Suppose a student measuring the boiling temperature of a certain liquid observes the reading (in degree Celsius) 102.5, 101.7, 103.1, 100.9, 100.5 and 102.2 on 6 different samples of the liquid. If he knows that the standard deviation for this procedure is 1.2°C, what is the interval estimation for the population mean at a 95% confidence level?

(100.86, 102.78)

The correct answer is Option (1) → $(100.86, 102.78)$

Given, $n = 6, σ = 1.2$, confidence level = 95% and observations are 102.5, 101.7, 103.1, 100.9, 100.5, 102.2

Sample mean $\bar x=\frac{102.5+ 101.7 + 103.1 + 100.9 + 100.5+102.2}{6}$

$=\frac{610.9}{6}= 101.816 = 101.82$

Confidence level = 95%

$⇒(1-α) = 0.95⇒ α = 0.05⇒\frac{α}{2}=0.025$

$∴Z_{α/2} = Z_{0.025} = 1.96$  (Using table)

Margin of error = $Z_{0.025}·\frac{1.2}{\sqrt{6}}= 1.96 × 0.49 = 0.96$

$∵μ=\bar x$ ± margin of error

$= 101.82 ± 0.96$

So, confidence interval is $(101.82-0.96, 101.82 +0.96)$ i.e. $(100.86, 102.78)$.