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Chapters 23 – 25 Homework answers

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Problem 8, p.598 Percent of 15-year-olds drunk at least twice. We need a 95% confidence interval for the difference between males and females. Plan: This is matched-pairs sample, paired off by country. A histogram of the differences shows no outliers or strong skewness, so we can safely say that the distribution is plausibly normal.

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More problem 8 Plan: It is not clear how the individuals in the sample were selected, or if the same method of selection was used in each country, but we shall assume that the samples are at least representative of the populations they come from. In each case, we’ll assume that the samples are less than 10% of all 15-year-olds in the country.

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More problem 8 Mechanics and conclusion: We are 95% confident that the percentage of boys drunk at least twice is between 4.5% and 11.4% more than girls in these countries.

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Problem 10, p.598 The petition problem – how many valid signatures are there? 88.6% of the sample signatures were valid. In order to have the initiative certified, 82.2% of the 304,266 signatures must be valid.

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More problem 10 We start with the hypothesis that there are not enough signatures. A type I error would mean that they certify the initiative when there are not enough signatures. d) A type II error would mean that the initiative is not put on the ballot when they actually had enough signatures.

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More problem 10 e) Complete a significance test here: what should we use? f) To increase the power of the test, the commission could increase the sample size.

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**Problem 34 Home burglary losses The margin of error is $373.50**

They are 95% confident that the average loss in a home burglary is between $1644 and $2391. In repeated samples, about 95% of our confidence intervals will contain/capture/hold the true mean loss.

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**Problem 28 Petal length of irises**

From the min, lower quartile, median, upper quartile, and max, you can make box plots… From the boxplots, it’s clear that versicolor has longer petals than virginica. The 95% C.I is (10.14, 14.46) for the difference

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**More problem 28 Petal length of irises**

We are 95% confident that the average petal length for versicolor is between and mm longer than that of virginica irises. Since the confidence interval does not include 0, we have strong evidence of a difference in petal length.

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**Problem 30 Speeding and racial bias**

This question calls for a one-proportion z-test, one-tailed. You should get a test-statistic of z = There is a strong evidence that proportion of tickets given to blacks on this section of the New Jersey Turnpike is unusually high. Significance tests NEVER prove anything.

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More problem 30 Speeding and racial bias c) One statistic that would be useful is the proportion of drivers on that section of turnpike who are black.

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**Problem 32 Frito bags Hypotheses: Ho: μ = 35.4 Ha: μ < 35.4**

b) There is one high value, but it is not technically an outlier. Otherwise, it looks roughly normal. We don’t have information on random sampling, but it’s probably a representative group. Clearly, 6 bags are less than 10% of the population.

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**More problem 32 Frito bags**

t = and p-value = With such a large p-value, we do NOT reject Ho. There is no evidence that the bags are underweight. Without the high “outlier,” we have t = and a p-value of , still not significant. The stated weight seems legitimate on these 6 bags.

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Problem 36 Hamsters We are 90% confident that the true mean litter size for hamsters is between 7.11 and 8.33 pups. To be more confident we need a larger interval. For a margin of error of 1, about 27 litters should be tested, based on t(24).

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CHAPTER 18: Inference about a Population Mean

CHAPTER 18: Inference about a Population Mean

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