Question

try it calculate a margin of error

a survey of 225 students showed the mean number of hours spent studying per week was 20.6 and the standard deviation was 2.7.

assuming a 90% confidence level, the margin of error is approximately dropdown

confidence level (%)	90	95	99

Explanation:

Identify the given values

Using the Z-Score Formula and Confidence Level Relationship knowledge points

$$
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Calculate the standard error

Using the Z-Score Formula knowledge point

$$ SE = \frac{s}{\sqrt{n}} = \frac{2.7}{\sqrt{225}} = \frac{2.7}{15} = 0.18 $$

Calculate the margin of error

Using the Z-Score Formula knowledge point

$$ ME = z^* \times SE = 1.645 \times 0.18 = 0.2961 \approx 0.30 $$

Answer:

A survey of 225 students showed the mean number of hours spent studying per week was 20.6 and the standard deviation was 2.7.

Assuming a 90% confidence level, the margin of error is approximately <blank>0.30</blank>