Question

12.22 The paper "The Effects of Adolescent Volunteer Activities on the Perception of Local Society and Community Spirit Mediated by Self-Conception" (Advanced Science and Technology Letters [2016]: 19-23) describes a survey of a large representative sample of middle school children in South Korea. One question in the survey asked how much time per year the children spent in volunteer activities. The sample mean was 14.76 hours and the sample standard deviation population mean even though the population distribution is not approximately normal. c. Calculate and interpret a confidence interval for the mean was 16.54 hours. number of hours spent in volunteer activities per year for South Korean middle school children. (Hint: See Example 12.7.) a. Based on the reported sample mean and sample standard deviation, explain why it is not reasonable to think that the distribution of volunteer times for the population of South Korean middle school students is approximately normal. b. The sample size was not given in the paper, but the sample size was described as "large." Suppose that the sample size was 500. Explain why it is reasonable to use a one-sample t confidence interval to estimate the

Answer 1

a) Given the std deviation, 16.54, is higher than the mean, 14.76, it suggests that the data isn't tightly spread around the mean (since std deviation is a measure of spread around the mean). In a normal distribution, we would normally expect the data to be more tightly clustered around the mean. In the given distribution, the tails are much longer and skewed than one would expect in a normal distribution.

b) Even though the population distribution is not approximately normal, it is reasonable to use 1-sample t confidence interval as we do not expect any significant outliers in the data given it is the no. of hours per year spent in volunteering activities, which shouldn't have any unreasonably extreme values

c) Confidence interval (say 95%) can be given by:

14.76 +/- Margin of Error

(where Margin of Error = T-critical * Std Deviation = T_{0.975, 499} * (16.54/ $\sqrt{500}$ ) = 1.97 * 0.739 = 1.457)

= (13.303, 16.217)