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dybincka [34]
2 years ago
6

PLEASE HELP I HAVE 45 min to SUBMIT WILL GIVE BRAINLY

Mathematics
1 answer:
bija089 [108]2 years ago
8 0

Answer:

Step-by-step explanation:

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

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

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

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.886*0.114}}{0.04})^{2}

n = 242.5

Rounding up

We need at least 243 stores.

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