Lesson 12: Inference for Proportions

October 7, 2025

Review:

• Inference for a population mean
• Confidence Intervals
• Lesson 11 problem solutions

Presentation:

• How to Estimate a Population Proportion
  • Conduct a Simple Random Sample (SRS) of size n
  • Record the count, x, of some attribute attributed to a portion of the population (e.g., number of voters favoring a candidate)
  • Calculate the sample proportion, p-hat = x/n
  • If n is sufficiently large (≥30), we can assume p-hat is Normally distributed
  • Estimate of the population proportion mean, μ = p-hat
  • Estimate of the population proportion std dev, σ = √(p*(1-p)/n)
  • Estimate margin of error, m = z*σ  (use z = 1.96 for 95% confidence)
  • Estimate with 95% confidence interval is p-hat ± m
  • Same procedure for a small sample size (<30) using the t distribution and substituting t* for z*

• Binge Drinking Survey
  • n = 13,819
  • x = 3,140
  • p-hat = 3140/13819 = 0.227
  • standard deviation = √(p-hat*(1-p-hat)/n) = 0.00356
  • p-hat ± z*√(p-hat*(1-p-hat)/n) = 0.227 ± 1.96*(0.00356) = 0.227 ± 0.007
  • {0.220,0.234}
• Video – Against All Odds Unit 28
• Consumer Survey – Online vs in-store shopping.
  • A marketing firm surveyed 200 consumers, and 126 said they prefer shopping online to in-store.
  • Construct a 95% confidence interval for the true proportion of consumers who prefer online shopping.

Activity:

1. A random sample of 2,454 12th-grade students were asked the following question: Taking all things together, how would you say things are these days – would you say you’re happy or not too happy? Of the responses, 2,098 students selected happy. Determine the sample proportion of students who responded they were happy and calculate a 95% confidence interval for the population proportion of 12th-grade students who are happy.
2. A phone survey contacted 1,910 households in which a computer was owned and respondents were asked if they could access the Internet from their home. A total of 1,816 of the households responded yes. Calculate a 95% confidence interval to estimate the proportion of American households with internet access.
3. Currently, mothers in North America are advised to put babies to sleep on their backs. This recommendation has reduced the number of cases of sudden infant death syndrome (SIDS). However, it is a likely cause of another problem, i.e., flat spots on babies’ heads. A study of 440 babies aged 7 – 12 weeks found that 46.6% had flat spots on their heads. Calculate a 95% confidence interval for the proportion of babies in this age group that have flat spots.

Assignment:

1. A marketing firm surveyed 200 consumers, and 126 said they prefer shopping online to in-store. Construct a 95% confidence interval for the true proportion of consumers who prefer online shopping.
2. A start-up wants to estimate how many potential customers would buy its new smartwatch. Out of a random sample of 300 adults, 174 said they would consider buying one. Construct a 95% confidence interval for the true proportion of adults likely to buy the smartwatch.
3. A factory produces smartphone screens. Out of 600 screens sampled, 18 were defective. Find a 95% confidence interval for the true defect rate.
4. In a statewide poll of 1,200 registered voters, 516 said they support Candidate A. Compute a 95% confidence interval for the true proportion supporting Candidate A.
5. Out of 250 students enrolled in an online statistics course, 205 finished all assignments. Estimate the true course-completion rate with a 95% confidence interval.
6. A retail store examines 400 purchases and finds 28 items were later returned. Construct a 95% confidence interval for the true return proportion.
7. A data-analytics firm records that 310 of 500 Instagram posts received at least one comment. Compute a 95% confidence interval for the population proportion of posts with engagement.
8. A company offers a new health-benefit plan. Out of 180 employees, 132 enroll. Calculate a 95% confidence interval for the true proportion who will enroll company-wide. Can the company claim that “over 70% of employees will sign up.”