$a.\\g=\frac{w}{m}\ \ \ \ |multiply\ both\ sides\ by\ m\neq0\\\\w=gm\ \boxed{c.}\\\\therefore:\boxed{a.}\ is\ equivalent\ to\ \boxed{c.}\\\\b.\\g=\frac{m}{w}\ \ \ \ |multiply\ both\ sides\ by\ w\neq0\\\\gw=m\ \ \ \ |divide\ both\ sides\ by\ g\neq0\\\\w=\frac{m}{g}\ \boxed{d.}\\\\therefore:\boxed{b.}\is\ equivalent\ to\ \boxed{d.}$

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3 years ago

In a sample of 70 stores of a certain company, 62 violated a scanner accuracy standard. It has been demonstrated that the condi

uysha [10]

Answer:

We need at least 243 stores.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error of the interval is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$n = 70, p = \frac{62}{70} = 0.886$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Determine the number of stores that must be sampled in order to estimate the true proportion to within 0.04 with 95% confidence using the large-sample method.

We need at least n stores.

n is found when M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.886*0.114}}{n}}$