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.
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
To estimate the population parameter, we can calculate the standard error of the mean (SEM) and construct confidence intervals. The formula for SEM is:
SEM = standard deviation / square root of sample size
Given the standard deviation (3.30) and the sample size (25), we can calculate the SEM as follows:
SEM = 3.30 / sqrt(25) = 3.30 / 5 = 0.66
Next, we can construct confidence intervals for the population mean at different confidence levels.
For a 95% confidence level, we can use the following formula:
Confidence Interval = sample mean ± (critical value * SEM)
The critical value is based on the desired confidence level and the sample size. For a 95% confidence level and a sample size of 25, the critical value is approximately 2.064 (obtained from a t-table or statistical software).
Confidence Interval (95%) = 3.70 ± (2.064 * 0.66) = 3.70 ± 1.36 = (2.34, 5.06)
Therefore, we can estimate that the average number of assaults among the population of students is between 2.34 and 5.06, with 95% confidence.
Similarly, for a 99% confidence level, the critical value is approximately 2.787. We can use this value to construct the confidence interval.
Confidence Interval (99%) = 3.70 ± (2.787 * 0.66) = 3.70 ± 1.84 = (1.86, 5.54)
Thus, with 99% confidence, we can estimate that the average number of assaults among the population of students is between 1.86 and 5.54.