1 2 3 4 5 6 7 8 9 10 49:3 Which of the following \( z \)-values, standard deviations, and sample sizes produce a margin of error of 0.95 ? \( M E=\frac{z \cdot s}{\sqrt{n}} \) \( z=2.14 ; s=4 ; n=9 \) \( z=2.14 ; s=4 ; n=81 \) \( z=2.14 ; s=16 ; n=9 \) \( z=2.14 ; s=16 ; n=81 \)

To find which of the given options produces a margin of error (ME) of 0.95, we can use the margin of error formula: \[ M E = \frac{z \cdot s}{\sqrt{n}} \] Plugging in the values for each option to see which one equals 0.95: 1. **Option 1**: \( z=2.14 ; s=4 ; n=9 \) \[ M E = \frac{2.14 \cdot 4}{\sqrt{9}} = \frac{8.56}{3} = 2.8533 \] 2. **Option 2**: \( z=2.14 ; s=4 ; n=81 \) \[ M E = \frac{2.14 \cdot 4}{\sqrt{81}} = \frac{8.56}{9} \approx 0.9511 \] 3. **Option 3**: \( z=2.14 ; s=16 ; n=9 \) \[ M E = \frac{2.14 \cdot 16}{\sqrt{9}} = \frac{34.24}{3} \approx 11.4133 \] 4. **Option 4**: \( z=2.14 ; s=16 ; n=81 \) \[ M E = \frac{2.14 \cdot 16}{\sqrt{81}} = \frac{34.24}{9} \approx 3.8156 \] The only combination that gives us a margin of error close to 0.95 is the second option: \( z=2.14 ; s=4 ; n=81 \) with \( M E \approx 0.9511 \).