Survey Sampling Exercises

Mean Estimation for a Simple Random Sample

Example 1: In the early stages of planning a school restructuring effort, school district board members are considering a year round schooling program. For the moment, the board is interested in the degree to which parents/legal guardians favor such a change. A simple random sample (n = 300) of parents/legal guardians was drawn from 1,850 families (only one adult per household) and given a questionnaire. On the questionnaire, one item addressed the 12-month schooling issue in question. The Likert item ranged from 1 to 5, with 5 being strongly agree, 3, neutral, and 1, strongly disagree. The mean response to the question was 2.8 and the standard deviation was .3. Calculate a 95% confidence interval around the mean of the item responses.

Example 2: A simple random sample (n = 200) of UCF Undergraduates who have received financial aid in 2001 (N = 20,000) are given a 10 item questionnaire regarding their satisfaction with the services provided by the financial aid department. Each Likert item on the questionnaire ranged from 1 to 5, with 5 being strongly agree. A score of 50 therefore suggested complete satisfaction, a score of 25, moderate satisfaction. Student responses were summed for a total score, and the mean (37) and standard deviation (5) of the scores calculated so that the average satisfaction of financial aid recipients could be estimated with 95% confidence. Calculate the confidence interval around the mean satisfaction score.

Example 3: A psychologist wishes to estimate the average reaction time to a stimulus among 200 patients in a hospital specializing in nervous disorders. A simple random sample of n = 20 patients was selected, and their reaction times were measured, with the following results: sample mean = 2.1 seconds, sample standard deviation = .4 second. Estimate the population mean, and calculate a 95% confidence interval around the estimate.

Total Estimation for a Simple Random Sample

Example 4: A Public School Superintendent wants to know the total amount of money spent on experimental academic programs in her school district quickly. Instead of having all of the 300 principals in the district respond to a telephone inquiry, she decides to draw a simple random sample of 25 principals. With respect to the key question motivating the survey, the superintendent discovers that principals, on average, spent $1,530.00 in 2001 experimental academic programs

(S² = $330.00).

Example 5: A simple random sample of 100 water meters within a community is monitored to estimate the average daily water consumption per household over a specified dry spell. The sample mean and sample variance are found to be 12.5 and 1252, respectively. If we assume there are N = 10,000 households within the community, estimate the total number of gallons of water consumed daily during the dry spell.

Proportion Estimation for a Simple Random Sample

Example 6: The principal of a large, southwestern high school, recognizing that many of his students come from bilingual homes, decides to determine more precisely the proportion of his students to which this applies. This information would guide his decision regarding how much more money to allocate to the school's ESOL program. With 4000 high school student families falling within his school zone, the principal decide to survey 200 students, using simple random sampling. It turns out 56 of the 200 students surveyed come from bilingual homes. Build a 95% confidence interval around the estimated proportion of high school students in the school zone who come from bilingual homes.

Example 7: A simple random sample of n = 40 college students was interviewed to determine the proportion of students in favor of converting the semester system to the quarter system. Twenty-five of the students answered affirmatively. Estimate the proportion of students on campus in favor of the change. (Assume N = 2000). Place a bound on the error of estimation.

Example 8: A sociological study conducted in a small town calls for the estimation of the proportion of households that contain at least one member over 65 years of age. The city has 621 households according to the most recent city directory. A simple random sample of n = 60 households was selected from the directory. At the completion of the fieldwork, out of the 60 households sampled, 11 contained at least one member over 65 years of age. Estimate the true population proportion, and place a bound on the error of estimation.

Mean Estimation for a Stratified Random Sample

Example 9: A school desires to estimate the average score that may be obtained on a reading comprehension exam for students in the sixth grade. The school’s students are grouped into three tracks, with the fast learners in track I and slow learners in track III. The school decides to stratify on tracks since this method would reduce the variability of test scores. The sixth grade contains 55 students in track I, 80 students in track II, and 65 students in track III. A stratified random sample of 50 students is proportionately allocated and yields simple random samples of 14, 20, and 16 for tracks I, II, and III, respectively. The test is administered to the sample of students, with the results shown in the table. Estimate the average for the sixth grade, and place a confidence interval around the estimate. Then, place a bound around each sample mean, and summarize your findings (i.e., Are population differences likely among the groups?).

Track I		Track II		Track III
80 68 72 85 90 62 61	92 85 87 91 81 79 83	85 48 53 65 49 72 53 68 71 59	82 75 73 78 69 81 59 52 61 42	42 36 65 43 53 61 42 39	32 31 29 19 14 31 30 32