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

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

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

Answer

Explanation:

Step1: Recall margin - of - error formula

$E = z^*\frac{\sigma}{\sqrt{n}}$

Step2: Identify values

$z^* = 1.645$ (from 90% confidence level), $\sigma=2.7$, $n = 225$.

Step3: Calculate $\sqrt{n}$

$\sqrt{225}=15$

Step4: Calculate $\frac{\sigma}{\sqrt{n}}$

$\frac{2.7}{15}=0.18$

Step5: Calculate margin of error

$E=1.645\times0.18 = 0.2961\approx0.30$

Answer:

$0.30$