Question

You are a psychologist who is taking some time away from the city for a while and write a textbook. To do so, you've moved to a very remote, rural area. It is so rural, in fact, that nobody in the entire town has indoor plumbing. As you start to unpack, you are overwhelmed by the stench from your outhouse, and it is such a horrible experience that you begin to rethink every decision you've ever made. Before moving again, you decide to survey the town about their experiences. You survey 12 of your neighbors and ask them to keep track of how many times they are overcome by the stench of an outhouse over the course of a week Now, after the week is over, your neighbors report their results: #Times Overwhelmed Neighbor by Outhouse Stench 1 2 15 3 17 4 11 11 6 7 4 9 13 10 16 11 10 12 5) Which measure of central tendency seems most appropriate for this dataset? Based on your measures of central tendency, what is your best estimate of the number of times you personally will be overwhelmed by outhouse stench in 1 week? Justify your answer 12) Now interpret the standard deviation in this sample. What does it tell you about the variation in the number of times people in this area get overwhelmed by outhouse stench? Based on the information you have now, would you still live in this rural area? Why or why not? n

Answer 1

5. Mean should be the most appropriate measure of central tendency. Mean of the the number of times neighbours were overwhelmed by outside stench should be best estimate of the number of times I personally shall be overwhelmed by outhouse stench in 1 week. o can think the number of times any individual from any has been large population overwhelmed by (here, outside stench myself) as best is population mean. So, sample mean will be estimate. Hence, the mean of the dataset the best estimate. the answer is = 30, q+15+ 17+11+ 9 + 11+44 7+13+ 16+ 1047) = 10.75 Therefore, the hast estimate of number of times shall I personally be overwhelmed by outside stench is 10.75.

12. standard We can deviation of think this the sample distribution as is a 3.934 Gaussian. Hence, x: number of times one will be overwhelmed by outside stench. XN NEM,82 where Momeana 10.45 ro standard deviation = 3.934 N(10.75, 13.48) Hence, neighbors will be overwhelmed by outside stench with variation of 3.934 ~ 4 times. 0 My decision is live in that 7 will mural not area. The reason is estimated number of times a shall be overwhelmed by outside stench i itself quite high le 10.75 c 11 times) Again, PCM. T (XLMAO) = 0.68 1. P (7{x{15) ~ 0.7 Hence by the outside probability stench that I between shall be overnshelmed 7 times to is in times is also mite high le. 0.7.