Suppose we have two groups of students given the same difficult test. However, one group of students was given an extra training session before the test.

normal students = 123

students with extra training = 119

18 of the normal students pass the test, but 25 of the students with extra training pass.

Did the extra training actually make a difference? What are the best tools to use for this? Hypothesis testing? Power analysis?

  • $\begingroup$ Use Fisher's exact test or Pearson Chi-square test. Both of them do not need the sample sizes are the same in two groups. $\endgroup$
    – user158565
    Jun 21, 2019 at 0:41

2 Answers 2


You're going to want a chi-square test for this.

First, you make a table of your observed values:

        Pass   Fail   Total
Normal  18     105    123
Extra   25     94     119
Total   43     199    242

The numbers in the "Total" row and column are called "marginals" - they represent the total number (out of all 242 students) who passed or failed, or who had normal or extra training.

Next, you calculate percents based on your marginals. 43/242 = overall, 17.8% of students passed. 123/242 = overall, 50.8% of students received extra training. Etc. Then you max a table of expected values, based on these marginals. For example, if the extra training had no effect on passing, then you would expect the same percent of students (17.76%) to pass in both groups. So for how many students you would expect to pass in the "normal" group, you would calculate 242*0.508*0.178 = you expect 21.9 students from the normal group to pass.

Here is the full expected table calculated using that method:

        Pass   Fail   Total
Normal  21.9   101.2  123
Extra   21.1   97.9   119
Total   43     199    242

Now for each cell, calculate (observed - expected)^2/expected, then add that together for all the cells. You should get 0.695 + 0.143 + 0.721 + 0.155 = 1.714.

You are almost ready to look up your test statistic in a chisq table - you just need to know the degrees of freedom. The degrees of freedom is (rows - 1)*(columns - 1) so here you have one degree of freedom.

Now, look up your chi-square test statistic (1.714) in a chi-square table. Find the row for one degree of freedom, then look in the cells for that row. You'll see in the column for p=0.2 the chi-square test statistic is 1.642, and in the column for p=0.1, the chi-square test statistic is 2.706. Your test statistic was in between those two values, so your p-value is somewhere between 0.1 and 0.2. Eg. not significant at the standard p=.05 cutoff.

  • 1
    $\begingroup$ Nice (+1). Comparison of your approach and mine. You are essentially doing a two-sided test because of the squaring in the the computation of the chi-squared statistic. The square $(-1.30)^2 = 1.69$ of my Z-statistic is essentially the same is your $\chi^2 = 1.71.$ Your exact P-value is 0.19 and the double of mine is 0.194. (I can see how this could be either a one- or a two-sided test. Although, someone must be hoping that the extra training helps.) $\endgroup$
    – BruceET
    Jun 21, 2019 at 1:56
  • $\begingroup$ @BruceET great explanation of the different results! I knew in theory that the chi-square distribution was based on the square of a Z, but this is the first time that's really made sense! $\endgroup$ Jun 21, 2019 at 4:19

You want to test whether the population rates of passing the exam are the same for the two groups. So your null hypothesis is $H_0: p_1=p_2$ vs $H_a: p_x < p_2.$

A test of the equality of binomial proportions is discussed in the NIST handbook and implement in Minitab statistical software (among other software programs). Minitab output is as follows:

Test and CI for Two Proportions 

Sample   X    N  Sample p
1       18  123  0.146341
2       25  119  0.210084

Difference = p (1) - p (2)
Estimate for difference:  -0.0637426
95% upper bound for difference:  0.0170092
Test for difference = 0 (vs < 0):  
    Z = -1.30  P-Value = 0.097

This Minitab procedure also includes the P-value for the Fisher Exact Test:

Fisher’s exact test: P-Value = 0.129

While it is true that the students who had extra training showed a somewhat higher pass rate than those who did not (21.0% vs. 14.6%), both P-values exceed 0.05, so at the 5% level of significance that pass rate for the students with additional training did not have a higher pass rate that is statistically significant.

Addendum: Here is the intuitive rationale for Fisher's Exact Test.

In Group 1 and Group 2 combined, you have $123 + 119 = 242$ students. Out of the $18+25=43$ who passed the exam, only 18 are from Group 1. If all students are equally likely to pass, what is the probability that such a small number of passes in Group 1 would occur at random.

Specifically, let $X$ be the number of Group 1 passes out of 43. The P-value of Fisher's Exact Test is $P(X \le 18).$

Symbolically, this is $$P(X \le 18) = \sum_{i=1}^{18} \frac{{123 \choose i}{119 \choose 43-i}}{{242 \choose 43}} \approx 0.129,$$

which agrees with Minitab's P-value for the Fisher test.

In R statistical software, this is computed as follows:

 phyper(18, 123, 119, 43)
[1] 0.129473

In the plot of the relevant hypergeometric distribution below, the P-value is the sum of the heights of the bars to the left of the vertical dotted line.

enter image description here


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