Answer:
The margin of error for this study is approximately 0.030 when rounded to the nearest thousandth.
Step-by-step explanation:
To calculate the margin of error for this study, we will use the formula for the margin of error for a proportion. The margin of error (ME) for a proportion is given by:
\(\[ME = Z \times \sqrt{\frac{p(1-p)}{n}}\]\)
where:
\(- \( Z \) is the Z-score corresponding to the desired confidence level.\\- \( p \) is the sample proportion.\\- \( n \) is the sample size.\\\)
In this case, we need to determine the proportion of teenagers with a TV in their bedroom and then calculate the margin of error at a given confidence level (commonly 95%).
First, let's find the sample proportion p:
\(\[p = \frac{642}{1000} = 0.642\]\)
For a 95% confidence level, the Z-score Z is approximately 1.96.
Now, we can plug the values into the margin of error formula:
\(\[ME = 1.96 \times \sqrt{\frac{0.642 \times (1 - 0.642)}{1000}}\]\[ME = 1.96 \times \sqrt{\frac{0.642 \times 0.358}{1000}}\]\[ME = 1.96 \times \sqrt{\frac{0.229836}{1000}}\]\[ME = 1.96 \times \sqrt{0.000229836}\]\[ME = 1.96 \times 0.01516\]\[ME \approx 0.0297\]\)
Thus, the margin of error for this study is approximately 0.030 when rounded to the nearest thousandth.
