Answer :
Given :
Sample mean $(x\bar) = 38$
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 :
$$\text{The margin of error, ME} = t_c \times \frac{s}{\sqrt{n}}$$ $$= 2.145 \times \frac{10}{\sqrt{15}}$$ $$= 5.5384$$
