Exercise 1

Suppose that a researcher is interested in the effect of caffeine on typing speed. A group of nine individuals are administered a typing test. The following day, they repeat the typing test, this time after taking 400 mg of caffeine. (Note: This is not recommended.) The data gathered, measured in words per minute, is

decaf = c(98,  124, 107, 105, 80, 43, 73, 68, 69)
caff  = c(104, 128, 110, 108, 86, 53, 72, 73, 72)
##   decaf caff
## 1    98  104
## 2   124  128
## 3   107  110
## 4   105  108
## 5    80   86
## 6    43   53
## 7    73   72
## 8    68   73
## 9    69   72

Note that these are paired observations.

Use the sign test with a significance level of 0.05 to assess whether or not caffeine has an effect on typing speed. That is, test

\[ H_0\colon \ m_D = m_C - m_N = 0 \quad \text{vs} \quad H_A\colon \ m_D = m_C - m_N \neq 0 \]

where

Since it is possible that the caffeine makes typing speed worse, use a two-sided test. (Also note that this is a silly experience, we aren’t considering typing accuracy!)

Report:

Solution

# the "test statistic" for the sign test
sum(caff - decaf > 0)
## [1] 8
# the expected value of the test stat under the null
# this is used to determine "extreme" values of the test statistic
# values that are equal distance from the expected are equally extreme
length(caff - decaf) / 2
## [1] 4.5
# add up the probabilities for test stat values that are as extreme or more extreme
sum(dbinom(c(0, 1, 8, 9), size = 9, prob = 0.5))
## [1] 0.0390625
  • Test statistic: \(X = 8\).
  • Distribution: \(X \sim \text{binom}(n = 9, p = 0.5)\)
  • p-value: 0.03906
  • Decision: Reject \(H_0\).
  • Conclusion: The data does provide evidence at the 0.05 level that caffeine has an effect on typing speed.

Exercise 2

Does meditation have an effect on blood pressure. A group of six college aged individuals were given a routine physical examination including a measurement of their systolic blood pressure. (Measured in millimeters of mercury.) A week after their physicals, the same six individuals returned for a guided meditation session. Immediately afterwords there (systolic) blood pressure was measured. The data gathered is

physical    = c(125, 108, 185, 135, 112, 133)
meditation  = c(120, 114, 160, 131, 124, 125)
##   physical meditation
## 1      125        120
## 2      108        114
## 3      185        160
## 4      135        131
## 5      112        124
## 6      133        125

Note that these are paired observations.

Use the sign test with a significance level of 0.10 to assess whether or not meditation has an effect on blood pressure. That is, test

\[ H_0\colon \ m_D = m_M - m_P = 0 \quad \text{vs} \quad H_A\colon \ m_D = m_M - m_P \neq 0 \]

where

Since it is possible that the meditation makes blood pressure worse, use a two-sided test.

Report:

Solution

# the "test statistic" for the sign test
sum(meditation - physical > 0)
## [1] 2
# the expected value of the test stat under the null
# this is used to determine "extreme" values of the test statistic
# values that are equal distance from the expected are equally extreme
length(meditation - physical) / 2
## [1] 3
# add up the probabilities for test stat values that are as extreme or more extreme
sum(dbinom(c(0, 1, 2, 4, 5, 6), size = 6, prob = 0.5))
## [1] 0.6875
  • Test statistic: \(X = 2\).
  • Distribution: \(X \sim \text{binom}(n = 6, p = 0.5)\)
  • p-value: 0.6875
  • Decision: Fail to reject \(H_0\).
  • Conclusion: The data does not provide evidence at the 0.10 level that meditation has an effect on systolic blood pressure.

Exercise 3

Return to the sleep data in Exercise 2. This time test

To do so, use a permutation test that permutes the statistic

\[ \bar{x}_D \]

where \(\bar{x}_D\) is the sample mean difference. Assume that the distribution of blood pressure with and without meditation has the same shape, but may have different locations. Use at least 10000 permutations.

physical    = c(125, 108, 185, 135, 112, 133)
meditation  = c(120, 114, 160, 131, 124, 125)

Solution

# create difference data
bp_diff = meditation - physical

# function to shuffle data and calculate statistic
permute_x_bar = function(data) {
  sample_size = length(data)
  permuted_data = sample(c(-1, 1), size = sample_size, replace = TRUE) * data
  mean(permuted_data)
}

# generate permuted statistics for sleep data
set.seed(42)
bp_x_bars = replicate(n = 10000, permute_x_bar(data = bp_diff))

# calculate statistic on observed data
bp_x_bar_obs = mean(bp_diff)
hist(bp_x_bars, col = "darkgrey",
     xlab = "t", probability = TRUE,
     main = "Permutation Test, Sample Mean, Blood Pressure Data")
box()
grid()
abline(v = c(-1, 1) * bp_x_bar_obs, col = "firebrick", lwd = 2)

mean(bp_x_bars > abs(bp_x_bar_obs)) + mean(bp_x_bars < -abs(bp_x_bar_obs))
## [1] 0.5003
  • This histogram looks real weird… This is mostly due to the choice of statistic to permute. Had we used some like the usual \(t\) statistic, the histogram would be more “normal” looking.

Example 4

Which profession pays more? Data Scientist or Actuary? A (far too small) survey of junior (less than three years experience) data scientists and actuaries resulted in the following data:

data_sci = c(88000, 121000, 91000, 50000, 78000, 95000)
actuary = c(63000, 75000, 81000, 75000, 85000)

Use a permutation test that permutes the statistic

\[ t = \frac{(\bar{x} - \bar{y}) - 0}{s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]

to test

Assume that the distribution of salaries for both has the same shape, but may have different locations. Use at least 10000 permutations.

Solution

# function to shuffle data and calculate statistic
permute_two_t_stat = function(data_1, data_2) {
  
  # determine samples sizes of both groups
  sample_size_1 = length(data_1)
  sample_size_2 = length(data_2)
  
  # create variable for group structure
  groups = c(rep(TRUE, sample_size_1), rep(FALSE, sample_size_2))
  
  # shuffle the groups
  shuffled_groups = sample(groups)
  
  # merge the data into a single group (null hypothesis)
  all_data = c(data_1, data_2)
  
  # create new groups
  shuffled_data_1 = all_data[shuffled_groups]
  shuffled_data_2 = all_data[!shuffled_groups]  
  
  # calculate statistics on permuted data
  t.test(x = shuffled_data_1, y = shuffled_data_2, var.equal = TRUE)$statistic
}

# generate t statistics for exam data
set.seed(42)
salary_t_stats = replicate(n = 10000, permute_two_t_stat(data_1 = data_sci, 
                                                         data_2 = actuary))

# calculate t statistic on observed data
salary_t_obs = t.test(x = data_sci, y = actuary, var.equal = TRUE)$statistic
hist(salary_t_stats, col = "darkgrey",
     xlab = "t", probability = TRUE,
     main = "Permutation t-Test, Salary Data")
box()
grid()
abline(v = c(-1, 1) * salary_t_obs, col = "firebrick", lwd = 2)