In this exercise, test \[ H_{0}: p_{1}=p_{2} \text { vs } H_{a}: p_{1}>p_{2} \] with \( \hat{p}_{1}-\hat{p}_{2}=0.8-0.7=0.10 \) for different sample sizes. In parts (a) and (b), use StatKey or other technology to find the \( p \) - value. Click here to access StatKey. (a) \( \hat{p}_{1}=\frac{24}{30}=0.8 \) and \( \hat{p}_{2}=\frac{14}{20}=0.7 \) Round your answer to two decimal places.

We are testing   H₀: p₁ = p₂  versus  Hₐ: p₁ > p₂ with the observed proportions   p̂₁ = 24/30 = 0.80  and  p̂₂ = 14/20 = 0.70 so that the observed difference is   p̂₁ – p̂₂ = 0.80 – 0.70 = 0.10. To evaluate the p-value (using StatKey or similar technology), we can first compute the test statistic as if we were performing a standard z-test. (Keep in mind that StatKey does a randomization procedure which approximates this calculation, and in many cases the p-value will be very similar.) Step 1. Pool the sample proportions under the null:   p̂_pooled = (24 + 14) / (30 + 20) = 38/50 = 0.76. Step 2. Compute the standard error for the difference in proportions:   SE = √[p̂_pooled(1 − p̂_pooled) (1/n₁ + 1/n₂)]    = √[0.76 × 0.24 × (1/30 + 1/20)]    = √[0.1824 × (0.03333 + 0.05)]    = √[0.1824 × 0.08333]    ≈ √0.0152    ≈ 0.1233. Step 3. Compute the z-statistic:   z = (0.80 − 0.70) / 0.1233 ≈ 0.10 / 0.1233 ≈ 0.81. Step 4. Find the p-value for a one-sided test (Hₐ: p₁ > p₂). This is the probability of observing a z-value at least as large as 0.81. Using the standard normal distribution:   p-value = 1 − Φ(0.81) ≈ 1 − 0.7910 ≈ 0.209. Thus, the rounded p-value is about 0.21. Using StatKey, you would simulate the sampling distribution under H₀, and the p-value computed from that randomization test should come out close to 0.21 (to two decimal places). Answer for part (a): 0.21.