Question

O 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 14.9% 7.6% 0.033% or 3.3%

Answer 1

To calculate the margin of error (ME) for a proportion at a specified confidence level, you can use the formula:

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

Where:
- \( z \) is the z-score corresponding to the desired confidence level.
- \( p \) is the sample proportion (the number of students in favor divided by the total number of students surveyed).
- \( n \) is the sample size.

In this case:
- The total number of students surveyed (\( n \)) = 250
- The number of students wanting year-round school = 19
- Thus, the sample proportion (\( p \)) = \( \frac{19}{250} = 0.076 \)

Now, let's calculate the margin of error using the given z-score of 1.96.

1. Calculate \( p(1-p) \):
\[
p(1-p) = 0.076(1 - 0.076) = 0.076 \times 0.924 = 0.070224
\]

2. Now calculate the standard error (SE):
\[
SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.070224}{250}} = \sqrt{0.000280896} \approx 0.01677
\]

3. Now calculate the margin of error (ME):
\[
ME = z \times SE = 1.96 \times 0.01677 \approx 0.0328
\]

4. To express this as a percentage, we multiply by 100:
\[
ME \approx 0.0328 \times 100 \approx 3.28\%
\]

So, the margin of error, rounded to one decimal point, is approximately **3.3%**.