which of the following z - values, standard deviations, and sample sizes produce a margin of error of…

which of the following z - values, standard deviations, and sample sizes produce a margin of error of 0.95?\n$me = \\frac{z\\cdot s}{\\sqrt{n}}$\n$z = 2.14, s = 4, n = 9$\n$z = 2.14, s = 4, n = 81$\n$z = 2.14, s = 16, n = 9$\n$z = 2.14, s = 16, n = 81$
Answer
Answer:
D. $z = 2.14, s = 16, n = 81$
Explanation:
Step1: Recall margin - of - error formula
$ME=\frac{z\cdot s}{\sqrt{n}}$
Step2: Check option A
For $z = 2.14, s = 4, n = 9$, $ME=\frac{2.14\times4}{\sqrt{9}}=\frac{8.56}{3}\approx2.85$
Step3: Check option B
For $z = 2.14, s = 4, n = 81$, $ME=\frac{2.14\times4}{\sqrt{81}}=\frac{8.56}{9}\approx0.951$
Step4: Check option C
For $z = 2.14, s = 16, n = 9$, $ME=\frac{2.14\times16}{\sqrt{9}}=\frac{34.24}{3}\approx11.41$
Step5: Check option D
For $z = 2.14, s = 16, n = 81$, $ME=\frac{2.14\times16}{\sqrt{81}}=\frac{34.24}{9}\approx3.80$
The closest value to $0.95$ is obtained in option B. So the answer is B. (There may be a small rounding - off difference, and if we consider the most accurate value among the given options to 0.95, it is B).