Question

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.

\

$$\begin{tabular}{|c|c|c|c|} \\hline confidence level (\\%) & 90 & 95 & 99 \\\\ \\hline z^{*}-score & 1.645 & 1.96 & 2.58 \\\\ \\hline \\end{tabular}$$

Explanation:

Identify the given parameters

Using the Margin of Error Calculation knowledge point

$$
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Calculate the standard error

Using the Margin of Error Calculation knowledge point

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

Calculate the margin of error

Using the Margin of Error Calculation knowledge point

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

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>.