Please help me!!!!!!!!!!!

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar

Mamont248 [21]

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

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

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

=> $n = 125 \ stores$

4 0

3 years ago

Help me with the first one please

natulia [17]

D- 2x-1
plug in a number for x and you'll see

6 0

3 years ago

It is known that a certain function is an inverse proportion. Find the formula for this function if it is known that the functio

katrin2010 [14]

Answer:

$y=\frac{24}{x}$

Step-by-step explanation:

We have been given that a certain function is an inverse proportion. We are asked to find the formula for the function if it is known that the function is equal to 12 when the independent variable is equal to 2.

We know that two inversely proportional quantities are in form $y=\frac{k}{x}$ , where y is inversely proportional to x and k is constant of variation.