AP Statistics · Lesson 1

Intervals and tests for one proportion

Know Know it

A confidence interval for a population proportion p is p^±z∗p^(1−p^)n\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. The part after ±\pm is the margin of error: the critical value z∗z^* times the standard error. Common critical values are 1.645 for 90 percent, 1.96 for 95 percent and 2.576 for 99 percent.

Conditions. Random: the data come from a random sample (or a randomized experiment). 10 percent: when sampling without replacement, n is less than 10 percent of the population. Large counts: np^≥10n\hat{p} \geq 10 and n(1−p^)≥10n(1-\hat{p}) \geq 10, so the sampling distribution is approximately normal.

A wider interval comes from a higher confidence level or a smaller sample. To plan a sample size for a margin of error ME, solve z∗p∗(1−p∗)n≤MEz^*\sqrt{\frac{p^*(1-p^*)}{n}} \leq ME for n, using p∗=0.5p^* = 0.5 when you have no prior estimate, and round up.

A significance test for p starts from H0:p=p0H_0: p = p_0 and an alternative HaH_a of p<p0p < p_0, p>p0p > p_0 or p≠p0p \neq p_0, chosen before seeing the data. The test statistic is z=p^−p0p0(1−p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}, and the large counts check uses np0np_0 and n(1−p0)n(1-p_0), because the test assumes H0H_0 is true.

The P-value is the probability, if H0H_0 were true, of a sample proportion at least as extreme as the one observed, in the direction of HaH_a. A two-sided alternative counts both tails. If the P-value is less than the significance level α\alpha, reject H0H_0; otherwise, fail to reject it.

Calculator On the exam

A graphing calculator with statistical capabilities is expected on both sections of the exam, and the exam provides a reference booklet of formulas and tables. The calculator can build a one-proportion interval or run a one-proportion test, but write the procedure name, the conditions, the formula with your numbers and the result, so a reader can follow every step.

Worked example Easy

In a random sample of 400 registered voters in a fictional city, 156 say they support a park bond, so the sample proportion is 0.39.

Which is a 95 percent confidence interval for the proportion of all registered voters in the city who support the bond?

  1. 0.350 to 0.430
  2. 0.366 to 0.414
  3. 0.342 to 0.438Answer
  4. 0.327 to 0.453

How to solve it

  1. Conditions: a random sample; 400 is less than 10 percent of the city's voters; 156 and 244 are both at least 10.
  2. Standard error: 0.39(0.61)400≈0.0244\sqrt{\frac{0.39(0.61)}{400}} \approx 0.0244.
  3. Margin of error: 1.96(0.0244)≈0.0481.96(0.0244) \approx 0.048, so the interval is 0.39±0.0480.39 \pm 0.048, from 0.342 to 0.438.

Why each choice is right or wrong

  • A. Incorrect. 0.350 to 0.430 uses z∗=1.645z^* = 1.645, which gives a 90 percent interval, narrower than 95 percent.
  • B. Incorrect. 0.366 to 0.414 adds and subtracts one standard error, leaving out the critical value of 1.96.
  • C. Correct. The standard error is 0.39(0.61)400≈0.0244\sqrt{\frac{0.39(0.61)}{400}} \approx 0.0244, and 0.39±1.96(0.0244)0.39 \pm 1.96(0.0244) gives 0.342 to 0.438.
  • D. Incorrect. 0.327 to 0.453 uses z∗=2.576z^* = 2.576, which gives a 99 percent interval, wider than 95 percent.

Trap The 95 percent is about the method

Once an interval is built, it either captures p or it does not. The confidence level describes the method: about 95 percent of intervals made this way from many random samples would capture p. Also, the interval is about the population proportion p, not the sample proportion, which is its known center.

Explain Explain it

Task verbs: interpret, calculate

To interpret an interval, name the confidence level, the endpoints and the parameter in context: "We are 95 percent confident that the interval from 0.342 to 0.438 captures the proportion of all registered voters in the city who support the bond."

To finish a test, compare the P-value with α\alpha and then say what that means for the parameter in context. Reject: "there is convincing evidence that" the alternative is true. Fail to reject: "there is not convincing evidence that" the alternative is true. Never say the data prove H0H_0.

Worked example Medium

A fictional streaming service says 20 percent of its subscribers watch mostly on a TV. An analyst suspects the share is higher. In a random sample of 250 subscribers, 62 watch mostly on a TV, a sample proportion of 0.248. She tests H0:p=0.20H_0: p = 0.20 against Ha:p>0.20H_a: p > 0.20.

Which gives the test statistic and P-value?

  1. z = 1.90 and P-value = 0.029Answer
  2. z = 1.90 and P-value = 0.058
  3. z = 1.76 and P-value = 0.039
  4. z = 1.90 and P-value = 0.971

How to solve it

  1. Large counts with the null value: 250(0.20)=50250(0.20) = 50 and 250(0.80)=200250(0.80) = 200, both at least 10.
  2. Standard deviation under H0H_0: 0.20(0.80)250≈0.0253\sqrt{\frac{0.20(0.80)}{250}} \approx 0.0253.
  3. z=0.248−0.200.0253≈1.90z = \frac{0.248 - 0.20}{0.0253} \approx 1.90, and since HaH_a is greater than, the P-value is P(Z>1.90)≈0.029P(Z > 1.90) \approx 0.029.

Why each choice is right or wrong

  • A. Correct. z=0.248−0.200.20(0.80)/250=0.0480.0253≈1.90z = \frac{0.248 - 0.20}{\sqrt{0.20(0.80)/250}} = \frac{0.048}{0.0253} \approx 1.90, and the area to the right of 1.90 is about 0.029.
  • B. Incorrect. A P-value of 0.058 doubles the tail area, which fits a two-sided alternative, but HaH_a here is one-sided.
  • C. Incorrect. z = 1.76 uses the sample proportion 0.248 in the standard deviation; a test uses the null value 0.20.
  • D. Incorrect. A P-value of 0.971 is the area to the left of z = 1.90, but Ha:p>0.20H_a: p > 0.20 points to the right tail.

Interpret Your turn

A fictional food bank surveyed a random sample of 500 households in its county. A 90 percent confidence interval for the proportion of all households in the county that used the food bank last year is 0.114 to 0.166. Interpret the interval in context, then say what the 90 percent confidence level means.

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A full answer does this

  • Interprets the interval: 90 percent confident it captures the proportion of all households in the county that used the food bank last year
  • Explains the confidence level as the capture rate of the method over many random samples

Model answer

We are 90 percent confident that the interval from 0.114 to 0.166 captures the true proportion of all households in the county that used the food bank last year. The 90 percent means that if the food bank took many random samples of 500 households and built an interval from each one in the same way, about 90 percent of those intervals would capture the true proportion.

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