# Non-Parametric Statistics , Rank test Question , Need help

Professor Inarb claims that 50% of his students in a large class achieve a final score 90 points or and higher. A suspicious student asks 17 randomly selected students from Professor Inarb’s class and they report the following scores. "80 81 87 94 79 78 89 90 92 88 81 79 82 79 77 89 90"

Test the hypothesis that the Professor Inarb‘s claim is not consistent with the evidence. i.e., that the 50%-tile (0.5-quantile, median) is not equal to 90. Use a = 0.05.

I order the 17 number from lower to higher number : 77 78 79 79 79 80 81 81 82 87 88 89 89 90 90 92 94 - - - - - - - - - - - - - + + + +

then I define the hypothesis H0 : final score of the class >= 90 H1 : final score of the class !>= 90

number of + = 2 (without 90)
number of - = 13 >>>> 15

Test statistics = 2

X ~ B (15 , 0.5)

X : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

5% Significance level

from the binomial table "using p=0.5" we observe that [0 1 2 3 ] is CR Accept H0 and the rest CR reject H0. So in the end, our TS=2 and number 2 is in the acceptance region.

Is my way of solving the question is right?

• You do not give totally clear details how your got your P-value, but this is a reasonable approach. See my Answer for more on this approach to this one-sided 'median test'. Also, a couple of speculative alternative approaches also leading to rejection of the null hypothesis. Mar 30 '20 at 19:35

Assume that your 17 students are chosen at random from the "large" class. Call it a Success when one of your observations is 90 or above. If the professor's claim is true, then $$P(Success) = .5$$

In your sample, you have only $$X=4$$ Successes among 17. If $$X \sim \mathsf{Binom}(17, .5),$$ then $$P(X \le 4) = 0.0245,$$ as calculated in R below. So the null hypothesis that the success probability is 1/2 or more, is rejected with P-value 0.0245.

pbinom(4, 17, .5)
[1] 0.02452087


Notes: It is clear that you have chosen to do a one-sided median test, but the power of that test is notoriously low. Here are notes on two possible alternative tests.

(1) If test scores are nearly normal, then you might say that the claim amounts to saying that the population mean is 90 or above. (The data are just barely consistent with a normal sample according to the Shapiro-Wilk test, so a strict normality assumption might not be warranted.)

A one-sample t test rejects $$H_0: \mu \ge 90$$ against $$H_1: \mu < 90)$$ with P-value 0.004. (There are no outliers, so maybe $$n=17$$ is large enough to rely on the well-known robustness of t tests against non-normality.)

shapiro.test(x)

Shapiro-Wilk normality test
data:  x
W = 0.89259, p-value = 0.05114

t.test(x, mu=90, alt="less")

One Sample t-test

data:  x
t = -4.0967, df = 16, p-value = 0.0004212
alternative hypothesis: true mean is less than 90
95 percent confidence interval:
-Inf 86.7933
sample estimates:
mean of x
84.41176


(2) A nonparametric one-sample Wilcoxon signed-rank test does not assume normality. It also rejects the null hypothesis that the median is at least 90, against the alternative that the median is smaller. (The exact P-value may not be exactly correct because of ties in the data; the signed rank test in R is programmed to give reasonable approximations.)

wilcox.test(x, mu = 90, alt="less")

Wilcoxon signed rank test with continuity correction

data:  x
V = 9.5, p-value = 0.002229
alternative hypothesis: true location is less than 90


(By randomly 'jittering' the data to break ties artificially, we got P-values between 0.001 and 0.002. Results of one such jittering are shown below.)

xj = x + runif(17, -.1, .1)  # jitter data with UNIF(-.1,.1) displacements
wilcox.test(xj, mu = 90, alt="less")

Wilcoxon signed rank test

data:  xj
V = 17, p-value = 0.001579
alternative hypothesis: true location is less than 90