← Interval Estimates for Proportions › Confidence Intervals for Proportions
Confidence Intervals for Proportions
Key Terms
- A confidence interval for p is a range of plausible values for the true population proportion p
- The interval is centred on p̂ (sample proportion) and extends by the margin of error (ME)
- Formula: p̂ ± z* × √(p̂(1−p̂)/n)
- z* is the critical value: 1.645 (90%), 1.96 (95%), 2.576 (99%)
- A 95% CI means: if we repeated sampling many times, 95% of intervals would contain the true p
- Conditions for validity: np̂ ≥ 5 and n(1−p̂) ≥ 5
p̂ ± 1.96 × √(p̂(1−p̂)/n)
General CI:
p̂ ± z* × √(p̂(1−p̂)/n)
Critical values z*:
90% CI: z* = 1.645
95% CI: z* = 1.96
99% CI: z* = 2.576
p̂ = 110/200 = 0.55
Check conditions: np̂ = 110 ≥ 5 ✓; n(1−p̂) = 90 ≥ 5 ✓
SE = √(0.55 × 0.45 / 200) = √(0.0012375) ≈ 0.03518
ME = 1.96 × 0.03518 ≈ 0.06895
95% CI: (0.55 − 0.069, 0.55 + 0.069) = (0.481, 0.619)
Interpretation: We are 95% confident the true proportion who prefer Brand A is between 48.1% and 61.9%.
What is a Confidence Interval for a Proportion?
A confidence interval is an interval estimate that gives a plausible range for an unknown population parameter. For a population proportion p, the interval is based on the sample proportion p̂ and uses the approximate normality of p̂ for large samples.
Constructing the Interval
The approximate 95% confidence interval for p is:
p̂ ± 1.96 × √(p̂(1−p̂)/n)
The term √(p̂(1−p̂)/n) is the estimated standard error of p̂. Multiplying by z* = 1.96 gives the margin of error, which determines the half-width of the interval.
Interpreting the Confidence Level
A 95% confidence level does not mean there is a 95% chance the true p is in this particular interval. Once the interval is computed, either p is in it or it isn’t. Instead, the 95% refers to the procedure: if we took many random samples and built a CI from each, about 95% of those intervals would contain the true p.
Conditions for Validity
The normal approximation is valid when:
- np̂ ≥ 5 (at least 5 successes in the sample)
- n(1 − p̂) ≥ 5 (at least 5 failures in the sample)
- The sample is random and independent
Comparing Confidence Levels
Higher confidence requires a wider interval for the same sample size:
- 90% CI: z* = 1.645 (narrower, less sure)
- 95% CI: z* = 1.96 (standard)
- 99% CI: z* = 2.576 (wider, more sure)
Mastery Practice
- Fluency A sample of 100 people finds 62 support a policy. Find p̂ and construct a 95% CI for the population proportion.
- Fluency A sample of 400 voters has p̂ = 0.45. Construct a 90% CI for the true proportion p.
- Fluency In a batch of 250 items, 15 are defective. Find the 99% CI for the population defect rate.
- Fluency Check the normality conditions for building a CI when n = 60 and p̂ = 0.08. Are conditions met?
- Understanding A study of 500 teenagers finds 345 own a smartphone. Construct a 95% CI. Does the interval suggest the majority (more than 50%) own a smartphone? Explain.
- Understanding Two pollsters survey the same population about the same issue. Pollster A uses n = 100 and gets p̂ = 0.52. Pollster B uses n = 1000 and gets p̂ = 0.52. Both construct 95% CIs. Which interval is narrower and why?
- Understanding A 95% CI for p is (0.38, 0.52). State: (a) the sample proportion p̂ (b) the margin of error (c) the sample size n (given σ = standard error formula applies).
- Understanding Explain why changing the confidence level from 95% to 99% (same n and p̂) produces a wider interval. Use the formula to illustrate.
- Problem Solving A manufacturer claims its product has a 5% defect rate. A quality inspector tests 300 products and finds 21 defectives. Construct a 95% CI. Does the interval support the manufacturer’s claim? Explain your reasoning.
- Problem Solving In an election, candidate A receives 54% of votes in a random exit poll of 800 voters. Candidate B receives 46%.
- (a) Construct 95% CIs for both candidates’ true proportions.
- (b) Do the intervals overlap? What does this mean for declaring a winner?
- (c) What sample size would give candidate A a CI entirely above 0.50?