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Question Determine the sample size needed to estimate the proportion of a population with a genetic marker, with 99% confidence and 1.5% margin of error, given the expected proportion is 80%.
n=zα/22p∗(1−p∗)E2n = \frac{z^{2}_{\alpha/2} p^{*} (1 - p^{*})}{E^{2}}

Studdy Solution

STEP 1

1. We are estimating a population proportion, which can be modeled using a binomial distribution.
2. The desired confidence level is 99%99\%, which will determine the critical value (zz-score) used in the sample size calculation.
3. The margin of error (E) is 1.5%1.5\% or 0.0150.015 in decimal form.
4. The estimated population proportion (p∗p^*) is 80%80\% or 0.80.8 in decimal form.
5. The formula for calculating the minimum sample size for estimating a population proportion with a given confidence level and margin of error is: n=(z⋅p∗(1−p∗)E)2 n = \left(\frac{z \cdot \sqrt{p^*(1-p^*)}}{E}\right)^2
6. We will use the zz-score for a 99%99\% confidence level, which can be found in standard statistical tables or calculated using statistical software. For this problem, we will use a zz-score accurate to three decimal places.

STEP 2

1. Determine the critical value (zz-score) for a 99%99\% confidence level.
2. Calculate the estimated standard deviation of the sample proportion using p∗p^*.
3. Calculate the required sample size using the formula and the values obtained in the previous steps.

STEP 3

Determine the critical value (zz-score) for a 99%99\% confidence level. The zz-score corresponding to a 99%99\% confidence level is approximately 2.5762.576 (accurate to three decimal places).

STEP 4

Calculate the estimated standard deviation of the sample proportion, which is the square root of the product of p∗p^* and (1−p∗)(1 - p^*).
p∗(1−p∗)=0.8(1−0.8)=0.8⋅0.2=0.16 \sqrt{p^*(1-p^*)} = \sqrt{0.8(1-0.8)} = \sqrt{0.8 \cdot 0.2} = \sqrt{0.16}

STEP 5

Finish calculating the estimated standard deviation of the sample proportion.
0.16=0.4 \sqrt{0.16} = 0.4

STEP 6

Use the formula to calculate the required sample size, substituting in the values for zz, p∗p^*, and EE.
n=(z⋅p∗(1−p∗)E)2=(2.576⋅0.40.015)2 n = \left(\frac{z \cdot \sqrt{p^*(1-p^*)}}{E}\right)^2 = \left(\frac{2.576 \cdot 0.4}{0.015}\right)^2

STEP 7

Calculate the numerator of the fraction inside the square.
2.576⋅0.4=1.0304 2.576 \cdot 0.4 = 1.0304

STEP 8

Divide the result by the margin of error EE.
1.03040.015=68.6933… \frac{1.0304}{0.015} = 68.6933\ldots

STEP 9

Square the result to get the sample size.
n=(68.6933…)2=4722.7689… n = (68.6933\ldots)^2 = 4722.7689\ldots

STEP 10

Since we cannot have a fraction of a sample, round up to the nearest whole number to get the minimum required sample size.
n=4723 n = 4723
The required sample size is 4723.

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