Question

A resident of Bayport claims to the City Council that the proportion of Westside residents

Answer 1

Answer:

The test statistics is  $z = -1.56$  

The p-value is   $p-value = 0.05938$

Step-by-step explanation:

From the question we are told  

   The West side sample  size is $n_1 = 578$

    The  number of residents on the West side with income below poverty level is $k = 76$

    The East side sample size  $n_2=688$

  The  number of residents on the East side with income below poverty level is $u = 112$

   The null hypothesis is  $H_o : p_1 = p_2$

    The alternative hypothesis is  $H_a : p_1 < p_2$

Generally the sample proportion of  West side is  

     $\^{p} _1 = \frac{k}{n_1}$

=>   $\^{p} _1 = \frac{76}{578}$

=>   $\^{p} _1 = 0.1315$

Generally the sample proportion of  West side is  

     $\^{p} _2 = \frac{u}{n_2}$

=>   $\^{p} _2 = \frac{112}{688}$

=>   $\^{p} _2 = 0.1628$

 Generally the pooled sample proportion is mathematically represented as

    $p = \frac{k + u}{ n_1 + n_2 }$

=>  $p = \frac{76 + 112}{ 578 + 688 }$

=>  $p =0.1485$

Generally the test statistics is mathematically represented as

$z = \frac{\^ {p}_1 - \^{p}_2}{\sqrt{p(1- p) [\frac{1}{n_1 } + \frac{1}{n_2} ]} }$

=> $z = \frac{ 0.1315 - 0.1628 }{\sqrt{0.1485(1-0.1485) [\frac{1}{578} + \frac{1}{688} ]} }$  

=> $z = -1.56$  

Generally the p-value  is mathematically represented as

          $p-value = P(z < -1.56 )$

From z-table  

         $P(z < -1.56 ) = 0.05938$

So

     $p-value = 0.05938$