STUDENT VERSION Weighing the Evidence: Misconceptions about measles-mumps-rubella (MMR) vaccine and autism Kristina Simeonsson, MD, MSPH Jeffrey…

Simeonsson, K., Bethel, J., & Lea, S. (n.d.). Weighing the evidence: Misconceptions about measles-mumps-rubella (MMR) vaccine and autism. Retrieved from Association for Prevention Teaching and Research website:http://c.ymcdn.com/sites/www.aptrweb.org/resource/resmgr/prevention_modules/module_2_student_case.pdf Using the information provided in the case study, complete Sections B, C, and D (pp. 5–10), typing your answers into a separate document. Be sure that your completed document contains your responses to all questions from Sections B, C, and D of the case study. I have also attached a PDF of the case study of the website to facilitate if you can not open the website. Please take your time to follow the […]

le choice With fewer degrees of freedom and larger critical values to exceed, how can the dependent samples t be more powerful than the independent t?…

 From PSY 325 Statistics Multiple choice With fewer degrees of freedom and larger critical values to exceed, how can the dependent samples t be more powerful than the independent t? 1, The larger critical value makes it easier to find statistical significance? 2. The  dependent group tests tend to have larger treatment effects? 3.  The  dependent group tests can be repeated without affecting type 1 error? 4. The larger critical value is offset by lower amounts of error variance?

In a population survey of patients in a rehabilitation hospital, the mean length of stay in the hospital was 12.0 weeks with a standard deviation…

In a population survey of patients in a rehabilitation hospital, the mean length of stay in the hospital was 12.0 weeks with a standard deviation equal to 1.0 week. The distribution was normally distributed. What is the length of stay at the 90th percentile?

The average spending at Neco’s salad bar is $8.22 with a standard deviation of $2. The distribution follows Normal distribution.

The average spending at Neco’s salad bar is $8.22 with a standard deviation of $2.53. The distribution follows Normal distribution. The management is interested in the middle 85% of the customers (spending wise) as it believes that they represent the customer base. What will be the difference between the upper and lower spending cut-offs which define the middle 85% of the customers? Selected Answer:7.0864 Got this wrong. I used M+/-z*sigma. UL=8.22+1.44(2.53)=11.8632 LL+8.22-3.6432=4.75768 The difference is 7.0864, but I got it wrong. Can you help me understand where I went wrong on my approach?