1 Page Inferences for paired data Groundwater quality was measured along a river to determine whether differences in quality existed for the east and west sides of the river. Data were obtained from 24 pairs of wells where each pair consisted of two wells across the river from each other. Assume these 24 pairs of wells are at a random sample of locations along the river. The results for chlorine (in milliequivalents) were:
Pair 1 2 3 4 5 6 7 8 9 10 11 12 West 0.58 0.29 0.67 0.50 1.56 0.22 0.61 0.63 0.56 0.54 0.41 0.39 East Difference 0.39 0.19 0.24 0.05 0.67 0.00 0.44 0.06 1.53 0.03 1.14 -0.92 0.45 0.16 0.46 0.17 0.54 0.02 0.40 0.14 0.33 0.08 0.37 0.02 Pair West East 13 0.24 0.18 14 0.44 0.34 15 0.55 0.62 16 1.28 1.05 17 0.31 0.30 18 0.44 0.33 19 0.51 0.48 20 0.46 0.36 21 0.27 0.20 22 0.22 0.22 23 0.24 0.23 24 0.23 0.19 Difference 0.06 0.10 -0.07 0.23 0.01 0.11 0.03 0.10 0.07 0.00 0.01 0.04
Summary statistics: West: mean = 0.506, standard deviation = 0.318 East: mean = 0.477, standard deviation = 0.330 Paired differences(West-East): mean = 0.029, standard deviation = 0.214
Paired Samples Test Paired Differences 95% Confidence Interval of the Difference Lower Upper -.06156 .11906
Mean Pair 1 West - East .02875
Std. Deviation .21387
Std. Error Mean .04366
t .659
df 23
Sig. (2-tailed) .517
2 Sign test Tests null hypothesis that median difference (along the whole river) is 0. No assumptions about distribution of differences are necessary. The test statistic is K = number of positive differences. If the null hypothesis is true, then the number of positive differences has a binomial distribution with probability of success .5. The sample size n is the number of non-zero differences. K= number of positive differences = Number of negative differences = Number of zero differences = n = number of non-zero differences = If the null hypothesis is true, then K is Binom(n,.5) (why?) and E(K) = n/2. For a one-sided test that the median difference is greater than 0, the P-value is the probability of K or more success on n flips of a fair coin, i.e., the probability that a Binomial(n,.5) is greater than or equal to K. For a two-sided test, we want the probability that that the number of positive differences is as different from the expected value in either direction which means we double the smaller tail area probability.
Exact P-value =
If n 20, we can also use the normal approximation to the binomial distribution. Since E(K) = n/2 and standard deviation of K is n(. 5)(. 5) = n / 4 , the test statistic Z= K (n / 2) n/4
can be compared to a standard normal distribution. Since we are approximating a discrete random variable with a continuous distribution, a more exact approximation comes from using ta continuity correction: if the numerator is positive, subtract from it; if the numerator is negative, subtract . We can also compute a confidence interval for the median difference in chlorine levels by using the correspondence between two-sided tests and confidence intervals. We did this last semester using a table I handed out. We now include the zero values. For n =24, we use 7th and 18th ordered values as the limits for a 97.7% CI (we cant get the exact confidence level we want because of the discreteness of the binomial distribution).
3 Wilcoxon signed rank test The sign test doesnt take into account the magnitude of the differences and so is a fairly crude test. A nonparametric test which does take into account the sizes of the differences...