load('data/brfss2.RData')
head(brfss)# A tibble: 6 × 5
genhlth gender age wtdesire weight
<fct> <fct> <int> <int> <int>
1 very good m 45 225 265
2 excellent m 24 150 150
3 excellent m 47 150 137
4 good f 26 125 159
5 very good f 33 125 145
6 very good f 28 120 1251. Observational data. Researchers measured existing health characteristics from a sample of U.S. adults; no treatment was assigned.
table(brfss$genhlth)
excellent very good good fair or poor
17 18 17 8 2. Ordinal: it is categorical because its values are qualitative, and the values are ordered because, e.g., good exceeds fair/poor.
3. The point estimate and standard error are:
\[ \begin{align*} \hat{p} &= \frac{17}{60} = 0.2833 \\ SE(\hat{p}) &= \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.2833\times 0.7167}{60}} = 0.0582 \end{align*} \]
Then, the CI is \(0.2833 \pm 2\times 0.0582\), which gives \((0.1669, 0.3997)\).
With 95% confidence, an estimated 16.7–40.0% of U.S. adults consider themselves to be in excellent health.
chi.out <- table(brfss$gender, brfss$genhlth) |> chisq.test()
chi.out
Pearson's Chi-squared test
data: table(brfss$gender, brfss$genhlth)
X-squared = 8.4002, df = 3, p-value = 0.03843chi.out$residuals
excellent very good good fair or poor
m 1.6687175 -0.2373222 -0.7733081 -0.9492889
f -1.6139905 0.2295390 0.7479468 0.91815624. The data provide evidence of an association between self-reported general health and gender (\(\chi^2\) = 8.4 on 3 degrees of freedom, p = 0.0384). The residuals indicate that more men and fewer women than expected under independence report themselves as being in excellent health.
table(brfss$gender, brfss$genhlth)
excellent very good good fair or poor
m 13 8 6 2
f 4 10 11 65. \(\frac{13}{29}\) men and \(\frac{4}{31}\) women, or 44.8% and 12.9%, respectively, report being in excellent health.
6. Rewrite the table as a 2×2: excellent vs. other rating.
| excellent | other.rating | |
|---|---|---|
| m | 13 | 16 |
| f | 4 | 27 |
The odds for men are \(\frac{13}{16}\), the odds for women are \(\frac{4}{27}\), so the odds ratio is \(\hat{\omega} = \frac{13/16}{4/27} = 5.48\).
For the interval:
With 95% confidence, men are between 1.48 and 20.23 times more likely than women to report being in excellent health.
fit <- aov(weight - wtdesire ~ genhlth, data = brfss)
summary(fit) Df Sum Sq Mean Sq F value Pr(>F)
genhlth 3 7395 2465 2.353 0.0818 .
Residuals 56 58657 1047
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 17. No, the data do not provide evidence that mean desired weight loss differs by self-reported general health (F = 2.35 on 3 and 56 df, p = 0.0818).
emmeans(fit, specs = ~ genhlth) |>
confint(adjust = 'bonferroni') genhlth emmean SE df lower.CL upper.CL
excellent 5.53 7.85 56 -14.73 25.8
very good 16.56 7.63 56 -3.13 36.2
good 34.47 7.85 56 14.21 54.7
fair or poor 14.38 11.40 56 -15.16 43.9
Confidence level used: 0.95
Conf-level adjustment: bonferroni method for 4 estimates 8. The mean desired weight loss among U.S. adults who rate themselves as being in excellent general health is estimated to be 5.53 lbs; with 95% confidence, this figure is estimated to be between a weight gain of 14.73 lbs and a weight loss of 25.8 lbs.
9. Good only. The conclusion of a 5% level two-sided test of \(H_0: \mu_i = 0\) can be inferred from whether the 95% intervals span zero. Only the good category has an interval estimate for mean desired weight loss that is strictly positive.
kruskal.test(weight - wtdesire ~ genhlth, data = brfss)
Kruskal-Wallis rank sum test
data: weight - wtdesire by genhlth
Kruskal-Wallis chi-squared = 6.0081, df = 3, p-value = 0.111210. The “good” group exhibits more variability than the others with respect to desired weight loss and includes two very large outliers. While the conclusions do not differ at the 5% level, they would at the 10% level.
mean(brfss$weight - brfss$wtdesire)[1] 18.21667sd(brfss$weight - brfss$wtdesire)[1] 33.4593611. The interval is \(18.22 \pm 2\times\frac{33.46}{\sqrt{60}} = (9.58, 26.86)\).
12. Right-skewed and unimodal, with a few large outliers.
13. It’s questionable. The sample size is large enough that the \(t\)-test may work fine, but the outliers are concerning and may affect results.
t.test(brfss$weight - brfss$wtdesire,
mu = 0, alternative = 'greater')
One Sample t-test
data: brfss$weight - brfss$wtdesire
t = 4.2172, df = 59, p-value = 4.311e-05
alternative hypothesis: true mean is greater than 0
95 percent confidence interval:
10.99824 Inf
sample estimates:
mean of x
18.21667 wilcox.test(brfss$weight - brfss$wtdesire,
mu = 0, alternative = 'greater')
Wilcoxon signed rank test with continuity correction
data: brfss$weight - brfss$wtdesire
V = 867.5, p-value = 9.799e-08
alternative hypothesis: true location is greater than 014. We want the greater alternative since a positive value for \(\mu\) means that actual weight exceeds desired weight. The nonparametric test is the better choice here (see previous solution) but either is acceptable.
The data provide evidence that the typical U.S. adult wishes to lose weight (signed rank test, p < 0.0001).
t.test(weight - wtdesire ~ gender, data = brfss)
Welch Two Sample t-test
data: weight - wtdesire by gender
t = -2.048, df = 38.875, p-value = 0.04735
alternative hypothesis: true difference in means between group m and group f is not equal to 0
95 percent confidence interval:
-33.4685580 -0.2066366
sample estimates:
mean in group m mean in group f
9.517241 26.354839 wilcox.test(weight - wtdesire ~ gender, data = brfss)
Wilcoxon rank sum exact test
data: weight - wtdesire by gender
W = 288.5, p-value = 0.01495
alternative hypothesis: true location shift is not equal to 015. The nonparametric test is more appropriate owing to the outliers. The data provide evidence that typical desired weight loss differs by gender (rank sum test, p = 0.0159).
cor(brfss$age, brfss$weight)[1] 0.0950206716. There is essentially no linear relationship between weight and age (\(r = 0.095\)).
fit <- lm(weight ~ age, data = brfss)
summary(fit)
Call:
lm(formula = weight ~ age, data = brfss)
Residuals:
Min 1Q Median 3Q Max
-68.744 -29.545 -8.087 17.796 228.828
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 159.5339 19.9751 7.987 6.45e-11 ***
age 0.3145 0.4327 0.727 0.47
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 49.23 on 58 degrees of freedom
Multiple R-squared: 0.009029, Adjusted R-squared: -0.008057
F-statistic: 0.5284 on 1 and 58 DF, p-value: 0.4702
17. The data provide no evidence of a relationship between mean weight and age (T = 0.727 on 58 df, p = 0.47).
18. Age explains an estimated 0.9% of variation in weight.
19. The estimate and standard error are shown in the output:
\[ \hat{\beta}_1 = 0.3145, \quad SE\left(\hat{\beta}_1\right) = 0.4327 \]
So the interval is \(0.3145 \pm 2\times 0.4327 = (-0.551, 1.180)\).
20. The prediction is \(159.53 + 0.31\times 35 = 170.38\) lbs.
load('data/chd.RData')
tbl.chd <- table(chd$smoking, chd$chd)
tbl.chd
CHD no CHD
nonsmoker 87 4913
smoker 84 291621. Among smokers, \(\frac{84}{3000} = 0.028\) (2.8%) developed CHD. Among nonsmokers, \(\frac{87}{5000} = 0.0174\) (1.74%) developed CHD.
22. The point estimate and standard error are:
\[ \begin{align*} \hat{p} &= \frac{84}{3000} = 0.028 \\ SE(\hat{p}) &= \sqrt{\frac{0.028\times 0.972}{3000}} = 0.0030 \end{align*} \]
Then, the CI is \(0.028 \pm 2\times 0.0030\), which gives \((0.022, 0.034)\). With 95% confidence, between 2.2% and 3.4% of smokers develop CHD.
riskratio(chd$smoking, chd$chd, rev = 'columns')[1:2]$data
Outcome
Predictor no CHD CHD Total
nonsmoker 4913 87 5000
smoker 2916 84 3000
Total 7829 171 8000
$measure
risk ratio with 95% C.I.
Predictor estimate lower upper
nonsmoker 1.000000 NA NA
smoker 1.609195 1.196452 2.16432523. Smokers are an estimated 1.61 times as likely to develop CHD compared with nonsmokers (RR = 1.61, 95% CI: 1.20 to 2.16). With 95% confidence, the risk of CHD among smokers is between 1.20 and 2.16 times that of nonsmokers; because the interval excludes 1, the data provide evidence of an association between smoking and CHD risk.
24. Efficacy is \(1 - \frac{1}{RR} = 1 - \frac{1}{1.61} = 0.379\), or about 38%. Not smoking reduces the risk of CHD by an estimated 38% compared with smoking.
metab <- ex0826
25. There is a strong, positive, linear trend: mammal species with larger log body mass tend to have larger log metabolic rates.
fit.metab <- lm(log(Metab) ~ log(Mass), data = metab)
summary(fit.metab)
Call:
lm(formula = log(Metab) ~ log(Mass), data = metab)
Residuals:
Min 1Q Median 3Q Max
-1.14216 -0.26466 -0.04889 0.25308 1.37616
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5.63833 0.04709 119.73 <2e-16 ***
log(Mass) 0.73874 0.01462 50.53 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.4572 on 93 degrees of freedom
Multiple R-squared: 0.9649, Adjusted R-squared: 0.9645
F-statistic: 2553 on 1 and 93 DF, p-value: < 2.2e-1626. The fitted model is:
\[ \widehat{\log(\text{Metab})} = 5.638 + 0.739 \times \log(\text{Mass}) \]
The slope indicates that a one-unit increase in log body mass is associated with an estimated 0.739-unit increase in log metabolic rate.
27. The estimate and standard error from the output are \(\hat{\beta}_1 = 0.7387\) and \(SE(\hat{\beta}_1) = 0.0146\). Then:
\[ 0.7387 \pm 2\times 0.0146 = (0.710, 0.768) \]
The interval includes 0.75, so the data are consistent with Kleiber’s law.
28. Substituting \(\log(10) \approx 2.30\):
\[ \widehat{\log(\text{Metab})} = 5.638 + 0.739 \times 2.30 = 7.34 \]
Back-transforming: \(\hat{y} = e^{7.34} \approx 1537\) kJ/day.