As mentioned in Chapters 2 and 3 (pp. 38 and 81) Steven Schmidt (1994) reported a series of studies examining the effect of humor on memory.  In one part of the study, participants were presented with a list containing a mix of humorous and nonhumorous sentences, and were then asked to recall as many sentences as possible.  Schmidt recorded the number of humorous and the number of nonhumorous sentences recalled by each individual. Notice that the data consist of two memory scores for each participant.  Suppose that a difference score is computed for each individual in a sample of n = 16 and the resulting data show that participants recalled an average of MD = 3.25 more humorous sentences than nonhumorous, with SS = 135.  Are these results sufficient to conclude that humor has a significant effect on memory? Use a two-tailed test with α = 0.05. H0: µ = 0; HA: µ ≠ 0 α = 0.05. Two-tailed test à critical area of 0.025 on each side. D.F. = 16-1 = 15.  àCritical value of t = 2.13 s2D = 135 / 15 = 9 à sMD = SQRT(9/16) = 0.75     Test value of t = (3.25 – 0)/0.75 = 4.33 |4.33| > |2.13| à Reject H0.

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As mentioned in Chapters 2 and 3 (pp. 38 and 81) Steven Schmidt (1994) reported a series of studies examining the effect of humor on memory.  In one part of the study, participants were presented with a list containing a mix of humorous and nonhumorous sentences, and were then asked to recall as many sentences as possible.  Schmidt recorded the number of humorous and the number of nonhumorous sentences recalled by each individual. Notice that the data consist of two memory scores for each participant.  Suppose that a difference score is computed for each individual in a sample of n = 16 and the resulting data show that participants recalled an average of MD = 3.25 more humorous sentences than nonhumorous, with SS = 135.  Are these results sufficient to conclude that humor has a significant effect on memory? Use a two-tailed test with α = 0.05.

  1. H0: µ = 0; HA: µ ≠ 0
  2. α = 0.05. Two-tailed test à critical area of 0.025 on each side. D.F. = 16-1 = 15.  àCritical value of t = 2.13
  3. s2D = 135 / 15 = 9 à sMD = SQRT(9/16) = 0.75

    Test value of t = (3.25 – 0)/0.75 = 4.33

  1. |4.33| > |2.13| à Reject H0.
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