For a sample of 4 operating rooms taken in a hospital study, the mean noise level was 39 decibels and the standard deviation was 11.3. Find the 90% confidence interval of the true mean of the noise levels in the operating rooms. Assume the variable is normally distributed. Round your answers to at least two decimal places.
Answer:
Given Data:
Sample mean $(x\bar)=39$
Sample standard deviation $(s)=11.3$
Sample size $(n)=4$
Confidence interval level $=90%$
Solution:
The level of significance $(\alpha) = 1 – 0.9 = 0.1$
The degree of freedom $(\text{df}) = n – 1 = 4 – 1 = 3$
The critical value $(t_c):$
$t_c = t_{\frac{\alpha}{2}, \text{df}} = t_{0.1/2, 3} = 2.353$
The confidence interval $(CI):$
$$CI = \overline{x} \pm t_c \times \frac{s}{\sqrt{n}} $$ $$= 39 \pm 2.353 \times \frac{11.3}{\sqrt{4}} $$ $$= (25.71, 52.29)$$
