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 era 10% 47%, 24%, 0.096%

1 answer

To calculate the margin of error for the proportion of students who would benefit from summer school, you can use the following formula:

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

where:

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

First, let's calculate the sample proportion \( p \):

\[ p = \frac{18}{75} = 0.24 \]

Now we can plug the values into the margin of error formula:

  1. Calculate \( 1 - p \):

\[ 1 - p = 1 - 0.24 = 0.76 \]

  1. Now substitute \( p \), \( 1 - p \), \( n \), and \( z \) into the formula:

\[ \text{ME} = 1.96 \times \sqrt{\frac{0.24 \times 0.76}{75}} \]

  1. Calculate \( 0.24 \times 0.76 \):

\[ 0.24 \times 0.76 = 0.1824 \]

  1. Then calculate \( \frac{0.1824}{75} \):

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

  1. Now calculate the square root:

\[ \sqrt{0.002432} \approx 0.04932 \]

  1. Finally, calculate the margin of error:

\[ \text{ME} = 1.96 \times 0.04932 \approx 0.0967 \]

Thus, the margin of error (rounded) is approximately \( 0.097 \) or \( 9.7% \).

Given your options: 10%, 47%, 24%, 0.096%, the closest choice is 10%.