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)

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

[tex] n = 125 \ stores [/tex]

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

[tex]\^ p = \frac{ k }{ n }[/tex]

=> [tex]\^ p = \frac{ 52 }{ 60 }[/tex]

=> [tex]\^ p = 0.867[/tex]

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

[tex]\alpha = (100 - 95 ) \%[/tex]

=> [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is

[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]

Generally the margin of error is mathematically represented as

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

=> [tex]E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} } [/tex]

=> [tex]E = 0.0859 [/tex]

Generally 95% confidence interval is mathematically represented as

[tex]\^ p -E < p < \^ p +E[/tex]

=> [tex] 0.867 - 0.0859 < p < 0.867 + 0.0859 [/tex]

=> [tex] 0.7811 < p < 0.9529 [/tex]

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

[tex]\alpha = (100 - 95 ) \%[/tex]

=> [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is

[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]

Generally the sample size is mathematically represented as

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

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

=> [tex] n = 125 \ stores [/tex]