Find $\bar { p }$ and $\bar { q }$, if $X _ { 1 } = 11 , n _ { 1 } = 30 , X _ { 2 } = 28$, and $n _ { 2 } = 80$.
A) $\bar { p } = 0.35$ and $\bar { q } = 0.65$
B) $\bar { p } = 0.65$ and $\bar { q } = 0.35$
C) $\bar { p } = 0.64$ and $\bar { q } = 0.36$
D) $\bar { p } = 0.36$ and $\bar { q } = 0.64$

Question

Quizplus · Accepted Answer

$\bar { p }$ is the combined proportion of successes, calculated as the total number of successes divided by the total number of trials. Here, the total number of successes is $X_1 + X_2 = 11 + 28 = 39$, and the total number of trials is $n_1 + n_2 = 30 + 80 = 110$. Thus, $\bar { p } = \frac{39}{110} = 0.35$. $\bar { q }$ is the complement of $\bar { p }$, so $\bar { q } = 1 - \bar { p } = 0.65$.

Find pˉ\bar { p }pˉ​ And qˉ\bar { q }qˉ​ , If X1=11,n1=30,X2=28X _ { 1 } = 11 , n _ { 1 } = 30 , X _ { 2 } = 28X1​=11,n1​=30,X2​=28

Find $\bar { p }$ And $\bar { q }$ , If $X _ { 1 } = 11 , n _ { 1 } = 30 , X _ { 2 } = 28$