**1.** The doctor must have been figuring the probability of eight girls out of eight babies as roughly 1/100 (actually, (1/2)^8 = 1/256) or maybe the probability of the previous seven babies all being girls as (1/2)^7 = 1/128. The fallacious law of averages suggests that she is due for a boy, but it is wrong. The odds are still roughly 50-50. (If anything, this history of girl babies suggests that she may be biologically inclined to have girl babies.)

**2.** Under the null hypothesis, the P value is virtually zero, rejecting the null hypothesis decisively:

**3.** The calculation given is the probability of winning exactly once; the probability of winning at least once is

**4.** The vertical axis in the math graph runs from -2 to -12, while the vertical axis in the reading graph only goes from -1 to 5. By stretching the axis in the reading graph, the smaller gains are made to look comparable to those in math.

**5.** E[profit] = ($1000 - $10,000) P + $1,000(1 - P) = $1,000 - $10,000 implies P > 0 if P < 0.10.

**6.** Marilyn's correct answer is

There are four equally likely ways that a two-coin flip can land:

1) Heads & Heads | 3) Tails & Heads |

2) Heads & Tails | 4) Tails & Tails |

**7.** Converting to standardized z values:

**8.** Their z value is correct. For estimating the standard deviation of the difference in the sample proportions, they correctly used the pooled proportion, based on these sample data. However, they are not testing the null hypothesis that p = 0.74; they are testing the null hypothesis that p1 = p2.

**9.** The ANOVA F statistic is testing the null hypothesis that the population means are equal (that the average score on the midterm is equal to the average score on the final), while the simple regression model is testing the null hypothesis that there is not a linear relationship between the individual scores on the midterm and final (that a student who does well on the midterm is no more likely to do well on the final than is a student who does poorly on the midterm). Thus it might be the case that the average score on the midterm was 50 and the average score on the final was 90 (with small standard deviations), but that there was no relationship between who did well on one test and who did well on the other.

**10.** There is no reason why yes and no answers should be equally likely. Plus, this test doesn't get at the central question, whether women who are at women's colleges feel more successful than those at coeducational colleges. A more plausible test is whether the yes/no answers are independent of the college attended. The expected values, assuming independence, are as follows:

Women's College | Coed College | Total | |

Yes | 36.67 | 30.32 | 67 |

No | 15.32 | 12.67 | 28 |

Total | 52 | 43 | 105 |

**11.** An alternative explanation is that people who are healthy and have above-average life expectancies (and those who retire early) choose the annual-income option.

**12.** Using a contingency table with 1,000 people tested and quotation marks identifying the expert's assessment,

"Honest" | "Deceptive" | Total | |

Truthful | 456 | 344 | 800 |

Lying | 34 | 166 | 200 |

Total | 490 | 510 | 1000 |

A federal law enacted in 1988 prohibits the use of polygraph tests to screen job applicants or to test an employee unless there is other evidence of a specific misdeed by this employee. Exemptions are provided for businesses that involve public health and safety--e.g., pharmaceutical companies and security-guard firms. Polygraph results are not considered admissible evidence in federal courts and in about half of the states.

**13.** a. Here is the regression line (notice that the y-intercept is 0.2935):

c. There is not a perfect fit because the choice of college is affected by other factors (such as geography and special student interests) in addition to the U.S. News & World Report rank.

d. Yes the estimated coefficient of x have a plausible value; the 0.0293 coefficient implies that if the non-Pomona school's ranking increases by 1, say from 10 to 11, the fraction of these students who choose to attend Pomona will increase by 0.0293 (from y = 0.2935 + 0.0293(10) = 0.5865 to y = 0.2935 + 0.0293(11) = 0.6158 in our example), which is plausible.

e. Yes, the estimated coefficient of x is statistically significant at the 5 percent level: the t value is 0.0293/0.0065 = 4.51, which exceeds the cutoff for a two-tail test at the 5 percent level with 19 - 2 = 17 degrees of freedom. The null hypothesis is that the coefficient of x is 0; that is, that x has no effect on y.

f. The predicted value of y for x = 30 is 0.2935 + 0.0293(30) = 1.1725. This is incautious extrapolation, which is clearly inappropriate here since the value of y cannot possibly be larger than 1.

**14.** The 343 female Pomona graduates who obtained postgraduate degrees should be compared not to 1,191 (the total number of graduates), but to the number of female graduates, which is perhaps half this amount.

**15.** Their data are evidence against the null hypothesis that the population mean is zero; that is, that the average error is zero. The null hypothesis is not that every error is zero, but that the positive and negative errors balance out, giving an average error of zero.

**16.** a. The predicted value of y is 220.58 - 29.13(0) + 6.71(1) - 37.49(0) + 7.69(0) - 2.25(10) = 204.79.

b. Attendance on Thursday is estimated to be 37.49 persons lower than on Monday of that week.

**17.** The placebo effect that arises from hope for a cure evidently made many of these patients feel better.

**18.** Violent men may be less likely to find someone who will marry them.

**19.** P[accepted by at least one] = 1 - P[rejected by all three] = 1 - (2/3)^3 = 0.7037

**20.** We can use a chi-square test:

Observed | Expected | |

Females | 138 | (88/195)276 = 124.55 |

Males | 138 | (107/195)276 = 151.45 |

Total | 276 | 276 |