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


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.


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.

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----------
             25        50        75       100       125

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