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The mean number of hours spent traveling per year is 24 hours, and the standard deviation is 2.9. What is the margin of error, assuming a 90% confidence level? Round your answer to the nearest tenth.

[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline
\begin{tabular}{c}
Confidence \\
Level (\%)
\end{tabular} & 90 & 95 & 99 \\
\hline
$z^{\text {*}}$-score & 1.645 & 1.96 & 2.58 \\
\hline
\end{tabular}
\][/tex]


Sagot :

Absolutely, let's go through the steps needed to calculate the margin of error for a 90% confidence level given the provided data:

1. Identify the mean and standard deviation:
- Mean ([tex]\(\mu\)[/tex]): 24 hours
- Standard Deviation ([tex]\(\sigma\)[/tex]): 2.9 hours

2. Determine the z-score for the given confidence level:
For a 90% confidence level, the z-score ([tex]\(z^*\)[/tex]) is 1.645 (as given in the table).

3. Calculate the margin of error:
The formula for the margin of error (ME) is:
[tex]\[ \text{ME} = z^* \cdot \sigma \][/tex]

4. Substitute the given values:
- [tex]\(z^*\)[/tex]: 1.645
- [tex]\(\sigma\)[/tex]: 2.9

[tex]\[ \text{ME} = 1.645 \times 2.9 \][/tex]

5. Perform the multiplication:
[tex]\[ \text{ME} = 4.7705 \][/tex]

6. Round the margin of error to the nearest tenth:
[tex]\[ \text{ME} \approx 4.8 \][/tex]

Therefore, the margin of error, when rounded to the nearest tenth, is [tex]\(4.8\)[/tex] hours.