Setting Up a Two-Proportion Test · Configuración de una Prueba para Dos Proporciones
| English | Español |
|---|---|
| combined (pooled) proportion/kəmˈbaɪnd prəˈpɔːʃn/ | Proporción combinada (agrupada) |
Do two order systems have equal on-time rates?
- Independent random samples find 30 on-time orders among 100 with system 1 and 20 among 100 with system 2.
- State · Estado $H_0:p_1=p_2$ and a planned alternative, such as $H_a:p_1\ne p_2$. The parameters describe the two defined order populations.
Estimate the common rate under equality
- The combined (pooled) proportion 合并比例 estimates a common rate under the null: $\hat p_c=(x_1+x_2)/(n_1+n_2)$.
- Here $\hat p_c=(30+20)/(100+100)=0.25$. Pool successes and sample sizes, not just two percentages.
The null hypothesis for a two-proportion test is...
H0 says the two proportions are equal (no difference).
Under H0 the two proportions are assumed ___ (one word).
H0: p1 = p2 means equal proportions.
Use sample-size weighting when groups differ
- If the groups instead had 30 of 100 and 40 of 200, the pooled estimate would be $(30+40)/(100+200)=0.2333$.
- The simple average of 0.30 and 0.20 is 0.25, which is wrong for those unequal sample sizes.
Using the pooled-null expected-count check taught here, which counts are checked?
Check expected successes and failures in each group using the pooled proportion. Other accepted normality checks exist; this question specifies the pooled-null method.
Check the independent-sample model
- Check random design, independence between groups and each applicable 10% condition. A paired before-and-after design needs another method.
- Under the pooled null estimate in the 30/100 versus 20/100 example, the expected counts are 25, 75, 25 and 75, all at least ten.
With unequal group sizes, 30/100 and 40/200 pool to 70/300, approximately 0.2333. The simple mean 0.25 gives the groups the wrong weights.
Group 1: 30 of 100. Group 2: 20 of 100. Find the pooled proportion (x1+x2)/(n1+n2).
(30+20)/(100+100) = 50/200 = 0.25.
One group has 30/100 successes and another 40/200. Calculate the pooled proportion to four decimal places.
Add successes and sample sizes: (30 + 40)/(100 + 200) = 70/300 ≈ 0.2333. Averaging 0.30 and 0.20 equally would give the wrong weights.
Keep the null variance pooled
- Use $SE_0=\sqrt{\hat p_c(1-\hat p_c)(1/n_1+1/n_2)}$. Here $SE_0=\sqrt{0.25(0.75)(1/100+1/100)}\approx0.06124$.
- The unpooled interval and pooled equality test estimate variability differently. Pooling follows the test’s equality assumption, not a rule to pool every two-group calculation.
The pooled test standard error answers a null-model question. Do not automatically use it for a confidence interval.
The two-proportion TEST uses a pooled proportion, while the interval does not.
Under H0 the proportions are equal, so the test pools.
Match each procedure here to its standard error.
Estimation allows different underlying proportions; the test standard error is calculated under equality.
Prepare the test without changing the question
- The observed difference is 0.10 and the null difference is zero. The next step is $z=(\hat p_1-\hat p_2)/SE_0$.
- Retain the planned alternative and group order. Large counts do not correct a nonrandom or dependent design.
The observed difference is 0.10 and the null difference is zero. The next step is $z=(\hat p_1-\hat p_2)/SE_0$.
Which must be considered before using this independent-samples test?
Exact sample equality is not a prerequisite. The procedure tests population equality under checked design and approximation assumptions.