Answer to Question #179035 in Statistics and Probability for Ralph

Question

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 H₀: μ_s=μ_n vs. H_a: μ_s >μ_n, where μ_s and μ_n 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 4.91–4.12 = 0.79. Use a normal distribution to find and interpret a p-value for this test.

1. What is the standardized test statistic? Round your answer to two decimal places.
2. Z = 
3. What is the p-value? Round your answer to three decimal places.
4. p-value = 

Answer 1

(a) The randomization distribution depends only on $H_o$ so it would not change for $H_A : µ_s 6= µ_n.$

For a two-tailed alternative we need to double the proportion in one tail, so the p-value is 2(0.027)

= 0.054.

(b) $Z=\dfrac{0.79}{0.393}=2.01$

(c) This is an upper tail test, so the p-value is the proportion of randomization samples with differences more than the observed D = 0.79. There are 27 dots to the right of 0.79 in the plot, so

the p-value is $\dfrac{27}{1000} = 0.027.$

(d) p-value= 0.027