Two different Universities record data on students who are unable to attend classes due to illness. University 1 recorded absences over ten consecutive days. This data is recorded as N1 below. University 2 recorded absences over six consecutive days. This data is recorded as N2 in below.

$N1: 9, 9, 5, 5, 5, 6, 16, 8, 8, 7$ (10 points)

$N2: 13, 11, 14, 13, 12, 11$ (6 points)

a) Assuming equal variances, using a two sampled t-test, determine whether there is a difference between means.

I have done this one and found that: The absolute value of the test statistic in this example, 3.2066, is greater than the critical value of 2.1448, so we reject the null hypothesis and conclude that the two population means are different at the 0.05 significance level.

b) Could you please comment on the appropriateness of the test?

c) In a related study, you observe that when students are ill, they take on average five days off. How would this affect the appropriateness of using a t-test described above.

Could anyone give me some advices/insights on solving the b, c questions?


b) The t-test (or z-test for that matter) are generally only used if the samples are coming from a population that is normally distributed, or if the samples are relatively large. In this case, there is no indication from the question that the number of absences per day follows a normal distribution, and the sample sizes are rather small, so it may be more appropriate to use a non-parametric test (such as the Wilcoxon Rank Test).

c) If a student who is sick takes an average of 5 days off, then the number of absences from one day to the next are not independent. This further makes the t-test inappropriate, as the t-test is designed for cases where the sample items are independent from one another.


Whether the two samples differ significantly. There is a clear difference between the two samples. Every observation in N2 is greater than all but one observation in N1. So the probability such a pattern of almost-complete separation occurred by chance alone is very small.

enter image description here

There is one outlier in the N1 sample, so some people would hesitate to do a t test. But I would want to see results from a Welch (separate variances) t test. If the result were near the critical value for a 5% level test, I would wonder if the t statistic really has the supposed t distribution, but that is not the case here. Results from a Welch t test done with R statistical software are shown below:

 N1 = c(9, 9, 5, 5, 5, 6, 16, 8, 8, 7) 
 N2 = c(13, 11, 14, 13, 12, 11)
 t.test(N1, N2, alt="two.sided")

 ##        Welch Two Sample t-test

 ## data:  N1 and N2 
 ## t = -3.9325, df = 12.382, p-value = 0.001877
 ## alternative hypothesis: true difference in means is not equal to 0 
 ## 95 percent confidence interval:
 ##  -7.036441 -2.030226 
 ## sample estimates:
 ## mean of x mean of y 
 ##   7.80000  12.33333 

I'm guessing your t statistic is from a pooled t test, but either way the p-value is minuscule. And so would be the p-value of tests that don't assume normal populations, such as a permutation test (p-value 0.007) or a Mann-Whitney-Wilcoxon rank sum test. (The MWW test has approximate p-value 0.01. There are enough ties in the data not to take the exact value seriously, but it is well below 5%.)

Whether the conclusion can be generalized. The issue of students staying out for several days might pose a difficulty. Specifically, neither sample might consist of independent observations. At the worst, maybe N2 data come at a time when a few students with the flu get counted over and over again. It would be a lot better if each observation were taken in a different week rather than on consecutive days. The issue of independence would apply just as well to the nonparametric permutation and MWW tests mentioned above.

Furthermore, we do not have much data from either university. Any serious conclusion whether one of these universities has a higher sick rate during exams should extend over a longer period of time and involve more subjects.

Note: Results from a simulated permutation test are shown below:

 N1 = c(9, 9, 5, 5, 5, 6, 16, 8, 8, 7) 
 N2 = c(13, 11, 14, 13, 12, 11)
 All = c(N1, N2)
 m = 10000;  d = numeric(m)
 d.obs = mean(N2) - mean(N1)
 for (i in 1:m) {
    perm = sample(All, 16)
    d[i] = mean(perm[1:10]) - mean(perm[11:16]) } 
 mean(abs(d) > abs(d.obs))
 ## 0.0068  # p-value (other runs gave 0.0064, 0.0065)

The p-value of the permutation test is represented by areas in the histogram outside the vertical dotted red lines.

enter image description here


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