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 done 0.05. 0.1. 0.2. 0.3. confidence level (%) 90 95 99 z*-score 1.645 1.96 2.58

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 done 0.05. 0.1. 0.2. 0.3. confidence level (%) 90 95 99 z*-score 1.645 1.96 2.58

Answer

Answer:

0.3

Explanation:

Step1: Identify the formula

The formula for the margin of error $E$ for a large - sample (when $n\geq30$) confidence interval for the population mean when the population standard deviation $\sigma$ is unknown (we use the sample standard deviation $s$ instead) is $E = z^*\frac{s}{\sqrt{n}}$.

Step2: Identify the values

We are given that $n = 225$, $s=2.7$, and for a 90% confidence level, $z^* = 1.645$.

Step3: Calculate $\sqrt{n}$

$\sqrt{n}=\sqrt{225}=15$.

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

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

Step5: Calculate the margin of error

$E=z^*\frac{s}{\sqrt{n}}=1.645\times0.18\approx 0.3$.