Question

The USDA conducted tests for salmonella in produce grown in California. In an independent sample of 252 cultures obtained from water used to irrigate the region, 18 tested positive for salmonella. In an independent sample of 476 cultures obtained from the region’s wildlife (e.g., birds), 20 tested positive for salmonella. Is this sufficient evidence for the USDA to conclude that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife? Assume the significance level of 0.01.

Answer 1

Answer:

 The decision rule is  

Fail to reject the null hypothesis

  The conclusion is  

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

Step-by-step explanation:

From the question we are told that

   The first  sample size is $n_1 = 252$

    The number that tested positive is  $k_1 = 18$

     The second sample size is  $n_2 = 476$

     The number that  tested positive is  $k_2 = 20$

     The level of significance is  $\alpha = 0.01$

Generally the first sample proportion is mathematically represented as

      $\^ p _1 = \frac{k_1 }{ n_1 }$

=>    $\^ p _1 = \frac{18 }{ 252 }$

=>    $\^ p _1 = 0.071$

Generally the second sample proportion is mathematically represented as

      $\^ p _2 = \frac{k_2 }{ n_2 }$

=>    $\^ p _2 = \frac{20 }{ 476}$

=>    $\^ p _2 = 0.042$

The  null hypothesis is            $H_o : p_1 - p_2 = 0$

The alternative hypothesis is  $H_a : p_1 - p_2 \ne 0$

Generally the test statistics is mathematically represented

       $z = \frac{ \^ p_1 - \^ p_2 - ( p_ 1 - p_2 )}{ \sqrt{\frac{\^ p_1 (1-\^ p_1)}{ n_1 } + \frac{\^ p_2 (1-\^ p_2)}{ n_2 } } }$

=>    $z = \frac{ 0.071 - 0.042 - 0 }{ \sqrt{\frac{0.071 (1-0.071)}{ 252 } + \frac{0.042 (1-\^ 0.042)}{ 476 } } }$

=>    $z = 1.56$

From the z table  the area under the normal curve to the right corresponding to  1.56   is  

        $P(Z > 1.56 ) =0.05938$

Generally the p-value is mathematically represented as

         $p-value = 2 * P(Z > 1.56 )$

=>      $p-value = 2 * 0.05938$

=>      $p-value = 0.1188$

From the value obtained we see that   $p-value > \alpha$ hence

  The decision rule is  

Fail to reject the null hypothesis

  The conclusion is  

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife