Question

43% of a sample of 100 visitors to a national park believes that additional funds should be used to preserve endangered species. What is the margin of sampling error for a 95% confidence interval?

A. 9.7%
B. 5.4%
C. 43%
D. 19.4%

Answer 1

Answer:

A. 9.7%

Step-by-step explanation:

margin of error:

z * $\sqrt{\frac{p(1-p)}{N} }$

n is sample size

z is the z score (for a 95% confidence interval you would use 1.96)

p is the sample proportion

to solve for 'p', remember that if a decimal value is given, use that. but if it is a whole number, then divide that number by the sample size, and substitute it into the formula.

in this case we are given 43% so 0.43.

in the formula :

1.96 * $\sqrt{\frac{0.43(1-0.43)}{100} }$

if you calculate this you get around 0.09703 this is 9.7%

Answer 2

Margin of error is 9.7% (A)

ρ = 43% which is equal to 0.430

$1 - P = 1 - 0.430 = 0.570$

The number of sample n = 100

We have to calculate Z at 95% confidence interval

∝ = $1 - 95%$%

∝ = $1 - 0.95$

∝ = $0.05$

∝/2 = 0.025

Z(∝/2) = Z(0.025) = 1.96

Standard error = $\sqrt{ ((1-P)xP / n)}$

Standard error = $\sqrt{(1-0.43)x0.43 / 100)$

Standard error = $\sqrt{(0.002451)}$

Standard error = 0.04950758

Standard error = 0.0495

Margin of error = Z*SE. z = 1.96 and standard error = 0.0495

Margin of error = $1.96x0.0496$

Margin of error = $0.097216$

Margin of error = $9.7216%$%

In conclusion, the margin of sampling error for a 95% confidence interval is 9.7%.

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