ECO220_20191008_Midterm1_AnswerKey.pdf - ECO220Y L0101 L0501 Midterm Test#1 – Answer Key Tuesday 8 October 2019 P Blanchenay 1[60pts a[4pts The

ECO220_20191008_Midterm1_AnswerKey.pdf - ECO220Y L0101...

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ECO220Y L0101 & L0501 Midterm Test #1 – Answer Key Tuesday 8 October 2019 P. Blanchenay 1. [60pts] a) [4pts] The distribution is unimodal and (very) skewed/ positively-skewed / right-skewed, since most participants had very few clicks (between 0-10) and decreasing after that. b) [3pts] 50% participants clicked 14 times or less before making their choice. c) [5pts] Since the distribution is positively skewed, we expect the mean to be higher than the median, therefore higher than 14. d) [4pts] The distribution of lnchoiceclicks will be less skewed (it could even be negatively/left-skewed). Observations with 0 clicks would be excluded by logging the variable, but since there are only 2, logging the variable does not dramatically change the sample. e) [8pts] | choicetime choiceclicks -------------+------------------------ choicetime | 1.000 choiceclicks | 0.575 1.000 Denote choicetime as F and choiclicks as G . We first note that a variable is always perfectly correlated with itself, so diagonal terms are 1 . To find the correlation @ FG , let’s first note from the covariance table that A F = p A 2 F = 5993 . 14 = 77 . 4154 . And likewise A G = 360 . 805 = 18 . 9948 . So @ FG = A F G A F A G = 845 . 005 77 . 4154 × 18 . 9948 = 0 . 575 . f) [5pts] The correlation is higher than 0.5, and that the scatter plot does exhibit a linear pattern. There exists a strong positive linear relationship between the two variables. Participants who click on more information links also take longer to choose a card.
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