Question

Consider a study about the amount of screen time spent by children aged 11 - 15. Suppose you want to determine the average screen time, knowing that the standard deviation of screen time is 13 minutes. A sample of 80 children is randomly selected, and the average screen time is 53 minutes. Find the 95% confidence interval of the average screen time.

Answer 1

To calculate the 95% confidence interval for the average screen time, we will use the formula for the confidence interval for the population mean when the population standard deviation is known. The formula is given by:

\[
\bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right)
\]

Where:
- \(\bar{x}\) = sample mean
- \(z\) = z-score corresponding to the desired confidence level
- \(\sigma\) = population standard deviation
- \(n\) = sample size

Given:
- Sample mean (\(\bar{x}\)) = 53 minutes
- Population standard deviation (\(\sigma\)) = 13 minutes
- Sample size (\(n\)) = 80

**Step 1: Determine the z-score for the 95% confidence level.**
For a 95% confidence level, the z-score (z) is approximately 1.96.

**Step 2: Calculate the standard error (SE).**
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{13}{\sqrt{80}} \approx \frac{13}{8.9443} \approx 1.455
\]

**Step 3: Calculate the margin of error (ME).**
\[
ME = z \times SE = 1.96 \times 1.455 \approx 2.855
\]

**Step 4: Calculate the confidence interval.**
\[
\text{Confidence Interval} = \bar{x} \pm ME = 53 \pm 2.855
\]

Lower limit:
\[
53 - 2.855 \approx 50.145
\]

Upper limit:
\[
53 + 2.855 \approx 55.855
\]

**Final 95% Confidence Interval:**
\[
(50.15, 55.86)
\]

Thus, the 95% confidence interval for the average screen time is approximately \((50.15, 55.86)\) minutes.