In: Statistics and Probability
In an attempt to increase business on Monday nights, a restaurant offers a free dessert with every dinner order. Before the offer, the mean number of dinner customers on Monday was 150. The following data represents the number of diners on a random sample of 12 days while the free dessert offer was in effect.
206 169 191 152 212 139 142 151 174 220 192 153
Using the Wilcoxon signed rank test, it is possible to conclude that the typical number of diners on a Monday night increased while the free dessert offer was in effect?
The Wilcoxon signed rank test is used on medians of a sample to test a claim. However, I'm told the mean prior and then asked to use the rank test on the sample above, but I don't know what my null hypothesis should be.
It is ok if you are asked for the mean, we have to took it as measure of central value both mean and median are the measure of central value or average. so we can put forward our hypothesis as
X | X-150 | |X-150| | Rank | Signed Rank | Zi |
206 | 56 | 56 | 10 | 10 | 1 |
169 | 19 | 19 | 6 | 6 | 1 |
191 | 41 | 41 | 8 | 8 | 1 |
152 | 2 | 2 | 2 | 2 | 1 |
212 | 62 | 62 | 11 | 11 | 1 |
139 | -11 | 11 | 5 | 5 | 0 |
142 | -8 | 8 | 4 | 4 | 0 |
151 | 1 | 1 | 1 | 1 | 1 |
174 | 24 | 24 | 7 | 7 | 1 |
220 | 70 | 70 | 12 | 12 | 1 |
192 | 42 | 42 | 9 | 9 | 1 |
153 | 3 | 3 | 3 | 3 | 1 |
normal approximation can be used with and
corresponding P.Value = 0.021
Thus we will reject our null hypothesis and accept our alternative hypothesis at 0.05 level of significance.