1. MSU Sports Interests Objectives Measure student participation in various sports on and off -campus Measure student interests in participation for different activities & levels of competition Compare partic & interests of men and women

2. Population/Sample All MSU students enrolled spring semester 1978 Stratify sample by class standing seek sample of 200 men, 200 women –assume 67% response rate –mail out 600 surveys

3. Variables/Info needed Gender, age, class Participation in sports –ever, last year, on campus – 40 activities Interests in sports partic on campus –40 activities –levels of competition

4. Data gathering Self-administered mailed 4 page questionnaire no follow-ups

5. 1. Define problem and study objectives 2. Identify information needs & study population(s) 3. Determine basic design/approach - cross sectional vs longitudinal - on-site vs household vs other - self-admin. vs personal interview vs phone - structured or unstructured questions 4. Questionnaire design 5. Choose sample (frame, size, sampling design) 6. Estimate time, costs, manpower needs, etc. STEPS IN A SURVEY

6. 7. Proposal & “Human subjects” review 8. Line up necessary resources 9. Pre-test instruments and field procedures 10. Data gathering and follow-up procedures 11. Coding, cleaning and data processing 12. Analysis: preliminary, then final. 13. Communication and presentation of results. Survey Implementation

7. Sampling error formula n = Z 2  2 / e 2 1. Solve for e to express error as a function of sample size, confidence level, and variance: e = (Z *  ) / SQRT ( n ) 2. For binomial,  = sqrt (p(1-p)), where p is proportion for “yes” in the population Generate numbers in binomial sampling error table as: [1.96 *sqrt( p * (1-p)) ]/ sqrt (n)

8. Sampling errors for binomial (95% confidence interval) percent distribution in population

9. Sample problems From binomial table: Survey with sample of 400 estimates 32% of sample have visited park X. Find 95% confidence interval for population estimate. Using formula: Same survey (n=400) estimates average number of visits per year to be 5.6. If standard deviation in sample is 4, find 95% CI for this estimate.