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3 years ago

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Which of the following SRS designs will give the most precision (smallest standard error) for estimating a population mean? Assu

me that each population has the same value of the population variance S^2?
1) An SRS of size 400 from a population of size 4000
2) An SRS of size 30 from a population of size 300
3) An SRS of size 3000 from a population of size 300,000,000

1 answer:

Serjik [45]3 years ago

7 0

Answer:

Design 3: An SRS of size 3000 from a population of size 300,000,000

Step-by-step explanation:

To check the SRS designs will give the most precision (smallest standard error) for estimating a population mean, we'll make use of the following formula:

V(y) = S²/n( 1 - n/N)

Where S² is a constant for the three SRS designs

Check the first design

n = 400

N = 4000

So, V(y) = S²/400 (1 - 400/4000)

V(y) = S²/400(1 - 0.1)

V(y) = 0.0025S²(0.9)

V(y) = 0.00225S²

V(y) = 2.25S²E-3

The second design

n = 30

N = 300

So, V(y) = S²/30 (1 - 30/300)

V(y) = S²/30(1 - 0.1)

V(y) = S²/30(0.9)

V(y) = 0.03S²

V(y) = 3S²E-2

The third design

n = 3,000

N = 300,000,000

So, V(y) = S²/3,000 (1 - 3,000/300,000,000)

V(y) = S²/3,000(1 - 0.00001)

V(y) = S²/3,000(0.99999)

V(y) = 0.00033333

V(y) = 3.33S²E-4

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Step-by-step explanation:

Question a:

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.

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The margin of error is of:

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

A sample of n is required, and n is found for M = 0.009. So

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

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$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

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c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

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$n = 20464.9$

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