If $n=15,x̄(x-bar)=38,ands=10$, find the margin of error at a 95% confidence level

Answer :

Given :

Sample mean $(x\overline{)}=38$

Sample standard deviation $(s)=10$

Sample size $(n)=15$

Confidence level $=95$

Solution :

The significance level :

$$\alpha =1–0.95=0.05$$

The degrees of freedom :

$$\text{df}=n–1=15–1=14$$

Critical value :

$$t_c=t_{\frac{\alpha }{2},\text{df}}$$ $$=t_{\frac{0.05}{2},14}$$ $$=t_{0.025,14}$$ $$=2.145$$

The margin of error :

$$\text{The margin of error, ME}=t_c\times \frac{s}{\sqrt{n}}$$ $$=2.145\times \frac{10}{\sqrt{15}}$$ $$=5.5384$$