Solutions — Confidence Intervals for Proportions

Fluency A sample of 100 people finds 62 support a policy. Find p̂ and construct a 95% CI for the population proportion.

p̂ = 62/100 = 0.62

Check conditions: np̂ = 62 ≥ 5 ✓; n(1−p̂) = 38 ≥ 5 ✓

SE = √(0.62 × 0.38 / 100) = √(0.002356) ≈ 0.04854
ME = 1.96 × 0.04854 ≈ 0.09513

95% CI: (0.62 − 0.095, 0.62 + 0.095) = (0.525, 0.715)

We are 95% confident the true proportion who support the policy is between 52.5% and 71.5%.

Fluency A sample of 400 voters has p̂ = 0.45. Construct a 90% CI for the true proportion p.

Given: n = 400, p̂ = 0.45, z* = 1.645 (90% CI)

Check conditions: np̂ = 180 ≥ 5 ✓; n(1−p̂) = 220 ≥ 5 ✓

SE = √(0.45 × 0.55 / 400) = √(0.00061875) ≈ 0.02488
ME = 1.645 × 0.02488 ≈ 0.04093

90% CI: (0.45 − 0.041, 0.45 + 0.041) = (0.409, 0.491)

We are 90% confident the true voter proportion is between 40.9% and 49.1%.

Fluency In a batch of 250 items, 15 are defective. Find the 99% CI for the population defect rate.

p̂ = 15/250 = 0.06

Check conditions: np̂ = 15 ≥ 5 ✓; n(1−p̂) = 235 ≥ 5 ✓

SE = √(0.06 × 0.94 / 250) = √(0.0002256) ≈ 0.01502
ME = 2.576 × 0.01502 ≈ 0.03869

99% CI: (0.06 − 0.039, 0.06 + 0.039) = (0.021, 0.099)

We are 99% confident the true defect rate is between 2.1% and 9.9%.

Fluency Check the normality conditions for building a CI when n = 60 and p̂ = 0.08. Are conditions met?

Conditions required: np̂ ≥ 5 and n(1−p̂) ≥ 5

np̂ = 60 × 0.08 = 4.8 < 5 ✗
n(1−p̂) = 60 × 0.92 = 55.2 ≥ 5 ✓

Conditions are NOT met. The expected number of successes (4.8) is less than 5, so the normal approximation is not valid. A confidence interval based on the normal approximation should not be used with this sample. A larger sample is needed.

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.

p̂ = 345/500 = 0.69

Check conditions: np̂ = 345 ≥ 5 ✓; n(1−p̂) = 155 ≥ 5 ✓

SE = √(0.69 × 0.31 / 500) = √(0.0004278) ≈ 0.02068
ME = 1.96 × 0.02068 ≈ 0.04053

95% CI: (0.69 − 0.041, 0.69 + 0.041) = (0.649, 0.731)

Yes, the interval strongly suggests a majority own a smartphone. The entire interval (64.9% to 73.1%) lies above 50%. We are 95% confident the true proportion exceeds 50% — in fact, it is well above that threshold.

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?

Pollster A (n = 100):
SE = √(0.52 × 0.48 / 100) ≈ 0.04999
ME = 1.96 × 0.04999 ≈ 0.0980
CI: (0.422, 0.618) — width ≈ 0.196

Pollster B (n = 1000):
SE = √(0.52 × 0.48 / 1000) ≈ 0.01580
ME = 1.96 × 0.01580 ≈ 0.0310
CI: (0.489, 0.551) — width ≈ 0.062

Pollster B’s interval is narrower because the standard error SE = √(p̂(1−p̂)/n) decreases as n increases (it is proportional to 1/√n). A larger sample provides more information about the population, producing a more precise estimate.

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 the standard error formula applies).

(a) p̂ is the midpoint of the interval:
p̂ = (0.38 + 0.52) / 2 = 0.45

(b) Margin of error = half the width of the interval:
ME = (0.52 − 0.38) / 2 = 0.07

(c) ME = 1.96 × √(p̂(1−p̂)/n)
0.07 = 1.96 × √(0.45 × 0.55 / n)
0.07/1.96 = √(0.2475/n)
0.035714 = √(0.2475/n)
0.035714² = 0.2475/n
0.001276 = 0.2475/n
n = 0.2475 / 0.001276 ≈ 194

Understanding Explain why changing the confidence level from 95% to 99% (same n and p̂) produces a wider interval. Use the formula to illustrate.

The formula is: p̂ ± z* × √(p̂(1−p̂)/n)

When n and p̂ are fixed, the SE is constant. The only thing that changes is z*:
• 95% CI: z* = 1.96
• 99% CI: z* = 2.576

A larger z* multiplies the same SE by a larger factor, giving a larger margin of error and thus a wider interval.

Example (p̂ = 0.5, n = 100, SE = 0.05):
95% CI: ME = 1.96 × 0.05 = 0.098 → width = 0.196
99% CI: ME = 2.576 × 0.05 = 0.129 → width = 0.258

To be more confident that the interval captures the true p, we must make the net wider. Higher confidence always comes at the cost of precision (wider interval) unless we also increase n.

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.

p̂ = 21/300 = 0.07

Check conditions: np̂ = 21 ≥ 5 ✓; n(1−p̂) = 279 ≥ 5 ✓

SE = √(0.07 × 0.93 / 300) = √(0.0002170) ≈ 0.01473
ME = 1.96 × 0.01473 ≈ 0.02887

95% CI: (0.07 − 0.029, 0.07 + 0.029) = (0.041, 0.099)

The manufacturer’s claimed defect rate of 5% (p = 0.05) is contained within the 95% CI (0.041, 0.099). Therefore, the inspector’s result is consistent with the manufacturer’s claim — we cannot conclude the true defect rate is higher than 5% based on this sample alone. However, the interval is wide and the upper end (9.9%) is considerably higher than 5%, so monitoring with a larger sample would be prudent.

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?

(a) Candidate A: p̂_A = 0.54, n = 800
SE_A = √(0.54 × 0.46 / 800) = √(0.00031050) ≈ 0.01762
ME_A = 1.96 × 0.01762 ≈ 0.03454
95% CI for A: (0.505, 0.575)

Candidate B: p̂_B = 0.46, n = 800
SE_B = √(0.46 × 0.54 / 800) ≈ 0.01762 (same)
95% CI for B: (0.425, 0.495)

(b) The intervals do not overlap (A’s lower bound 0.505 > B’s upper bound 0.495). This is strong evidence that candidate A is genuinely ahead — the entire plausible range for A’s true proportion lies above the entire plausible range for B’s true proportion at the 95% confidence level.

(c) We need the lower bound of A’s CI to exceed 0.50:
p̂ − 1.96 × √(p̂(1−p̂)/n) > 0.50
0.54 − 1.96 × √(0.54 × 0.46/n) > 0.50
1.96 × √(0.2484/n) < 0.04
√(0.2484/n) < 0.02041
0.2484/n < 0.0004166
n > 0.2484 / 0.0004166 ≈ 596

So n ≥ 597 would give a 95% CI for A entirely above 0.50 (assuming p̂ remains at 0.54). The current sample of 800 already achieves this.