Find the minimum sample size that should be chosen to assure that the proportion estimate p will be within the required margin of error, .06. Use a 95% confidence interval and a population proportion of .7. The critical value for a 95% confidence level is 1.96

Try this formula:

n = [(z-value)^2 * p * q]/E^2
= [(1.96)^2 * .7 * .3]/.06^2
= ? (round to the next highest whole number)

I'll let you finish the calculation.

Note: n = sample size needed. Use .7 for p and .3 for q (q = 1 - p). E = maximum error, which is .06 and z-value is 1.96.

I hope this helps.

To determine the minimum sample size needed, we can use the formula:

n = (z^2 * p * (1-p)) / E^2

Where:
n = sample size
z = critical value for a given confidence level
p = estimated population proportion
E = margin of error

In this case, the critical value for a 95% confidence level is 1.96, the estimated population proportion is 0.7, and the margin of error is 0.06.

Plugging in these values into the formula:

n = (1.96^2 * 0.7 * (1-0.7)) / 0.06^2

n ≈ (3.8416 * 0.7 * 0.3) / 0.0036

n ≈ (0.80844) / 0.0036

n ≈ 224.567

Rounding up to the nearest whole number, the minimum sample size that should be chosen is 225.

To find the minimum sample size required to assure that the proportion estimate p will be within the required margin of error, we can use the following formula:

n = [(z^2) * p * (1-p)] / (E^2)

Where:
- n is the sample size
- z is the critical value for the desired confidence level (1.96 for a 95% confidence level)
- p is the estimated population proportion (0.7 in this case)
- E is the desired margin of error (0.06 in this case)

Substituting the given values into the formula, we get:

n = [(1.96^2) * 0.7 * (1-0.7)] / (0.06^2)

Simplifying the equation further:

n = [3.8416 * 0.7 * 0.3] / 0.0036

n ≈ 26.35

Rounding up to the nearest whole number, the minimum sample size required is 27.