Find the margin of error
-95% confidence interval; $\mathrm { n } = 91 ; \overline { \mathrm { x } } = 72 , \mathrm {~s} = 11.4$
A) $2.13$
B) $2.37$
C) $2.03$
D) $4.57$

Question

Quizplus · Accepted Answer

The margin of error (ME) for a 95% confidence interval when the population standard deviation is unknown and the sample size is less than 30 or the population is not normally distributed can be calculated using the formula: $ME = t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}$, where $t_{\frac{\alpha}{2}}$ is the t-score corresponding to the desired confidence level, $s$ is the sample standard deviation, and $n$ is the sample size. For a 95% confidence level and degrees of freedom $df = n - 1 = 90$, the t-score (from a t-distribution table or calculator) is approximately 1.986. Plugging in the given values: $ME = 1.986 \times \frac{11.4}{\sqrt{91}} \approx 2.37$.

Find the Margin of Error -95% Confidence Interval; n=91;x‾=72, s=11.4\mathrm { n } = 91 ; \overline { \mathrm { x } } = 72 , \mathrm {~s} = 11.4n=91;x=72, s=11.4

Find the Margin of Error
-95% Confidence Interval; $\mathrm { n } = 91 ; \overline { \mathrm { x } } = 72 , \mathrm {~s} = 11.4$