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Stat 205 Final Exam

Write clearly and show steps for partial credit. This exam is open book. There are six problems.

1. (20 points) A case-control study was designed to examine risk factors for cervical dysplasia(Becker et al., 1994); the women in the study were patients were aged 18 to 40. There weren1 = 175 cases with cervical dysplasia andn2 = 308 controls without. Each woman wastested for human papilloma virus (HPV) and either positive (HPV+) or negative (HPV):

dysplasia no dysplasia

HPV+ 164 130HPV 11 178total 175 308

Let p1 be the probability of HPV+ among the cases andp2 be the probability of HPV+among the controls.

(a) Estimate the differencep1 p2, the relative riskp1/p2 and the odds ratio = p1/(1p1)p2/(1p2)for these data and interpret.Answer:

p1 =164

175= 0.937, p2 =

130

308= 0.422,

p1 p2 = 0.515, p1p2

= 2.22, =p1/(1 p1)p2/(1 p2) = 20.4.

The probability of being HPV+ increases byone halfwhen dysplasia is present. Oneis twice as likely to have HPV when dysplasia is present. The odds of being HPV+increase by a factor of 20 when dysplasia is present.

(b) Find a 95% CI for the differencep1 p2, and testH0 : p1 = p2 at the 5% level.Answer:

p1 =164 + 1

175 + 2= 0.932, p2 =

130 + 1

308 + 2= 0.423,

SEp1p2 =

p1(1 p1)

n1 + 2+

p2(1 p2)n2 + 2

= 0.0338.

p1 p2 1.96SEp1p2 = (0.443, 0.576).We reject thatH0 : p1 = p2 at the 5% level because the 95% confidence interval doesnot include zero.

(c) Obtain the 95% CI for the odds ratio and interpret it in the context of these data. Doyou rejectH0 : = 1 at the 5% level? Why or why not? Is dysplasia associated withHPV?Answer:

log = log164 178130 11 = log 20.4 = 3.02.

SElog =

1

y1+

1

n1 y1 +1

y2+

1

n2 y2=

1

164+

1

11+

1

130+

1

178= 0.332.

The 95% confidence interval forlog is given by

3.02 1.96 0.332 = (2.37, 3.67).

The 95% confidence interval for is given by

(e2.37, e3.67) = (10.6, 39.1).

The odds of dysplasia increase by 10 to 40 times for HPV+ versus HPV. Also, theodds of HPV+ increase 10 to 40 times for those with dysplasia versus those without.We rejectH0 : = 1 at the 5% level because the 95% confidence interval does notinclude one. Therefore there is a significant association between dysplasia and HPV.

2. (20 points) Dental measurements were made onn1 = 11 girls andn2 = 16 boys, all 16years old. Each measurement is the distance (mm) from the center of the pituitary to thepterygomaxillary fissure. We want to test whether the distributions of these measurementsare the same for boys and girls. Here are the sorted data:

Girls Boys0 19.5 25.0 6.50 21.5 25.0 6.50 22.5 25.5 7.50 23.0 26.0 8.50 23.5 26.0 8.50 24.0 26.0 8.51 25.0 26.5 9.5

2.5 25.5 26.5 9.54.5 26.0 27.0 107 26.5 27.5 10

10.5 28.0 28.0 10.528.5 1129.5 1130.0 1131.0 1131.5 11

25.5 150.5

(a) Compute the Wilcoxin-Mann-Whitney countsK1 andK2, and the test statisticUs.Answer: Remember to add 0.5 for tied values.K1 = 25.5 andK2 = 150.5. Us =150.5, the bigger of the two.

(b) Check thatK1 + K2 = n1 n2.Answer:

K1 + K2 = 25.5 + 150.5 = 176 = 11 16 = n1 n2.(c) Compute the pvalue for the test, perhaps by bounding using Table 6 (pp. 680684).

Use a two-sided alternative.Answer:

0.001 < p-value < 0.002,

from Table 6 page 682.

(d) Do you reject that the distributions are the same at the 5% significance level? Why orwhy not?Answer:Yes, we reject that the distributions are the same at the 5% significance levelbecause the p-value is less than 0.05.

3. (20 points) A criminologist studying the relationship between level of education and crimerate in medium-sized U.S. counties collected data from a random sample ofn = 84 counties.Y is the crime rate (crimes per 100 people) andX is the percentage of individuals in thecounty having at least a high-school diploma. Heres a scatterplot:

60 65 70 75 80 85 90

24

68

1014

highschool

crim

es

For these data,

n = 84

y = 7.111

x = 78.60(xi x)(yi y) = 547.9

(xi x)2 = 3212SS(resid) = 455.3

(a) Compute the least squares estimatesb0 andb1 for regressing crime rate on percentageof high-school graduates.Answer:

b1 =

(xi x)(yi y)

(xi x)2 =547.93212

= 0.171.

b0 = y b1x = 7.111 (0.171)78.60 = 20.5.(b) Draw the regression line onto the scatterplot; describe the relationship.

Answer:Roughly linear, crime rates tend to decrease with more education.

(c) For Richland County,X = 85.2%. Predict Richlands crime rate.Answer:

20.5 0.171(85.2) = 5.9%.

(d) Compute a 95% confidence interval for1.Answer:

SEb1 =sy|x(xi x)2

=

SS(resid)/(n 2)

(xi x)2

=

455.3/(84 2)

3212= 0.0416.

b1 t0.975SEb1 = 0.171 1.99 0.0416 = (0.25,0.088).Note thatdf = 84 2 = 82 for the t distribution; I rounded down todf = 80 to usethe table in the back of the book.

(e) Do you rejectH0 : 1 = 0 at the 5% significance level? Why or why not? What doesthis tell you about the relationship between crime and education?Answer:We rejectH0 : 1 = 0 at the 5% significance level because a 95% confidenceinterval for1 does not include zero. There is a significant, linear association betweencrime rate and graduating high school.

4. (20 points) For the dental data in problem 2, assume that that at-test is appropriate and thatwe want to testH0 : 1 = 2 versusHA : 1 < 2 where1 is the mean for girls and2 isthe mean for boys. Formula (7.1) givesdf = 19.3.

Girls Boysn 11 16y 24.09 27.47s 2.44 2.09

(a) ComputeSEy1y2.Answer:

SEy1y2 =

s21n1

+s22n2

=

2.442

11+

2.092

16= 0.902.

(b) Find a 95% confidence interval for1 2.Answer:

y1 y2 t0.975SEy1y2 = 24.09 27.47 2.093(0.902) = (5.27,1.49).

I useddf = 19 in thet table in the back of the book.

(c) Compute the test statistic for testingH0 : 1 = 2.Answer:

ts =y1 y2SEy1y2

=24.09 27.47

0.902= 3.75.

(d) Find the pvalue for the alternativeHA : 1 < 2.Answer:Using thedf = 19 row in the back of the book,

0.0005 < p-value < 0.005.

(e) Do you rejectH0 : 1 = 2 in favor of HA : 1 < 2 at the 5% level? Why or whynot?Answer:We rejectH0 : 1 = 2 in favor ofHA : 1 < 2 at the 5% significance levelbecause the p-value is less than 0.05.

5. TRUE or FALSE. Each question is worth 2 points.

(a) TRUE or FALSE : A 95% confidence interval for1 2 contains 0 means we rejectH0 : 1 = 2 at the 5% level.

(b) TRUE or FALSE : The population median is always within one standard deviationfrom the population mean.

(c) TRUE or FALSE: For rare events, the odds ratio is approximately equal to the relativerisk.

(d) TRUE or FALSE: p-values are smaller for one-sided alternatives compared to two-sided alternatives.

(e) TRUE or FALSE : The central limit theorem tells us that sample means will be ap-proximately distributed binomial as the sample size increases.

(f) TRUE or FALSE : The sample proportionp always equals the true proportionp underrandom sampling.

(g) TRUE or FALSE: In a box-plot, the box contains 50% of the sample values.

(h) TRUE or FALSE : A andB are independent if and only ifPr(A andB) = Pr(A) +Pr(B).

(i) TRUE or FALSE : Alice rejectsH0. The only kind of error Alice could have made isa Type II error.

(j) TRUE or FALSE: The least squares regression line minimizes the residual sum ofsquares.

(k) TRUE or FALSE: (Extra credit) Thet test has its origins in the Guinness brewery.

6. (Extra credit): A study was performed to determine the effect of a carcinogen on the survivalof rats. Thirty rats were injected with varying levels of a carcinogenx (mg). For rats whosurvived one week,Yi = 1 was recorded; for rats who died within one week,Yi = 0 wasrecorded. A logistic regression was fit, and computer output is shown below:

Standard WaldParameter DF Estimate Error Chi-Square Pr > ChiSqIntercept 1 6.6656 2.3119 8.3129 0.0039x 1 -0.2506 0.0826 9.2070 0.0024

(a) (2 points) Does the carcinogen increase or decrease the odds of survival? Is the effectsignificant? Explain.Answer:The regression coefficient is negative, so increasing the carcinogen decreasesthe odds of survival. This makes intuitive sense. The effect is significant, we rejectH0 : 1 = 0 at the 5% level because0.0024 < 0.05.

(b) (3 points) Find a 95% confidence interval for how the odds of survival changes whenthe amount of carcinogen is increased by one milligram.Answer:A 95% confidence interval for the log odds ratio is

b1 1.96 SEb1 = 0.2506 1.96 0.0826 = (0.412,0.089).

Exponentiating gives the 95% confidence interval for how the odds change when in-creasing the carcinogen by 1 mg:

(e0.412, e0.089) = (0.66, 0.91).