Setting Up a Two-Proportion Test

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 $H_0:p_1=p_2$ 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.

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.

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.

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.

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.

Write from a pooled proportion of 0.24

• Under a null of equal proportions, group A has 15 successes out of 50.
• Group B has 9 successes out of 50. The lesson pooled value 0.25 is another pair of samples.