Two Proportions: Estimate Separately, Test a Common Null — Solutions and Rubrics

p_A and p_B are the proportions of adults favoring the expansion in towns A and B. Estimate A minus B with a two-sample z-interval for a difference of population proportions. The samples are independent and randomly selected. 250≤2,000 and 200≤1,500 meet the without-replacement bound. Observed counts are A:150 successes,100 failures; B:90 successes,110 failures, all at least 10. Use observed p̂ values for this estimation procedure.

p̂_A=150/250=0.60; p̂_B=90/200=0.45; point estimate=0.15. SE=√[0.60(0.40)/250+0.45(0.55)/200]=√0.0021975≈0.0468775. MOE=1.96SE≈0.0918799. Interval≈(0.0581201,0.2418799). No equality is assumed for estimation, so each sample supplies its own variance estimate. Reversing B−A negates the estimate and changes the endpoints to (−0.2418799,−0.0581201).

We are 95% confident that the proportion favoring expansion in town A is about 0.0581 to 0.2419 greater than in town B, a difference of about 5.81 to 24.19 percentage points. In repeated independent samples of these same sizes from these same populations, approximately 95% of constructed intervals capture p_A−p_B. This one interval may or may not capture the fixed difference. Since zero is excluded and all values are positive, it provides convincing evidence of a population difference with A higher. A difference of 0.15 is 15 percentage points; it is not a 15% relative increase.

p_A,p_B are each town population’s support proportions. Use a two-sample z-test for a difference between population proportions. H₀:p_A=p_B or p_A−p_B=0. For “differ,” Hₐ:p_A≠p_B or p_A−p_B≠0. Lower A support would use p_A<p_B or p_A−p_B<0; higher A would use p_A>p_B or p_A−p_B>0. These are population, not observed sample, proportions.

p̂_c=(240(0.60)+240(0.40))/(240+240)=240/480=0.50. The null assumes a common population proportion, estimated by pooling successes. Both SRSs are independent; 240≤4,000 in each population. Each group’s expected counts under the common estimate are 240(0.50)=120 successes and 120 failures, all at least 10. An interval checks observed counts using each p̂, while this equality-null test checks all four counts using p̂_c.

p̂_A−p̂_B=0.20. Null SE=√[0.50(0.50)]√(1/240+1/240)≈0.0456435465. Using the full-precision SE, z=(0.20−0)/SE≈4.381780460. Under H₀ it is approximately standard normal. Two-sided p=2P(Z≥4.381780460)≈0.0000117713390976 using the provided tail output or normal-distribution table/technology. The two equal theoretical tails follow the not-equal alternative; the pooled SE follows the common-proportion null.

If the two town population support proportions were equal, the probability of obtaining a standardized sample difference at least this extreme in either direction is about 0.00001177. Since p≤0.05, reject H₀:p_A=p_B. There is convincing statistical evidence that the town population support proportions differ. The evidence is statistical, not proof that every adult or every town pattern differs; the sampled population parameters and response remain the subject of the claim.

Use a two-sample proportion z-interval for p_A−p_B and a two-sample proportion z-test for H₀:p_A=p_B against Hₐ:p_A>p_B. Random assignment supports the experimental randomization. The 10% sampling-without-replacement condition is unnecessary for this randomized experiment; no population sample size was asserted. Interval observed counts are 65/35 and 45/55, all≥10. Pooling gives p̂_c=110/200=0.55, so each group’s null expected counts are 55/45, also all≥10. The ordered difference concerns success under A minus success under B among comparable volunteers.

Observed difference=0.65−0.45=0.20. Interval SE=√[0.65(0.35)/100+0.45(0.55)/100]=0.0689202, MOE≈0.1350837, interval≈(0.0649163,0.3350837). Null SE=√[0.55(0.45)(1/100+1/100)]=0.0703562; z≈2.8426762 and upper p≈0.00223682. The interval estimates separate unknown proportions; the test models an assumed common proportion. Different assumptions produce different variance estimates.

We are 95% confident that success under A exceeds B by about 6.49 to 33.51 percentage points for comparable volunteers. Repeated experiments of these sizes using this interval method would capture the fixed difference approximately 95% of the time; this interval may or may not capture it. If success proportions under A/B were equal, a standardized difference at least this high has probability about 0.002237. Since p≤0.05, reject equality; there is convincing evidence A gives higher success under the experiment’s design. The positive interval also supports a difference. Random assignment supports a causal message comparison for these participants under the design; volunteering does not support blanket generalization to all city residents.

First: we are 95% confident X’s proportion is between 11 percentage points lower and 3 points higher than Y’s. Zero is plausible, so it supplies insufficient evidence of a difference. Second: we are 95% confident X’s proportion is 4 to 20 percentage points higher; excluding zero supplies evidence of a difference with X higher. Approximately 95% of intervals from repeated same-size/same-population sampling capture the fixed difference, but each computed interval either does or does not. Reversing order gives (−0.03,0.11) and (−0.20,−0.04); it changes signs/wording, not information or whether zero is included.

Under equal town population proportions, probabilities of standardized sample differences at least as extreme in either direction are 0.18 in the first study and 0.008 in the second. First: 0.18>0.05, fail to reject, with insufficient evidence of a population difference; equality is not proven. Second: 0.008≤0.05, reject, with convincing evidence of a difference. Conclusions concern the town population transport-choice proportions and use evidence language; they do not assert identical behavior or a proven cause.