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There were 324 adults surveyed. Among the participants, the mean number of hours of sleep each night was 7.5, and the standard deviation was 1.6.

The margin of error, assuming a 95\% confidence level, is approximately [tex]$\qquad$[/tex]. Round to the nearest hundredth.

[tex]\[
\begin{array}{|c|c|c|c|}
\hline
\text{Confidence Level (\%)} & 90 & 95 & 99 \\
\hline
z^*\text{-score} & 1.645 & 1.96 & 2.5 \\
\hline
\end{array}
\][/tex]



Answer :

GinnyAnswer
To determine the margin of error for the mean number of hours of sleep with a 95% confidence level, we go through the following steps:

1. Identify the given information:
- Sample size ([tex]\(n\)[/tex]) = 324 adults
- Mean number of hours of sleep ([tex]\(\bar{x}\)[/tex]) = 7.5 hours
- Standard deviation ([tex]\(\sigma\)[/tex]) = 1.6 hours
- [tex]\(z^*\)[/tex]-score for a 95% confidence level = 1.96 (from the z-score table provided)

2. Calculate the standard error of the mean (SE):
The standard error is given by the formula:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]
Plugging in the values:
[tex]\[ SE = \frac{1.6}{\sqrt{324}} = \frac{1.6}{18} \approx 0.0889 \][/tex]

3. Calculate the margin of error (ME):
The margin of error is given by the formula:
[tex]\[ ME = z^* \cdot SE \][/tex]
Using the [tex]\(z^*\)[/tex]-score for a 95% confidence level:
[tex]\[ ME = 1.96 \cdot 0.0889 \approx 0.1742 \][/tex]

4. Round the margin of error to the nearest hundredth:
[tex]\[ ME \approx 0.17 \][/tex]

So, the margin of error for the mean number of hours of sleep, with a 95% confidence level, is approximately 0.17 hours.

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