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kozerog [31]
4 years ago
9

You are interested in estimating the the mean age of the citizens living in your community. In order to do this, you plan on con

structing a confidence interval; however, you are not sure how many citizens should be included in the sample. If you want your sample estimate to be within 5 years of the actual mean with a confidence level of 97%, how many citizens should be included in your sample
Mathematics
1 answer:
Assoli18 [71]4 years ago
4 0

Question:

You are interested in estimating the the mean age of the citizens living in your community. In order to do this, you plan on constructing a confidence interval; however, you are not sure how many citizens should be included in the sample. If you want your sample estimate to be within 5 years of the actual mean with a confidence level of 97% , how many citizens should be included in your sample? Assume that the standard deviation of the ages of all the citizens in this community is 18 years.

Answer:

61.03

Step-by-step explanation:

Given:

Standard deviation = 18

Sample estimate = 5

Confidence level = 97%

Required:

Find sample size, n.

First find the Z value. Using zscore table

Z-value at a confidence level of 97% = 2.17

To find the sample size, use the formula below:

n = (Z * \frac{\sigma}{E})^2

n = ( 2.17 * \frac{18}{5})^2

n = (2.17 * 3.6)^2

n = (7.812)^2

n = 61.03

Sample size = 61.03

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Line segment 19 units long running from (x,0) ti (0, y) show the area of the triangle enclosed by the segment is largest when x=
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The area of the triangle is

A = (xy)/2

Also,

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We solve this for y.

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Now we substitute this expression for y in the area equation.

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dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]

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Now we set the derivative equal to zero.

(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0

-2x^2 + 361 = 0

-2x^2 = -361

2x^2 = 361

x^2 = 361/2

x = 19/sqrt(2)

x^2 + y^2 = 361

(19/sqrt(2))^2 + y^2 = 361

361/2 + y^2 = 361

y^2 = 361/2

y = 19/sqrt(2)

We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
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