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Construct and interpret confidence intervals for a population proportion.
Learn step-by-step with practice exercises built right in.
A confidence interval for a population proportion p is:
In a random sample of 400 voters, 220 support a proposition. Construct a 95% confidence interval for the true proportion of voters who support the proposition.
Step 1: Identify the information n = 400 (sample size) x = 220 (number of successes) pฬ = 220/400 = 0.55 (sample proportion)
Confidence level: 95%
Step 2: Check conditions for proportion CI RANDOM: Sample is random โ NORMAL: npฬ โฅ 10 and n(1-pฬ) โฅ 10 400(0.55) = 220 โฅ 10 โ 400(0.45) = 180 โฅ 10 โ INDEPENDENT: n โค 0.10N 400 โค 0.10(all voters) - assume yes โ
All conditions met!
Step 3: Find critical value 95% confidence โ ฮฑ = 0.05 z* = 1.96 (from table for 95% CI)
Step 4: Calculate standard error SE = โ[pฬ(1-pฬ)/n] = โ[0.55(0.45)/400] = โ[0.2475/400] = โ0.00061875 โ 0.0249
Step 5: Calculate margin of error ME = z* ร SE = 1.96 ร 0.0249 โ 0.0488
Step 6: Construct confidence interval CI = pฬ ยฑ ME = 0.55 ยฑ 0.049 = (0.501, 0.599)
Avoid these 3 frequent errors
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where:
Common critical values:
Before constructing CI for proportions, verify:
If these fail, don't use the standard formula.
A poll of 400 likely voters finds 220 support Candidate A. Construct a 95% CI for the population proportion.
Step 1:
Step 2: Check conditions:
Step 3: Calculate SE:
Step 4: Calculate margin of error (ME):
Step 5: Confidence interval:
Interpretation: We are 95% confident that between 50.1% and 59.9% of likely voters support Candidate A.
The margin of error (ME) represents the uncertainty in the point estimate:
Larger ME means wider CI (less precise). Smaller ME means narrower CI (more precise).
To find the sample size needed for margin ME:
If p is unknown, use p = 0.5 (gives maximum sample size; most conservative).
Example: You want 95% CI with ME = 0.03. How large must the sample be?
Know the margin of error formula cold. Many free-response questions ask you to find the sample size needed to achieve a specific margin. Always verify conditions before calculating; if conditions fail, state which ones and explain why the interval is not valid.
Or: (0.50, 0.60) rounded
Step 7: Interpret the interval We are 95% confident that the true proportion of voters who support the proposition is between 0.50 and 0.60 (or 50% and 60%).
This means:
Answer: 95% CI for p: (0.50, 0.60)
We are 95% confident that between 50% and 60% of all voters support the proposition.
A quality control inspector finds 8 defects in a sample of 200 items. Construct a 90% confidence interval for the defect rate.
Step 1: Calculate sample proportion n = 200 x = 8 pฬ = 8/200 = 0.04
Step 2: Check conditions RANDOM: Assume random sample โ NORMAL: npฬ = 200(0.04) = 8 < 10 โ n(1-pฬ) = 200(0.96) = 192 โฅ 10 โ
Condition fails! But let's proceed with caution. (In practice, might use exact binomial method)
Step 3: Find z* for 90% confidence 90% confidence โ z* = 1.645
Step 4: Calculate SE SE = โ[pฬ(1-pฬ)/n] = โ[0.04(0.96)/200] = โ[0.0384/200] = โ0.000192 โ 0.0139
Step 5: Calculate ME ME = 1.645 ร 0.0139 โ 0.023
Step 6: Construct CI CI = 0.04 ยฑ 0.023 = (0.017, 0.063) = (1.7%, 6.3%)
Step 7: Interpret with caution We are 90% confident the true defect rate is between 1.7% and 6.3%.
Note: This interval may not be as reliable since npฬ < 10.
Answer: 90% CI: (0.017, 0.063) or (1.7%, 6.3%)
Caution: The success-failure condition is marginally violated (only 8 successes), so this normal-based interval may not be fully reliable.
A researcher wants to estimate the proportion of defective items with a margin of error no more than 0.03 at 90% confidence. How large a sample is needed if no prior estimate exists?
Step 1: Identify what we need ME = 0.03 Confidence level = 90% โ z* = 1.645 No prior estimate โ use pฬ = 0.5
Step 2: Use sample size formula n = (z*)ยฒpฬ(1-pฬ)/MEยฒ
Step 3: Calculate n = (1.645)ยฒ(0.5)(0.5)/(0.03)ยฒ = 2.706(0.25)/0.0009 = 0.6765/0.0009 โ 751.67
Step 4: Round UP Always round UP to ensure ME is no larger than desired n = 752
Step 5: Why use pฬ = 0.5? The product pฬ(1-pฬ) is maximized at pฬ = 0.5 This gives the most conservative (largest) sample size Guarantees ME โค 0.03 regardless of true p
Answer: n = 752
Need a sample of at least 752 items to achieve a margin of error no more than 0.03 at 90% confidence.
Two polls: Poll A (n=500, pฬ=0.52) and Poll B (n=1000, pฬ=0.51). Both use 95% confidence. Which poll has a smaller margin of error? Calculate both.
Step 1: Recall margin of error formula ME = z*โ[pฬ(1-pฬ)/n]
For 95% CI: z* = 1.96
Step 2: Calculate ME for Poll A pฬ = 0.52, n = 500
ME_A = 1.96โ[0.52(0.48)/500] = 1.96โ[0.2496/500] = 1.96โ0.0004992 = 1.96(0.0223) โ 0.044
Step 3: Calculate ME for Poll B pฬ = 0.51, n = 1000
ME_B = 1.96โ[0.51(0.49)/1000] = 1.96โ[0.2499/1000] = 1.96โ0.0002499 = 1.96(0.0158) โ 0.031
Step 4: Compare Poll A: ME โ 0.044 or 4.4% Poll B: ME โ 0.031 or 3.1%
Poll B has smaller margin of error!
Step 5: Why is Poll B better? Larger sample size (1000 vs 500) ME โ 1/โn Doubling n reduces ME by factor of โ2 โ 1.41
500 ร 2 = 1000 ME_A/ME_B = โ(1000/500) = โ2 โ 1.41 0.044/0.031 โ 1.42 โ
Step 6: Effect of pฬ Poll B also has pฬ closer to 0.5 But this increases ME slightly Effect of larger n dominates
Answer: Poll B has smaller ME (0.031 vs 0.044)
Poll B's larger sample size (1000 vs 500) gives more precision despite having pฬ closer to 0.5.
Explain why we can't construct a valid confidence interval for a proportion when the sample proportion is 0 or 1.
Step 1: Recall CI formula CI = pฬ ยฑ z*โ[pฬ(1-pฬ)/n]
SE = โ[pฬ(1-pฬ)/n]
Step 2: What happens when pฬ = 0? SE = โ[0(1)/n] = 0 CI = 0 ยฑ 0 = (0, 0)
This says we're 100% certain p = 0 Unreasonable from a sample!
Step 3: What happens when pฬ = 1? SE = โ[1(0)/n] = 0 CI = 1 ยฑ 0 = (1, 1)
This says we're 100% certain p = 1 Also unreasonable!
Step 4: Normal approximation fails Need: npฬ โฅ 10 AND n(1-pฬ) โฅ 10
When pฬ = 0: npฬ = 0 < 10 โ When pฬ = 1: n(1-pฬ) = 0 < 10 โ
Can't use normal-based method!
Step 5: What to do instead Use: Wilson score interval, Agresti-Coull, or exact binomial methods These give reasonable intervals even with extreme values
Answer: When pฬ = 0 or 1, SE = 0, giving a degenerate interval. Normal approximation conditions fail. Alternative methods should be used.