Question

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mart stores, 100 random items were scanned. The researchers found that 52 of the 60 stores had more than 2 items that were priced inaccurately (the national standards allow a maximum of 2 items out of 100 items priced accurately by the scanners). a) Construct a 95% confidence interval for proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned and give a practical interpretation of this interval. (3 Points) b) Wal-Mart claims that 99% of its stores comply with the national standard on accuracy of price scanners. Comment on the believability of Wal-Mart’s claim based on your results in part (a). (1 Point) c) Determine the number of Wal-Mart stores that must be sampled in order to estimate the true proportion to within 0.05 with 90% confidence. (3 Points)

Answer 1

Answer:

a

The 95% confidence interval is

   $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

    There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval  

b

  Generally  99% is outside the interval obtained in a  above then the claim of Wal-mart is not believable  

c

 $n = 125 \ stores$  

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  60  

    The number of stores that had more than 2 items price incorrectly is  k =  52  

   

Generally the sample proportion is mathematically represented as  

             $\^ p = \frac{ k }{ n }$

=>          $\^ p = \frac{ 52 }{ 60 }$

=>          $\^ p = 0.867$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

     $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=>   $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=>   $E = 0.0859$

Generally 95% confidence interval is mathematically represented as  

      $\^ p -E < p < \^ p +E$

=>    $0.867 - 0.0859 < p < 0.867 + 0.0859$

=>    $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

    There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval  

Considering question b

Generally  99% is outside the interval obtained in a  above then the claim of Wal-mart is not believable  

   

Considering question c

From the question we are told that

    The margin of error is  E = 0.05

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.645$

Generally the sample size is mathematically represented as  

    $n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n= [\frac{1.645 }}{0.05} ]^2 * 0.867 (1 - 0.867 )$

=>   $n = 125 \ stores$