An experiment has been conducted to study the effects of smiling on leniency in judging students accused of cheating. The figure below shows a dotplot of a randomization distribution of differences in sample means. The relevant hypotheses are Vs "H = rf: 0H Ha : H, > Hn ,where Hs and are the mean leniency scores for smiling and neutral expressions, respectively. This distribution is reasonably bell- shaped and we estimate the standard error of the differences in means under the null hypothesis to be about 0.393. For the actual sample in Smiles, the original difference in the sample means is D = x, - In = 4.91 4.12 = 0.79 .Use a normal distribution to find and interpret a p-value for this test. 8800080 -1.0 -0.5 0.0 0.5 1.0 Diff D = 0.79 Randomization distribution for 1000 samples testing "H = 1 : H using Smiles data Randomization distribution for 1000 samples testing Ho : Hs = Hn using Smiles data Click here for the dataset associated with this question. What is the value of the test statistic? Round your answer to two decimal places. i What is the p-value? Round your answer to two decimal places. p-value = i