Business, 12.03.2020 23:54 jraghubir50

In a sample of 7070 stores of a certain company, 6262 violated a scanner accuracy standard. It has been demonstrated that the conditions for a valid large-sample confidence interval for the true proportion of the stores that violate the standard were not met. Determine the number of stores that must be sampled in order to estimate the true proportion to within 0.050.05 with 9595% confidence using the large-sample method.

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Answer from: Hollywood0122

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.886(1-0.886)}{(\frac{0.05}{1.96})^2}=155.206$

And rounded up we have that n=156

Explanation:

Notation and definitions

X=62 number of companies that violated the scanner accuracy standard.

n=70 random sample taken

$\hat p=\frac{62}{70}=0.886$ estimated proportion of companies that violated the scanner accuracy standard

true population proportion of companies that violated the scanner accuracy standard

ME =0.05 represent the margin of error desired

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.886(1-0.886)}{(\frac{0.05}{1.96})^2}=155.206$

And rounded up we have that n=156

Answer from: Quest

answer; because many corporate watch groups will discover and publicize unethical behavior by suppliers, link it to the american company, and harm its reputation;

Answer from: Quest

Answer(c) no, because the ordinance is void on its face;

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In a sample of 7070 stores of a certain company, 6262 violated a scanner accuracy standard. It has b...

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