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VashaNatasha [74]
3 years ago
13

. Suppose that we want to estimate the true proportion of defectives in a very large shipment of adobe bricks, and that we want

to be at least 95% confident that the error is at most 0.04. How large a sample will we need if:
Mathematics
1 answer:
wlad13 [49]3 years ago
8 0

Answer:

A sample of 601 bricks will be needed.

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 is:

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

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.

How large a sample will we need?

A sample of n bricks will be needed.

n is found when M = 0.04.

We have no estimate for the proportion of defective bricks, so we work with the worst case scenario, which is M = 0.04

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

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.04})^{2}

n = 600.25

Rounding up

A sample of 601 bricks will be needed.

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