In a survey of 75 students, 18 said they would benefit from summer school.

Given a 95% level of confidence with a z-score of 1.96, what is the margin of error?

(1 point)
Responses

24%
24%

0.096%
0.096%

47%
47%

10%
10%

1 answer

To calculate the margin of error for a proportion, 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 proportion of respondents who said they would benefit from summer school,
  • \( n \) is the total number of respondents.

First, we need to calculate \( p \): \[ p = \frac{\text{number of favorable responses}}{\text{total number of responses}} = \frac{18}{75} = 0.24 \]

Now, we can substitute \( p \) and \( n \) into the margin of error formula:

\[ \text{Margin of Error} = 1.96 \times \sqrt{\frac{0.24(1-0.24)}{75}} \]

Calculating the inside of the square root:

\[ 1-0.24 = 0.76 \]

\[ 0.24 \times 0.76 = 0.1824 \]

Now, divide by \( n = 75 \):

\[ \frac{0.1824}{75} = 0.002432 \]

Next, take the square root:

\[ \sqrt{0.002432} \approx 0.0493 \]

Now we can calculate the margin of error:

\[ \text{Margin of Error} = 1.96 \times 0.0493 \approx 0.0967 \]

Converting this to a percentage:

\[ 0.0967 \approx 9.67% \]

Rounded to the nearest whole number, the margin of error is approximately 10%.

So, the correct response is:

10%