Question

a group of ninth-grade students is conducting a survey to estimate the average time spent on social media each day by their classmates. they randomly select 150 students from their school and record the time spent on social media. after collecting the data, they found that the average time spent is 120 minutes, with a standard deviation of 15 minutes.

using the formula for calculating the margin of error for a sample mean at a 90% confidence level, what is the margin of error for their estimate of the average time spent on social media by ninth-grade students?

\\\text{margin of error} = \text{critical value} \cdot \frac{\text{standard deviation}}{\sqrt{\text{sample size}}}\\

( ) 1.61
( ) 2.40
( ) 3.15
( ) 2.02

Explanation:

Identify the given parameters

We extract the parameters from the problem statement:

• Sample size \(n = 150\)
• Sample standard deviation \(s = 15\)
• Confidence level \(= 90\%\)

Determine the critical value

For a \(90\%\) confidence level, the critical value \(z^*\) (or \(t^*\) for large sample sizes, where \(z^* \approx t^*\)) is:

$$ z^* \approx 1.645 $$

Calculate the margin of error

Using the provided formula:

$$ \text{Margin of Error} = \text{Critical Value} \cdot \frac{\text{Standard Deviation}}{\sqrt{\text{Sample Size}}} $$

Substitute the values:

$$ \text{Margin of Error} = 1.645 \cdot \frac{15}{\sqrt{150}} $$

Compute the numerical value

Calculate the standard error and the final margin of error:

$$ \sqrt{150} \approx 12.2474 $$

$$ \text{Standard Error} = \frac{15}{12.2474} \approx 1.2247 $$

$$ \text{Margin of Error} = 1.645 \cdot 1.2247 \approx 2.0147 \approx 2.02 $$

Answer:

• 1.61
• 2.40
• 3.15
• 2.02 (Correct answer)