Relationship between Sign test and z-test for proportion

Question

$88$ of the respondents have salaries over the median ($\$13500$), and $112$ of the respondents have salaries below the median ($\$13500$).

Carry out the two-tailed “Sign” test as a one sample z-test for proportion, find the test value and p-value of the test.

That's the question I got What is the relationship between Sign test and z-test for proportion? How to calculate this question?

I did some calculation on it, but I'm not sure about it.

H$0$: median = $13500$

H$1$: median =/= $13500$

Since median is $0.5$, and there are $88$ positive signs. Such that

$$\text{test value}=\frac{(0.44-0.5)}{\sqrt{\frac{0.5\times0.5}{200}}}\approx1.7$$

Is it correct?

Answer 1

You want to test $H_0: \eta = 13500$ vs $Ha: \eta \ne 13500,$ where $\eta$ is the population median. Count the $X = 88$ salaries out of $n = 200$ as Successes. Fundamentally, the distribution under the null hypothesis is $X \sim \mathsf{Binom}(200, 1/2).$ If $H_0$ is true, you would 'expect' 100 Successes.

For an exact two-sided test, the P-value (probability of a result as or more extreme than the observed, in the directions of the alternative), would be the sum $P(X \le 88) + P(X \ge 112).$ That can be computed using software. From R statistical software, I get the P-value 0.1035 > .05, so you cannot reject $H_0$ at the 5% level of significance.

 pbinom(88, 200, .5)
 ## 0.05181952              # P(X <= 88)
 1 - pbinom(111, 200, .5)
 ## 0.05181952              # P(X >= 112) = 1 - P(X <= 111)

You have proposed using a normal approximation to the null binomial distribution of $X,$ which should work quite well for $n$ as large as 200.

Using $p = 88/200 = 0.44$ as the estimate of $\theta = P(\text{Success}),$ you have the test statistic $Z \sim \mathsf{Norm}(0,1)$. $$Z = \frac{p - \theta_0}{\sqrt{\theta_0(1-\theta_0)/n}} = \frac{.44 - .5}{\sqrt{.5(.5)/200}} = -1.697,$$

where $\theta_0 = .5$ is the hypothetical Success probability.

Roughly, the P-value is $P(|Z| \ge 1.697) = 0.0897 > .05,$ which you can evaluate using printed normal tables. A somewhat more accurate P-value (nearer to 0.1025) could be found by using a continuity correction, which may be discussed in your text. But the value we got is so much larger than .05, that the decision not to reject is not in doubt.

Put another way, you would reject $H_0$ at the 5% level if $|Z| > 1.96$ and you have $|Z| = 1.697.$

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The figure below shows black bars representing the probabilities in $\mathsf{Binom}(200, .5)$ and the blue density curve of the 'best fitting' normal distribution. The exact P-value is the sum of the heights of the bars outside the vertical red lines.