AP Statistics Significance Tests for a Proportion Worksheet Solutions (PDF)

(a) \(H_0: p=0.90\), \(H_a: p < 0.90\), p is the true proportion of packages arriving on time. (b) \(H_0: p=0.5\), \(H_a: p\neq 0.5\), p is the true proportion of heads. (c) \(H_0: p=0.60\), \(H_a: p > 0.60\), p is the true proportion of students supporting block scheduling.

Hypotheses as shown.

Random: given. 10%: 120 < 250. Large counts using p0 = 0.40: \(np_0=48\) and \(n(1-p_0)=72\), both at least 10.

All conditions are satisfied.

\(\hat{p}=\frac{54}{120}=0.45\). \(SE=\sqrt{\frac{0.4(0.6)}{120}}=0.0447\). \(z=\frac{0.45-0.40}{0.0447}=1.12\).

z = 1.12.

\(P(Z>1.12)=1-0.8686=0.1314\).

0.1314.

0.1314 > 0.05.

Because the p-value is greater than 0.05, we fail to reject H0. There is not convincing evidence that more than 40% of customers prefer self-checkout.

The p-value is computed assuming H0 is true.

If exactly 50% of voters favored the proposal, there would be a 2.1% chance of getting a sample proportion as far from 0.5 as the one observed, or farther, in a random sample of 400 voters.

\(z=\frac{0.62-0.5}{0.05}=2.40\). p-value \(=2P(Z > 2.40)=0.0164\).

0.0164 < 0.05: reject H0; convincing evidence the coin is unfair.

\(\hat{p}=0.92\). \(SE=\sqrt{\frac{0.95(0.05)}{200}}=0.0154\). \(z=-1.95\). p-value \(=0.0256\).

Reject H0: convincing evidence that fewer than 95% of bulbs last over 1,000 hours.

The p-value exceeds alpha.

We fail to reject H0. There is not convincing evidence that p differs from 0.5. We never accept H0, since the data could be consistent with other values.

0.03 < 0.05 but 0.03 > 0.01.

At alpha = 0.05, reject H0. At alpha = 0.01, fail to reject H0.

0.50 lies outside the interval.

Reject H0: there is convincing evidence that p is not 0.50. (Intervals and two-sided tests give consistent conclusions.)

A Type I error is rejecting a true H0.

Concluding the schedule increases attendance when it actually does not; the school might spend money on a schedule that has no effect.

A Type II error fails to reject a false H0.

Concluding there is no convincing evidence that the new schedule increases attendance when it really does; the school misses a beneficial change.

The probability of a Type I error equals the significance level.

0.10.

Larger n reduces the standard error, so the sampling distribution is narrower and a true difference is easier to detect.

Increase the sample size; increase alpha (or study a larger true effect).

Type I: tell a healthy patient they may be sick. Type II: tell a sick patient they are fine.

Type II is usually more serious because a sick patient would not receive treatment, though a Type I error causes worry and extra testing.

State: \(H_0: p=0.50\), \(H_a: p < 0.50\).

Plan: one-proportion z test; random, 10%, \(np_0=150\) and \(n(1-p_0)=150\) at least 10.

Do: \(\hat{p}=0.44,\ SE=\sqrt{0.25/300}=0.0289,\ z=-2.08,\ p=0.0188\).

Conclude: p-value < 0.05, so reject H0. There is convincing evidence that fewer than half of the city's voters favor the tax.

n = 100: \(z=\frac{-0.06}{0.05}=-1.20\). n = 1000: \(z=\frac{-0.06}{0.0158}=-3.79\).

The larger sample gives a much larger |z|, so the same difference is more convincing with more data.

Statistical significance tells us the difference is unlikely to be chance, not that it is large.

Statistically significant but very small (0.4 percentage points), so it has little practical importance.

\(H_0: p=0.60,\ H_a: p > 0.60\). \(\hat{p}=0.63,\ SE=\sqrt{0.24/400}=0.0245,\ z=1.22\). p-value \(=0.1112\).

Fail to reject H0 at alpha = 0.01: the evidence is not convincing at that stricter level.