Problem Statement
A random sample of $64$ light bulbs produced by a manufacturer had a sample mean lifetime of $\bar{x} = 1480\text{ hours}$. The population standard deviation is known to be $\sigma = 120\text{ hours}$.
(a) Find a $95\%$ confidence interval for the population mean lifetime $\mu$. [3 marks]
(b) The manufacturer claims that the mean lifetime of all light bulbs is $1500\text{ hours}$. Comment on this claim in light of the confidence interval. [1 mark]
(c) Find the minimum sample size required to estimate $\mu$ with a margin of error of at most $15\text{ hours}$ at the $95\%$ confidence level. [2 marks]
Verified Solution & Marking Scheme
Part (a): 95% Confidence Interval
For a $95\%$ confidence level, $z^* = 1.960$.
Standard error $SE = \frac{\sigma}{\sqrt{n}} = \frac{120}{\sqrt{64}} = \frac{120}{8} = 15\text{ hours}$.
$\text{Margin of Error } E = 1.960 \times 15 = 29.4\text{ hours}$
$\text{CI} = [\bar{x} - E, \, \bar{x} + E] = [1480 - 29.4, \, 1480 + 29.4] = [1450.6, \, 1509.4]$
Part (b): Comment on Claim
Since $1500\text{ hours}$ lies within the $95\%$ confidence interval $[1450.6, 1509.4]$, there is no significant evidence at the $5\%$ level to reject the manufacturer's claim. The claim is plausible.
Part (c): Minimum Sample Size for E ≤ 15
$z^* \frac{\sigma}{\sqrt{n}} \le 15 \implies 1.960 \times \frac{120}{\sqrt{n}} \le 15$
$\frac{235.2}{\sqrt{n}} \le 15 \implies \sqrt{n} \ge \frac{235.2}{15} = 15.68$
$n \ge (15.68)^2 = 245.86$
Since $n$ must be an integer, minimum $n = 246$.