The below is assuming a normal distribution. However this is not true.
Z = (score-mean)/SD = 0 - 3.7/3.3 = -1.12
This means that over 13% have never been assaulted. The distribution would be positively skewed.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.025 and .005) to get the Z score.
95% = mean ± 1.96 SEm
99% = mean ± 2.575 SEm
SEm = SD/√n
A criminologist is studying school violence, she selects a random sample of 25 adolescents aged 12 to 15 from a local middle school and asks the following,"How many times within the past school year have you been physically assaulted at school? She calculated the following stats based on the students answers: the average number of assaults was 3.70, with a standard deviation of 3.30. On the basis of this information, provide an estimate of the population parameter(average number of assaults among the population of students) by calculating the standard error of the mean and constructing confidence intervals for the 95% and 99% levels.
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