A scientist wishes to estimate the average depth of a river. He wants to be 99% confident that the estimate is accurate within 2 feet. From a previous study, the standard deviation of the depths measured was 4.33 feet. Find the Minimum Sample Size Needed for an Interval Estimate of the POPULATION MEAN

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08-07-2020
Mathematics

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A scientist wishes to estimate the average depth of a river. He wants to be 99% confident that the estimate is accurate within 2 feet. From a previous study, the standard deviation of the depths measured was 4.33 feet. Find the Minimum Sample Size Needed for an Interval Estimate of the POPULATION MEAN

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joaobezerra joaobezerra · Answer 1 · 2020-07-09T02:22:37+02:00

Answer:

The minimum sample size needed is 32

Step-by-step explanation:

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

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

Find the Minimum Sample Size Needed for an Interval Estimate of the POPULATION MEAN

The minimum sample size needed is n

n is found when M = 2.

We have that [tex]\sigma = 4.33[/tex]

So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]2 = 2.575*\frac{4.33}{\sqrt{n}}[/tex]

[tex]2\sqrt{n} = 2.575*4.33[/tex]

[tex]\sqrt{n} = \frac{2.575*4.33}{2}[/tex]

[tex](\sqrt{n})^{2} = (\frac{2.575*4.33}{2})^{2}[/tex]

[tex]n = 31.1[/tex]

Rounding up

The minimum sample size needed is 32

MrRoyal MrRoyal · Answer 2 · 2022-01-04T14:15:03+01:00

The Minimum Sample Size Needed for an Interval Estimate of the POPULATION MEAN is 32

The given parameters are:

[tex]\sigma = 4.33[/tex] ---the standard deviation
[tex]E = 2[/tex] --- the margin of error

At 99% confidence interval, the z value is 2.575.

So, we start by calculating the margin of error (E) using:

[tex]E = z \times \frac{\sigma}{\sqrt n}[/tex]

Substitute values for E, z and [tex]\sigma[/tex].

So, we have:

[tex]2 = 2.575 \times \frac{4.33}{\sqrt n}[/tex]

Multiply both sides by [tex]\sqrt n[/tex]

[tex]2 \sqrt n= 2.575 \times 4.33[/tex]

Divide both sides of the equation by 2

[tex]\sqrt n= \frac{2.575 \times 4.33}{2}[/tex]

Square both sides of the equation

[tex]n=( \frac{2.575 \times 4.33}{2})^2[/tex]

[tex]n=( 5.574875)^2[/tex]

Evaluate the exponent

[tex]n=31.0792312656[/tex]

The smallest integer greater than 31.0792312656 is 32.

So, we have:

[tex]n=32[/tex]

Hence, the minimum size is 32