Answer:
Given:
Sample mean $(x\bar)=2.8$
Sample standard deviation $(s)=0.64$
Sample size $(n)=10$
Confidence interval level $=95%$
Solution:
→ The significance level $\alpha = 1 – 0.95 = 0.05$
→ The degree of freedom $\text{df} = n – 1 = 10 – 1 = 9$
→ The critical value, $t_c = t_{\frac{\alpha}{2}, \text{df}} = t_{0.05/2, 9} = 2.262$
→ The confidence interval:
$$CI = \overline{x} \pm t_c \times \frac{s}{\sqrt{n}} $$ $$= 2.8 \pm 2.262 \times \frac{0.64}{\sqrt{10}} $$ $$= (2.3, 3.3)$$
