P-Value from Sign Test

Question

I'm so confused with this question and its answer provided as following:

The Question

The Answer provided

I totally don't understand why when we calculate the P-value, we need to calculate as $P(X \ge 1) = 1-P(X=0).$ Since we set the data less than 2 as positive sign here, then we only have 1 positive sign in the data set. Then what does $P(X=0)$ represent? And what does $P(X \ge 1)$ stand for?

Thank you for your help, I really can't get any answer directly from Google.

Answer 1

You have data $\{1,4,3,9,5\}$ from a symmetrical distribution with unknown mean $\mu.$ You wish to find the P-value for a sign test of $H_0: \mu=2$ vs. $H_1: \mu < 2.$

To begin, only one of the five observations is smaller than 2, so there hardly seems evidence that $\mu < 2.$ It seems that the person who decided to test $H_0: \mu=2$ vs. $H_1: \mu < 2$ did not correctly anticipate the nature of the population distribution.

Under $H_0$ the observed number $X$ of observations less than 2 has $X \sim \mathsf{Binom}(n=5, p=1/2).$ The P-value is the probability of a more extreme result (in the direction of the alternative) than what we have observed. If $H_1$ were true we would expect to see more observations below 2.

So the P-value is
$$P(X \ge 1) = 1 - P(X =0) = 1 - 1/32 = 31/32 = .96875.$$

The linked answer gives the P-value as about 0.97.

Ordinarily, one does not expect to see such a large P-value. So I see why this problem confused you. But the answer given is correct for the (somewhat strangely posed) problem.

Hote: There is a saying among applied statisticians: If the P-value is very small (say below 0.05), then doubt the null hypothesis; if the P-value is very large (say above 0.95), then doubt the model. It is a way of saying something may be wrong with assumptions or the statement of hypothesis and alternative, if you see such a large P-value. If the alternative had been posed as $H_1: \mu > 2,$ then the P-value would have been $1/32 = 0.0125 < 0.05$ and one would reject $H_O: \mu = 2.$

Addendum: Here is a sign test in Minitab statistical software of $H_0: \eta = 2$ vs $H_1: \eta < 2,$ were $\eta$ is the population median. But for your symmetrical population the mean $\mu$ and the median are the same.

Sign Test for Median: X 

Sign test of median =  2.000 versus < 2.000

    N  Below  Equal  Above       P  Median
 X  5      1      0      4  0.9688   4.000