Question 1: Bioavailability of Two Types of Aspirin

Each of the 10 subjects received both type A and type B aspirin (one week apart, with the order randomized). The two measurements therefore come from the same person and are dependent, which calls for a paired analysis.

The data

   person aspirinA aspirinB difference
1       1       15       13          2
2       2       26       20          6
3       3       13       10          3
4       4       28       21          7
5       5       17       17          0
6       6       20       22         -2
7       7        7        5          2
8       8       36       30          6
9       9       12        7          5
10     10       18       11          7
            Variable Mean       SD  n
1          Aspirin A 19.2 8.625543 10
2          Aspirin B 15.6 7.777460 10
3 Difference (A - B)  3.6 3.098387 10

The mean within-person difference is 3.6 mg% with a standard deviation of 3.098 mg%.

Eight of the ten subjects show a decrease from A to B, one shows no change, and one shows an increase — the paired structure is clearly informative.

(a) Statement of the hypotheses

Let \(d_i = (\text{Aspirin A concentration})_i - (\text{Aspirin B concentration})_i\) for subject \(i\), and let \(\mu_d\) denote the true mean difference.

\[H_0: \mu_d = 0 \qquad \text{(the mean concentrations of the two drugs are the same)}\] \[H_1: \mu_d \neq 0 \qquad \text{(the mean concentrations of the two drugs differ)}\]

Because the question asks whether the means are the same, this is a two-sided test. The test statistic is

\[t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} \sim t_{n-1} = t_{9} \quad \text{under } H_0.\]

(b) Paired t-test


    Paired t-test

data:  aspirinA and aspirinB
t = 3.6742, df = 9, p-value = 0.005121
alternative hypothesis: true mean difference is not equal to 0
95 percent confidence interval:
 1.383548 5.816452
sample estimates:
mean difference 
            3.6 

    One Sample t-test

data:  aspirin$difference
t = 3.6742, df = 9, p-value = 0.005121
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 1.383548 5.816452
sample estimates:
mean of x 
      3.6

Results.

  • \(\bar{d} = 3.6\) mg%, \(s_d = 3.098\) mg%
  • \(t = 3.6742\) with \(df = 9\)
  • p-value \(= 0.005121\)
  • 95% CI for \(\mu_d\): (1.384, 5.816)

Conclusion. Since the p-value (0.00512) is less than \(\alpha = 0.05\), we reject \(H_0\). There is statistically significant evidence that the mean 1-hour urine concentrations of the two aspirin types are not the same. On average, type A produces a concentration about 3.6 mg% higher than type B, and the 95% confidence interval excludes zero (it lies entirely above 0), which is consistent with rejecting the null hypothesis.


    Shapiro-Wilk normality test

data:  aspirin$difference
W = 0.91574, p-value = 0.3228

The Shapiro-Wilk test does not reject normality of the differences, so the paired t-test is appropriate.

(c) What if a two-sample t-test had been used instead?


    Two Sample t-test

data:  aspirinA and aspirinB
t = 0.9802, df = 18, p-value = 0.34
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -4.116103 11.316103
sample estimates:
mean of x mean of y 
     19.2      15.6

p-value \(= 0.34\) with \(t = 0.9802\) and \(df = 18\).

(For completeness, Welch’s unequal-variance version gives essentially the same answer:)

[1] 0.3401157

Comparison and interpretation.

Analysis t df p-value Decision at \(\alpha = 0.05\)
Paired t-test 3.674 9 0.005121 Reject \(H_0\)
Two-sample t-test 0.98 18 0.34 Fail to reject \(H_0\)

The two analyses lead to opposite conclusions. The estimated mean difference is identical in both cases (19.2 − 15.6 = 3.6 mg%), but the standard errors are very different. The two-sample test treats the 20 measurements as independent, so its standard error is driven by the large between-person variability (subject concentrations range from 5 to 36 mg%). The paired test removes that person-to-person variability by working with the within-person differences, whose standard deviation is only 3.1 mg% compared with roughly 8.6 and 7.8 mg% for the individual groups.

Ignoring the pairing therefore throws away the design’s main advantage and loses a great deal of power; it is also statistically invalid here because the independence assumption of the two-sample t-test is violated. The paired t-test is the correct analysis.

Question 2: Active Exercise and Age at Independent Walking

Twelve one-week-old male infants were randomly allocated to one of two groups, so the two samples are independent.

The data

    time  group
1   9.50 Active
2  10.00 Active
3   9.75 Active
4   9.75 Active
5   9.00 Active
6  13.00 Active
7  11.50   None
8  12.00   None
9  13.25   None
10 11.50   None
11 13.00   None
12  9.00   None
            Group n     Mean Median       SD Min   Max
1 Active Exercise 6 10.16667   9.75 1.428869   9 13.00
2     No Exercise 6 11.70833  11.75 1.520005   9 13.25

(a) Statement of the hypotheses

The question is directional (is the mean time to walk less for the Active Exercise group?), so a one-sided test is used. Because we will analyze the data with a rank-based method, the hypotheses are most properly stated in terms of the distributions / locations of walking times.

  • \(H_0\): the distribution of walking times is the same in the two groups (\(M_{\text{Active}} = M_{\text{None}}\); no shift in location)
  • \(H_1\): walking times are shifted lower in the Active Exercise group (\(M_{\text{Active}} < M_{\text{None}}\))

Stated in terms of means, \(H_0: \mu_{\text{Active}} = \mu_{\text{None}}\) versus \(H_1: \mu_{\text{Active}} < \mu_{\text{None}}\), with \(\alpha = 0.05\).

(b) Why use a non-parametric method?

  1. Very small sample size. With only \(n = 6\) per group there is not enough information to verify the normality assumption of the t-test, and the Central Limit Theorem offers no protection at this sample size.
  2. Evidence against normality / presence of outliers. The Active Exercise group contains five values clustered between 9.00 and 10.00 months plus a single value of 13.00; the No Exercise group similarly contains a low outlier of 9.00. Both groups look skewed, and a single extreme observation has a large influence on the mean and variance in samples this small.
  3. Unequal spread. The sample standard deviations (1.43 vs. 1.52 months) come from distributions of quite different shape, which also strains the pooled-variance assumption.
  4. Robustness. The Mann-Whitney U (Wilcoxon rank-sum) test uses only the ranks of the observations, so it requires only independent samples and continuous, similarly shaped distributions — no normality — and it is far less sensitive to the outliers noted above.

    Shapiro-Wilk normality test

data:  active
W = 0.72061, p-value = 0.01012

    Shapiro-Wilk normality test

data:  none
W = 0.88521, p-value = 0.2939

(c) Mann-Whitney U test in R


    Wilcoxon rank sum test with continuity correction

data:  time by group
W = 9, p-value = 0.08523
alternative hypothesis: true location shift is less than 0
   group rank
1 Active   30
2   None   48

Because there are tied observations (9.00, 9.75 and 13.00 each appear twice), R cannot compute an exact p-value and uses the normal approximation with a continuity correction. For comparison, the exact test obtained by ignoring the ties is:

[1] 0.08523293

Results.

  • \(W = 9\) (sum of ranks in the Active group \(= 30\); No Exercise \(= 48\))
  • p-value \(= 0.08523\) (exact version ignoring ties: 0.0898)
  • Median walking age: 9.75 months (Active) vs. 11.75 months (No Exercise)

Conclusion. The p-value (0.0852) is greater than \(\alpha = 0.05\), so we fail to reject \(H_0\). At the 5% level there is not sufficient evidence to conclude that the mean (median) time to walk independently is less for infants who receive active exercise than for those who do not. The observed medians do differ in the hypothesized direction (9.75 vs. 11.75 months), and the result would be significant at \(\alpha = 0.10\), but with only six infants per group the study has very limited power. A larger trial would be needed before drawing a conclusion about the effect of active exercise.


    Two Sample t-test

data:  time by group
t = -1.8102, df = 10, p-value = 0.05019
alternative hypothesis: true difference in means between group Active and group None is less than 0
95 percent confidence interval:
        -Inf 0.001957879
sample estimates:
mean in group Active   mean in group None 
            10.16667             11.70833

Note that the parametric t-test lands almost exactly on the boundary (\(p \approx 0.050\)) while the rank-based test gives \(p \approx 0.086\); the difference is driven almost entirely by the single 13.00-month observation in the Active group, which illustrates precisely why the non-parametric approach is the safer choice here.

knitr::opts_chunk$set(echo = TRUE, message = FALSE, warning = TRUE,
                      fig.align = "center", fig.width = 6, fig.height = 4)

person   <- 1:10
aspirinA <- c(15, 26, 13, 28, 17, 20,  7, 36, 12, 18)
aspirinB <- c(13, 20, 10, 21, 17, 22,  5, 30,  7, 11)

aspirin <- data.frame(person, aspirinA, aspirinB,
                      difference = aspirinA - aspirinB)
aspirin
summary_stats <- data.frame(
  Variable = c("Aspirin A", "Aspirin B", "Difference (A - B)"),
  Mean     = c(mean(aspirinA), mean(aspirinB), mean(aspirin$difference)),
  SD       = c(sd(aspirinA),   sd(aspirinB),   sd(aspirin$difference)),
  n        = c(10, 10, 10)
)
summary_stats

plot(rep(1:2, each = 10), c(aspirinA, aspirinB),
     xaxt = "n", xlim = c(0.8, 2.2),
     xlab = "", ylab = "1-hour urine concentration (mg%)",
     main = "Paired observations by subject", pch = 19, col = "steelblue")
axis(1, at = 1:2, labels = c("Aspirin A", "Aspirin B"))
segments(1, aspirinA, 2, aspirinB, col = "grey60")

paired_test <- t.test(aspirinA, aspirinB, paired = TRUE,
                      alternative = "two.sided", conf.level = 0.95)
paired_test

t.test(aspirin$difference, mu = 0)

shapiro.test(aspirin$difference)
qqnorm(aspirin$difference, main = "Normal Q-Q plot of the differences")
qqline(aspirin$difference, col = "red")

two_sample_test <- t.test(aspirinA, aspirinB, paired = FALSE,
                          var.equal = TRUE, alternative = "two.sided")
two_sample_test

t.test(aspirinA, aspirinB, paired = FALSE, var.equal = FALSE)$p.value

active <- c( 9.50, 10.00,  9.75,  9.75,  9.00, 13.00)   # Active Exercise
none   <- c(11.50, 12.00, 13.25, 11.50, 13.00,  9.00)   # No Exercise

walking <- data.frame(
  time  = c(active, none),
  group = factor(rep(c("Active", "None"), each = 6),
                 levels = c("Active", "None"))   # Active first => "less" is correct
)
walking

data.frame(
  Group  = c("Active Exercise", "No Exercise"),
  n      = c(length(active), length(none)),
  Mean   = c(mean(active),   mean(none)),
  Median = c(median(active), median(none)),
  SD     = c(sd(active),     sd(none)),
  Min    = c(min(active),    min(none)),
  Max    = c(max(active),    max(none))
)

boxplot(time ~ group, data = walking,
        ylab = "Age at independent walking (months)", xlab = "Group",
        main = "Walking age by treatment group",
        col = c("lightblue", "lightpink"))
stripchart(time ~ group, data = walking, vertical = TRUE, method = "jitter",
           pch = 19, col = "grey30", add = TRUE)

shapiro.test(active)
shapiro.test(none)

mwu <- wilcox.test(time ~ group, data = walking,
                   alternative = "less",   # Active < None
                   exact = FALSE,          # ties present => normal approximation
                   correct = TRUE)
mwu

walking$rank <- rank(walking$time)
aggregate(rank ~ group, data = walking, FUN = function(x) c(sum = sum(x)))

suppressWarnings(
  wilcox.test(active, none, alternative = "less", exact = TRUE)$p.value
)

t.test(time ~ group, data = walking, alternative = "less", var.equal = TRUE)