Question

among 320 randomly selected airline travelers, the mean number of hours spent travelling 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.

Explanation:

Identify the given values

Using the Margin of Error Calculation knowledge point

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Calculate the standard error

Using the Margin of Error Calculation knowledge point

$$ SE = \frac{s}{\sqrt{n}} = \frac{2.9}{\sqrt{320}} \approx \frac{2.9}{17.8885} \approx 0.1621 $$

Calculate the margin of error

Using the Margin of Error Calculation knowledge point

$$ ME = z^* \times SE = 1.645 \times 0.1621 \approx 0.2667 $$

Round to the nearest tenth

Using the Margin of Error Calculation knowledge point

$$ ME \approx 0.3 $$

Answer:

Among 320 randomly selected airline travelers, the mean number of hours spent travelling 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. <blank>0.3</blank>