AP Statistics · Free-response skill

A significance test for a proportion

The move. Run a one-proportion significance test from start to finish and state the conclusion about the population in context, linked to the significance level.

How to do it

  1. Hypotheses: write H0:p=p0H_0: p = p_0 and HaH_a with the direction the question suspects, and define p in words as the proportion of the whole population, not the sample.
  2. Conditions: say the sample is random, check that n is less than 10 percent of the population, and check large counts with the null value: np0np_0 and n(1−p0)n(1 - p_0) both at least 10.
  3. Mechanics: name the test, find p^\hat{p}, compute z=p^−p0p0(1−p0)/nz = \frac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}} with the null value in the standard deviation, and find the P-value in the direction of HaH_a.
  4. Conclusion: compare the P-value with alpha, say reject or fail to reject, then say whether there is convincing evidence for HaH_a, in context and about the population. Never say the test proves either hypothesis.

Practice prompt Try it

A transit official in the fictional city of Halden says that 60 percent of the city's adult residents support building a new bus line. A city council member suspects the true level of support is lower. The council takes a random sample of 250 of the city's 80,000 adult residents, and 135 of them say they support the bus line.

Use the situation described in the passage to answer each part. Show your work.

  1. Set up the council member's test: write the two hypotheses in symbols and say in words what your parameter stands for.
  2. Show that the conditions for the test are met, using numbers from the situation.
  3. Carry out a one-sample z test for a proportion: work out the test statistic and the P-value.
  4. Using a significance level of 0.05, say what the result tells the council about support for the bus line among Halden's adults.

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Checklist

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Model answer, annotated

(a) H0:p=0.60H_0: p = 0.60 and Ha:p<0.60H_a: p < 0.60, where p is the proportion of all adult residents of Halden who support building the new bus line.

  • Point 1 The null uses the official's 60 percent, and the alternative points down because the council member suspects lower support.
  • Point 2 Defines the parameter for the whole population of Halden adults, not for the sample of 250.

(b) Random: the council took a random sample of adults. 10 percent: 250 is less than 10 percent of the 80,000 adult residents (8,000), so the responses can be treated as independent. Large counts: np0=250(0.60)=150np_0 = 250(0.60) = 150 and n(1−p0)=250(0.40)=100n(1 - p_0) = 250(0.40) = 100 are both at least 10, so the sampling distribution of p^\hat{p} is approximately normal.

  • Point 3 States that the sample was random.
  • Point 4 Backs the 10 percent condition with the actual numbers, 250 against 8,000.
  • Point 5 Uses the null value 0.60, not the sample proportion, because the test assumes the null is true.

(c) This is a one-sample z test for a proportion. The sample proportion is p^=135250=0.54\hat{p} = \frac{135}{250} = 0.54. The standard deviation under the null is 0.60(0.40)250≈0.031\sqrt{\frac{0.60(0.40)}{250}} \approx 0.031, so z=0.54−0.600.031≈−1.94z = \frac{0.54 - 0.60}{0.031} \approx -1.94. The P-value is P(Z<−1.94)≈0.026P(Z < -1.94) \approx 0.026.

  • Point 6 Puts the null value 0.60 in the standard deviation, because the test assumes the null is true, then standardizes.
  • Point 7 Takes the lower tail because the alternative is less than.

(d) Because the P-value of about 0.026 is less than 0.05, we reject H0H_0. There is convincing evidence that the proportion of all adult residents of Halden who support the new bus line is less than 0.60, as the council member suspected.

  • Point 8 Links the decision to the significance level with an explicit comparison.
  • Point 9 States the conclusion about all Halden adults in the words of the context, as evidence rather than proof.