Confidence intervals for a proportion

About this lesson

Statistics (AP-aligned) › Confidence intervals

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Confidence intervals for a proportion

Warm-up

Three quick questions from earlier lessons.

Discover

In a sample of 400, p̂ = 0.5. The standard error is 0.5×0.5/400=0.025\sqrt{0.5 \times 0.5 / 400} = 0.025. For 95% confidence (z* ≈ 2), what is the margin of error?

  1. 0.05
  2. 0.025

Key idea

The one-proportion z interval.

p^±z∗p^(1−p^)n\hat{p} \pm z^*\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}, for a random sample with np^≥10n\hat{p} \ge 10 and n(1−p^)≥10n(1 - \hat{p}) \ge 10. For 95% confidence, z∗=1.96z^* = 1.96. Interpret: "we are 95% confident that the interval captures the population proportion".

Worked example

n=100n = 100, p^=0.3\hat{p} = 0.3, 95%.

SE≈0.0458SE \approx 0.0458
√(0.3 × 0.7/100).
ME≈0.090ME \approx 0.090
1.96 × SE.
(0.210,0.390)(0.210, 0.390)
0.3 ± 0.090.

Your turn, guided

Your turn: SE = 0.02 and z* = 2. The margin of error.

0.040.04
2 × 0.02.

Practice ★

  1. In a random sample of 400, p̂ = 5/10. Using z* = 1.96, find the margin of error of a 95% interval for p.
  2. In a random sample of 400, p̂ = 9/10. Using z* = 1.96, find the margin of error of a 95% interval for p.
  1. In a random sample of 100, p̂ = 3/10. Using z* = 2, find the upper end of the approximate 95% interval for p.

Watch out!

Practice ★★

  1. A 95% confidence interval for the proportion of voters who support a plan is (0.48, 0.52). Which interpretation is right?
    1. We are 95% confident that the interval from 0.48 to 0.52 captures the true proportion of supporters
    2. 95% of the sample supports the plan
    3. There is a 95% probability that p is between 0.48 and 0.52
  1. Boss problem. In a random sample of 100, p̂ = 3/10. Build a 95% interval with z* = 1.96.

Exit ticket

Three questions to check you have got it.

  1. In a random sample of 400, p̂ = 5/10. Using z* = 1.96, find the margin of error of a 95% interval for p.
  2. In a random sample of 400, p̂ = 9/10. Using z* = 1.96, find the margin of error of a 95% interval for p.
  3. In a random sample of 400, p̂ = 8/10. Using z* = 1.96, find the margin of error of a 95% interval for p.

Summary card

Interval for p

p^±z∗p^(1−p^)/n\hat{p} \pm z^*\sqrt{\hat{p}(1 - \hat{p})/n}; check random and large counts.

Watch out: "confident", not "probability", for a computed interval.

End of the lesson

Next lesson: Confidence level and critical values Back to the chapter

Answers: Confidence intervals for a proportion

A

  1. 491000\frac{49}{1000}
  2. 1475000\frac{147}{5000}

B

  1. 2150+310\frac{\sqrt{21}}{50} + \frac{3}{10}

C

  1. We are 95% confident that the interval from 0.48 to 0.52 captures the true proportion of supporters

D

  1. 310−49212500\frac{3}{10} - \frac{49 \sqrt{21}}{2500}

E

  1. 491000\frac{49}{1000}
  2. 1475000\frac{147}{5000}
  3. 491250\frac{49}{1250}

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