Answer:
Yes at the level of 0.02 significance
Step-by-step explanation:
we want to compare if P₁ = P₂
P1 = 9/142= 0.0634
P2 = 5/268 = 0.0187
P = 14/410 = 0.03414
significance level, α = 0.02
Test statistic, z = [tex]\frac{p1 - p2}{\sqrt{(p * (1 - p} )) * (\frac{1}{142} * \frac{1}{268})}}[/tex]
Test Statistic, z = [tex]\frac{0.0634 - 0.0187}{\sqrt{({0.0341 * ({1 - 0.0341}})) * ({\frac{1}{142} + \frac{1}{268} ) }} } = 2.373[/tex]
Test statistic, z = 2.373
p-value = 2*p(z<|z₀|) = 2*p(z<2.37) = 0 .0176
Answer: Since p-value (0.0176) is less than the significance level, α (0.02), the null hypothesis can not hold. we can therefore say that at 0.02 level of significance, there is sufficient evidence, statistically, that p₁ is different from p₂
