In a survey of 250 students, 19 said they want year-round school. Given a 95% level of confidence with a z-score of 1.96, what is the margin of error?(1 point)

Question

Responses

0.033%
0.033%

14.9%
14.9%

3.3%
3.3%

7.6%

Answer 1

To calculate the margin of error, we can use the formula:

\[ \text{Margin of Error} = z \times \sqrt{\frac{p(1-p)}{n}} \]

Where:

• \( z \) is the z-score (1.96 for 95% confidence),
• \( p \) is the sample proportion,
• \( n \) is the sample size.

1. Calculate the sample proportion \( p \):

   • \( p = \frac{x}{n} = \frac{19}{250} = 0.076 \)

2. Calculate \( 1 - p \):

   • \( 1 - p = 1 - 0.076 = 0.924 \)

3. Calculate the sample size \( n \):

   • \( n = 250 \)

4. Calculate the margin of error: \[ \text{Margin of Error} = 1.96 \times \sqrt{\frac{0.076 \times 0.924}{250}} \]

5. Calculating the inside of the square root:

   • \( 0.076 \times 0.924 = 0.070224 \)
   • \( \frac{0.070224}{250} = 0.000280896 \)

6. Taking the square root:

   • \( \sqrt{0.000280896} \approx 0.0168 \)

7. Final calculation: \[ \text{Margin of Error} = 1.96 \times 0.0168 \approx 0.0330 \]

8. Convert the margin of error into a percentage:

   • \( 0.0330 \times 100 = 3.3% \)

Therefore, the margin of error is 3.3%.

The correct response is:

3.3%