(a) Determine the standard error (in dollars) of the mean.
(b) With a 0.95 probability, determine the margin of error (in dollars).
(c) What is the 95% confidence interval of the population mean (in dollars)?
Answer :
Given information :
Sample mean, $\bar{x} = 130$
Population standard deviation $(σ) = 35$
Sample size $(n) = 100$
Confidence level = 95%
Solution :
→ The standard error of the mean :
$$\therefore \sigma_x = \frac{\sigma}{\sqrt{n}}$$ $$= \frac{35}{\sqrt{100}}$$ $$= 3.5$$
For a 0.95 probability, $α = 1−0.95 = 0.05$
Critical value $$Z_c = Z_{\frac{\alpha}{2}} = Z_{0.025} = 1.96$$
→ The 95% confidence interval :
$$\text{The 95\% confidence interval} = \bar{x} \pm \text{ME}$$ $$= 130 \pm 6.86$$ $$= (123.14, 136.86)$$
