Answered

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.