Find the margin of error.
-$n = 30 , \bar { x } = 86.5 , s = 10.3,90$ percent
A) $81.32

Question

Find the margin of error.
-$n = 30 , \bar { x } = 86.5 , s = 10.3,90$ percent
A) $81.32 < \mu < 91.68$
B) $83.31 < \mu < 89.69$
C) $83.32 < \mu < 89.68$
D) $82.65 < \mu < 90.35$

Quizplus · Accepted Answer

The margin of error (ME) for a confidence interval can be calculated using the formula $ME = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$, where $t_{\alpha/2}$ is the t-score corresponding to the desired confidence level (in this case, 90%), $s$ is the sample standard deviation, and $n$ is the sample size. For a 90% confidence level and a sample size of 30, the degrees of freedom (df) would be $n-1 = 29$. The t-score for a 90% confidence level and 29 df is approximately 1.699. Plugging in the values: $ME = 1.699 \times \frac{10.3}{\sqrt{30}} \approx 3.19$. The confidence interval is then calculated as $\bar{x} \pm ME = 86.5 \pm 3.19$, which gives $83.31 < \mu < 89.69$.

Find the Margin of Error n=30,xˉ=86.5,s=10.3,90n = 30 , \bar { x } = 86.5 , s = 10.3,90n=30,xˉ=86.5,s=10.3,90

Find the Margin of Error $n = 30 , \bar { x } = 86.5 , s = 10.3,90$