This ap statistics proportions worksheet answer key solves all 24 problems of the inference for proportions worksheet step by step.
Every solution shows the conditions, the formula, the arithmetic and a conclusion in context. Intervals use the table critical values 1.645, 1.960 and 2.576, and P-values come from the standard normal model.
AP Statistics Proportions Worksheet Answer Key: problems 1–5
Problem 1: Large Counts check
Here \(\hat{p}=\frac{9}{45}=0.20\). The counts are \(n\hat{p}=9\) and \(n(1-\hat{p})=36\).
Because \(9 < 10\), the condition fails. Large Counts is not met, so a z interval is not appropriate.
Problem 2: Standard error of a proportion
\(\hat{p}=\frac{90}{150}=0.60\).
\(SE=\sqrt{\frac{0.60(0.40)}{150}}=\sqrt{0.0016}=\)0.040.
Problem 3: Critical values
For confidence level \(C\), \(z^{*}\) leaves \(\frac{1-C}{2}\) in each tail.
- 90%: area \(0.95\) to the left gives 1.645.
- 95%: area \(0.975\) to the left gives 1.960.
- 99%: area \(0.995\) to the left gives 2.576.
Problem 4: One-proportion z interval
Large Counts. \(\hat{p}=\frac{85}{250}=0.340\); the counts 85 and 165 are both at least 10.
Interval. \(SE=\sqrt{\frac{0.34(0.66)}{250}}=0.0300\), so \(ME=1.960(0.0300)=0.059\).
\(0.340-0.059=0.281\) and \(0.340+0.059=0.399\). The 95% interval is (0.281, 0.399).
Problem 5: Interpreting an interval
We are 95% confident that the interval from 0.281 to 0.399 captures the true proportion of all players in the league who play a second sport.
The sentence names the confidence level, the interval, the population, and the parameter.
AP Statistics Proportions Worksheet Answer Key: problems 6–10
Problem 6: Meaning of the confidence level
The true proportion is a fixed number, not a random quantity, so it is either in the interval or it is not. The 95% describes the method.
If many random samples of the same size were taken and a 95% interval built from each, about 95% of those intervals would capture the true proportion.
Problem 7: Sample proportion and margin of error
The sample proportion is the midpoint: \(\hat{p}=\frac{0.412+0.508}{2}=\)0.460.
The margin of error is half the width: \(ME=\frac{0.508-0.412}{2}=\)0.048.
Problem 8: Effect of confidence level
\(\hat{p}=\frac{140}{400}=0.350\) and \(SE=\sqrt{\frac{0.35(0.65)}{400}}=0.0238\).
- 90%: \(ME=1.645(0.0238)=0.039\), interval (0.311, 0.389).
- 99%: \(ME=2.576(0.0238)=0.061\), interval (0.289, 0.411).
A higher confidence level needs a larger \(z^{*}\), so the interval is wider.
Problem 9: Sample size without a prior estimate
Use the conservative value \(\hat{p}=0.5\): \(n\ge(\frac{1.960}{0.03})^{2}(0.5)(0.5)=4268.4(0.25)=1067.1\).
Round up: n = 1,068.
Problem 10: Sample size with a prior estimate
\(n\ge(\frac{2.576}{0.02})^{2}(0.30)(0.70)=16589.4(0.21)=3483.8\).
Round up: n = 3,484.
AP Statistics Proportions Worksheet Answer Key: problems 11–15
Problem 11: Writing hypotheses
Let \(p\) be the true proportion of this company’s batteries that are defective.
\(H_0: p=0.08\) and \(H_a: p>0.08\).
Problem 12: Right-tailed z test
\(H_0: p=0.10\), \(H_a: p>0.10\). Large Counts: \(120(0.10)=12\) and \(120(0.90)=108\), both at least 10.
\(\hat{p}=\frac{18}{120}=0.150\), \(SE_0=\sqrt{\frac{0.10(0.90)}{120}}=0.0274\).
\(z=\frac{0.150-0.10}{0.0274}=\)1.83, and the P-value is \(P(Z>1.83)=\)0.034.
Problem 13: Two-sided z test
Large Counts: \(180(0.40)=72\) and \(180(0.60)=108\); both at least 10.
\(\hat{p}=\frac{63}{180}=0.350\), \(SE_0=\sqrt{\frac{0.40(0.60)}{180}}=0.0365\).
\(z=\frac{0.350-0.400}{0.0365}=\)-1.37.
Two-sided P-value \(=2P(Z < -1.37)=\)0.171.
Problem 14: Interpreting a P-value
If the true proportion of underweight muffins were 0.10, there would be about a 3.4% chance of getting a sample proportion of 0.150 or greater just by random sampling variability.
It is not the probability that \(H_0\) is true.
Problem 15: Drawing a conclusion
Since \(0.034 < 0.05\), reject \(H_0\). There is convincing evidence that more than 10% of the bakery’s muffins are underweight.
At \(\alpha=0.01\), \(0.034 > 0.01\), so we fail to reject \(H_0\): the conclusion changes, and the data would not be convincing.
AP Statistics Proportions Worksheet Answer Key: problems 16–20
Problem 16: Interval for a difference
\(\hat{p}_A=0.520\), \(\hat{p}_B=0.420\), difference \(0.100\).
\(SE=\sqrt{\frac{0.52(0.48)}{300}+\frac{0.42(0.58)}{250}}=\sqrt{0.000832+0.000974}=0.0425\).
\(ME=1.960(0.0425)=0.083\). The interval is \(0.100\pm0.083\): (0.017, 0.183).
Problem 17: Justifying a claim with an interval
No. The interval contains 0, so 0 is a plausible value for the difference. Negative values are also plausible, so the data do not give convincing evidence that group 1 has the higher proportion.
Problem 18: Two-proportion z test
\(H_0: p_A=p_B\), \(H_a: p_A>p_B\). \(\hat{p}_A=0.650\), \(\hat{p}_B=0.560\).
Pooled: \(\hat{p}_c=\frac{130+112}{400}=0.605\).
\(SE=\sqrt{0.605(0.395)(\frac{1}{200}+\frac{1}{200})}=0.0489\), so \(z=\frac{0.650-0.560}{0.0489}=\)1.84.
P-value \(=P(Z>1.84)=\)0.033. Since \(0.033 < 0.05\), there is convincing evidence that App A has the higher completion rate.
Problem 19: Describing a Type I error
A Type I error rejects a true \(H_0\). The inspector concludes that more than 8% of the batteries are defective when the true proportion is actually 8%. The company might stop production unnecessarily.
The probability of a Type I error equals the significance level: 0.05.
Problem 20: Describing a Type II error
A Type II error fails to reject a false \(H_0\). The inspector does not find convincing evidence that more than 8% of the batteries are defective, even though the true proportion is greater than 8%.
Consequence: defective batteries keep being shipped to customers.
AP Statistics Proportions Worksheet Answer Key: problems 21–24
Problem 21: Power and Type II error
\(\beta=1-\text{power}=1-0.72=\)0.28.
Power increases if the sample size is increased or if the significance level \(\alpha\) is increased (accepting a higher chance of a Type I error).
Problem 22: Computing power
Step 1: rejection cutoff. Under \(H_0\), \(SD=\sqrt{\frac{0.5(0.5)}{100}}=0.05\). Reject when \(\hat{p}>0.5+1.645(0.05)=0.5822\).
Step 2: power. If \(p=0.60\), \(SD=\sqrt{\frac{0.6(0.4)}{100}}=0.0490\), so \(P(\hat{p}>0.5822)=P(Z>\frac{0.5822-0.6}{0.0490})=P(Z>-0.36)=\)0.641.
About a 64% chance of detecting the change, so \(\beta\) is about 0.36.
Problem 23: Four-step significance test
State. \(H_0: p=0.70\), \(H_a: p < 0.70\), where \(p\) is the proportion of all seniors in the state who plan to attend college; \(\alpha=0.05\).
Plan. One-proportion z test. Conditions: random sample; the population is more than 10 times the sample; \(220(0.70)=154\) and \(220(0.30)=66\) are both at least 10.
Do. \(\hat{p}=\frac{143}{220}=0.650\), \(SE_0=\sqrt{\frac{0.70(0.30)}{220}}=0.0309\), \(z=\frac{0.650-0.700}{0.0309}=-1.62\), P-value \(=P(Z < -1.62)=0.053\).
Conclude. Since \(0.053 > 0.05\), fail to reject \(H_0\). There is not convincing evidence that fewer than 70% of the state’s seniors plan to attend college.
Problem 24: Interval and test agree
A two-sided test at \(\alpha=0.05\) rejects \(H_0\) exactly when the null value lies outside the 95% interval.
- 0.60 is outside \((0.43, 0.57)\): reject \(H_0\). There is convincing evidence that \(p\neq 0.60\).
- 0.50 is inside the interval: fail to reject \(H_0\). There is not convincing evidence that \(p\neq 0.50\).
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