Sign Test for Hypothesized Median

Question

Given the following data, apply the sign test to the hypothesis that the median of the underlying distribution is equal to 75.

Use the large-sample version of the sign test, two tails, and a probability of Type I error of 5%.

Find the test statistic, critical value, and whether you accept or reject the null hypothesis.

Observed Value

31
106
76
66
119
122
76
71
47
35
122
130
95
70
58
137
43
16
102
13

Attempted Solution:

So I think I found how to approach this problem but there are a bunch of different tests so I'm not sure I picked the right one. S=max{$S_1,S_2$} where $S_1$ is the number of data values less than the hypothesized median, and $S_2$ is the number of data values more than the hypothesized median. Both are 10, giving $S$ = 10. From the formula:

$Z$ = ${S-{n\over{2}}}\over\sqrt(20)*{1\over{2}}$ $\Rightarrow$ Z = 0.

So my test statistic is 0, critical value is 1.96, thus accepting the null hypothesis.

Answer 1

Your general approach and conclusion are valid. But the Z statistic should be $Z = \frac{S=10}{\sqrt{20*.5*.5}} = 0.$

Another possible method is to use the proportion $\hat p$ of values below 75, and use $Z = \frac{\hat p - 1/2}{\sqrt{(.5 \times .5)/20}} = 0.$

In both formulations $Z$ has approximately a standard normal distribution under $H_0.$ And in both the denominator is the standard error of the estimate ($S$ in the first; $\hat p$ in the second.) You would reject at the 5% level of significance against a two-sided alternative if $|Z| > 1.96.$

Either way, because half of the observations are above 75 and half are below, there is no possibility that $H_0$ is rejected.

Here is a sign test procedure from Minitab 17 software:

Sign Test for Median: Val 

Sign test of median =  75.00 versus ≠ 75.00

     N  N*  Below  Equal  Above       P  Median
Val 20   0     10      0     10  1.0000   73.50

Comment: Maybe an exercise that would be generally more instructive would be to use these data to test whether the population median is 100. Minitab (which uses an exact method, not a normal approximation) gives a P-value of about 0.26, so the null hypothesis that the population median is $\eta = 100$ can't be rejected at the 5% level. You might want to try the $Z$ test for that.

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Note: Another nonparametric test for the median is the Wilcoxon signed-rank test. Results of that test from Minitab 17 are also shown below, in case they are of any interest.

Wilcoxon Signed Rank Test: Val 

Test of median = 75.00 versus median ≠ 75.00

            N for   Wilcoxon         Estimated
     N  N*   Test  Statistic      P     Median
Val 20   0     20      109.0  0.896      76.25

Finally, here is an old-fashioned 'character graphics' boxplot from Minitab. It shows that the sample is roughly symmetrical (a condition of the Wilcoxon test) and that the median (+ within the box) is very nearly 75.

                      ----------------------------
         -------------I          +               I----------
                      ----------------------------
      --------+---------+---------+---------+---------+--------Val
             25        50        75       100       125