Non-parametric tests — practice questions

Let $m$ be the median response time. 1. Hypotheses: $H_0: m = 8$ $H_1: m \neq 8$ (two-tailed test)

2. Calculate Signs: We compare each value to the hypothesised median of 8. 7.5 (-), 9.1 (+), 8.2 (+), 10.5 (+), 7.9 (-), 8.0 (0), 9.5 (+), 11.2 (+), 7.1 (-), 8.9 (+) We discard the one value of 8.0, so our effective sample size is $n=9$. Number of '+': $N_+ = 6$ Number of '−': $N_- = 3$

3. Test Statistic: The test statistic is $S = \min(N_+, N_-) = 3$.

4. P-value Calculation: Let $X$ be the number of '−' signs. Under $H_0$, $X \sim B(9, 0.5)$. We are conducting a two-tailed test, so we find the probability of a result as or more extreme than $S=3$. The p-value is $2 \times P(X \le 3)$. $P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$ $P(X=k) = \binom{9}{k}(0.5)^9$ $P(X \le 3) = [\binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3}](0.5)^9 = [1+9+36+84](0.5)^9 = 130 \times 0.001953... = 0.2539$ p-value = $2 \times 0.2539 = 0.5078$

5. Conclusion: Since the p-value (0.5078) is greater than the significance level (0.05), we do not reject $H_0$. There is insufficient evidence to suggest that the median response time is different from 8 minutes.

Let $m$ be the median lifespan. 1. Hypotheses: $H_0: m = 250$ $H_1: m \neq 250$ (two-tailed test)

2. Calculate Differences and Ranks: Sample size $n=8$. No values are equal to 250. | Lifespan (x) | Difference (x-250) | |Difference| | Rank | |---|---|---|---|---| | 241 | -9 | 9 | 4 | | 262 | +12 | 12 | 6 | | 255 | +5 | 5 | 2 | | 238 | -12 | 12 | 6 | (Tie with 262) | 245 | -5 | 5 | 2 | (Tie with 255) | 271 | +21 | 21 | 8 | | 231 | -19 | 19 | 7 | | 258 | +8 | 8 | 3 |

Handling ties: The absolute differences 5 and 5 would be ranks 1 and 2. Average is $(1+2)/2 = 1.5$. Oh wait, I made a mistake in my scratchpad. Let's re-rank properly.

Corrected Ranking: Absolute differences in order: 5, 5, 8, 9, 12, 12, 19, 21. Ranks to be assigned: 1, 2, 3, 4, 5, 6, 7, 8.

• The two '5's occupy ranks 1 and 2. Their rank is $(1+2)/2 = 1.5$.
• The '8' gets rank 3.
• The '9' gets rank 4.
• The two '12's occupy ranks 5 and 6. Their rank is $(5+6)/2 = 5.5$.
• The '19' gets rank 7.
• The '21' gets rank 8.

| Lifespan (x) | Difference (x-250) | |Difference| | Rank | |---|---|---|---|---| | 241 | -9 | 9 | 4 | | 262 | +12 | 12 | 5.5 | | 255 | +5 | 5 | 1.5 | | 238 | -12 | 12 | 5.5 | | 245 | -5 | 5 | 1.5 | | 271 | +21 | 21 | 8 | | 231 | -19 | 19 | 7 | | 258 | +8 | 8 | 3 |

3. Sum Ranks and find Test Statistic: $W_+$ (sum of ranks for positive differences): $5.5 + 1.5 + 8 + 3 = 18$ $W_-$ (sum of ranks for negative differences): $4 + 5.5 + 1.5 + 7 = 18$ Check: $W_+ + W_- = 18 + 18 = 36$. Also $\frac{1}{2}n(n+1) = \frac{1}{2}(8)(9) = 36$. Correct. Test statistic $W = \min(W_+, W_-) = 18$.

4. Critical Value: From MF19 tables for Wilcoxon signed-rank test, for a two-tailed test at 5% significance with $n=8$, the critical value is 3.

5. Conclusion: We reject $H_0$ if $W \le$ critical value. Here, $18 > 3$. So we do not reject $H_0$. There is insufficient evidence at the 5% level to suggest the median battery lifespan is different from 250 hours.