sqrt{\frac{\left(0.344\right)\left(0.656\right)}{n}}\Rightarrow n={\left(\frac{1.6449}{0.02}\right)}^{2}\left(0.344\right)\left(0.656\right)\approx 1526.359$; thus, a sample of 1527 ought to do.

An airline wants to evaluate the depth perception of its pilots over the age of fifty. A random sample of fourteen airline pilots over the age of fifty were asked to judge the distance between two markers placed 20 feet apart at the opposite end of the laboratory. The sample data listed here are the pilots’ error (recorded in feet) in judging the distance.

2.7 |
2.4 |
1.9 |
2.6 |
2.4 |
1.9 |
2.3 |

2.2 |
2.5 |
2.3 |
1.8 |
2.5 |
2.0 |
2.2 |

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5. |
State and check the conditions for constructing a 99% confidence interval for the population mean error in judging the distance. |
(8) |

The sample must have been obtained randomly, the sample must be smaller than 10% of the population, and the sample data must show no signs of strong skew and outliers. I’m told that this was a random sample. I’ll have to assume that at least 140 pilots over the age of 50 work for this airline. Here is a graph of the data:

The data show no signs of strong skew and outliers.

6. |
Regardless of your answer to [5], construct and interpret a 99% confidence interval for the population mean error in judging the distance. |
(7) |

With 99% confidence and 13 degrees of freedom, ${t}^{*}\approx 3.0123$. The interval is $2.2643\pm 3.0123\frac{0.279}{\sqrt{14}}\approx \left(2.0396,2.4889\right)$. I am 99% confident that the population mean error in judging distance is between 2.0396 feet and 2.4889 feet.

7. |
What do we mean when we say that |
(5) |

If we took many samples and constructed an interval from each sample, then about 99% of those intervals ought to contain the population mean error in judging the distance.

8. |
How large of a sample would be required if we wanted the margin of error to be half of its current value? |
(2) |

To halve the margin of error, quadruple the sample size—thus, 56.

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