A group of 36 Canadian ice-hockey fans watches one hockey game while imbibing Molson Canadian and eating ketchup chips and then a second game (with an one-week interval) while imbibing Labatt's Blue and eating Timbits (counterbalanced order). If the estimated variance of the population of difference scores is 18.11, approximately what is the standard error of the distribution of sample difference scores? 87 0.71 3.02 2.04
Question
A group of 36 Canadian ice-hockey fans watches one hockey game while imbibing Molson Canadian and eating ketchup chips and then a second game (with an one-week interval) while imbibing Labatt's Blue and eating Timbits (counterbalanced order). If the estimated variance of the population of difference scores is 18.11, approximately what is the standard error of the distribution of sample difference scores? 87 0.71 3.02 2.04
Solution
The standard error of the distribution of sample difference scores can be calculated using the formula for the standard error of the mean, which is the standard deviation divided by the square root of the sample size.
In this case, the variance of the population of difference scores is given as 18.11. The standard deviation is the square root of the variance, so the standard deviation is sqrt(18.11) = 4.26 (rounded to two decimal places).
The sample size is 36 (the number of Canadian ice-hockey fans).
So, the standard error of the distribution of sample difference scores is 4.26 / sqrt(36) = 4.26 / 6 = 0.71 (rounded to two decimal places).
So, the answer is 0.71.
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