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kvv77 [185]
3 years ago
10

A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she

desires to be 90% confident that her estimate is within 4 ounces of the true mean
Mathematics
1 answer:
Nataly_w [17]3 years ago
4 0

Answer:

The minimum sample size needed is n = (\frac{1.96\sqrt{\sigma}}{4})^2. If n is a decimal number, it is rounded up to the next integer. \sigma is the standard deviation of the population.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.9}{2} = 0.05

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.05 = 0.95, so Z = 1.645.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample must she select if she desires to be 90% confident that her estimate is within 4 ounces of the true mean?

A sample of n is needed, and n is found when M = 4. So

M = z\frac{\sigma}{\sqrt{n}}

4 = 1.96\frac{\sigma}{\sqrt{n}}

4\sqrt{n} = 1.96\sqrt{\sigma}

\sqrt{n} = \frac{1.96\sqrt{\sigma}}{4}

(\sqrt{n})^2 = (\frac{1.96\sqrt{\sigma}}{4})^2

n = (\frac{1.96\sqrt{\sigma}}{4})^2

The minimum sample size needed is n = (\frac{1.96\sqrt{\sigma}}{4})^2. If n is a decimal number, it is rounded up to the next integer. \sigma is the standard deviation of the population.

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