Question

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) Responses 7.6% 7.6% 3.3% 3.3% 14.9% 14.9% 0.033%

Answer 1

To calculate the margin of error (ME) for a survey result, you can use the formula:

\[
ME = z \times \sqrt{\frac{p(1-p)}{n}}
\]

Where:
- \( z \) is the z-score corresponding to the confidence level (for a 95% confidence level, \( z = 1.96 \)),
- \( p \) is the sample proportion (the number of students who want year-round school divided by the total number of students surveyed),
- \( n \) is the total sample size.

In this case:
- The total number of students surveyed, \( n = 250 \).
- The number of students who want year-round school, \( x = 19 \).

First, we calculate the sample proportion \( p \):

\[
p = \frac{x}{n} = \frac{19}{250} = 0.076
\]

Next, we use the value of \( p \) to calculate the margin of error:

\[
ME = 1.96 \times \sqrt{\frac{0.076(1 - 0.076)}{250}}
\]

Calculating \( 1 - p \):

\[
1 - 0.076 = 0.924
\]

Now calculate the standard error (SE):

\[
SE = \sqrt{\frac{0.076 \times 0.924}{250}} = \sqrt{\frac{0.070176}{250}} = \sqrt{0.000280704} \approx 0.0167
\]

Now, substituting the SE back into the margin of error formula:

\[
ME = 1.96 \times 0.0167 \approx 0.0328
\]

To express it as a percentage:

\[
ME \times 100 \approx 3.28\%
\]

Since the options given were percentages, rounding may lead us to select **3.3%** as the margin of error for the study. Thus, the margin of error is:

**3.3%**