Consider the following hypothesis:
$H_0: μ = 295$
$H_a: μ ≠295$
A sample of 50 provided a sample mean of 297.6. The population standard deviation is 12 and level of significance $α$ = 0.05. Check the hypothesis test given above using interval estimation.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Confidence interval: (294.3, 300.9). Since 295 lies inside this interval, fail to reject $H_0$.
Given $μ_0=295, n = 50, \bar x = 297.6, σ = 12$ and $α = 0.05$
So, $Z_{α/2} = Z_{0.025} = 1.96$
Confidence interval = $\bar x + Z_{α/2}\frac{σ}{\sqrt{n}}$
$= 297.6 ± 1.96 ×\frac{12}{\sqrt{50}}$
$= 297.6 ± 3.3$
So, confidence interval is $(297.6-3.3, 297.6+ 3.3)$ i.e. $(294.3, 300.9)$.
Since the hypothesized value of population mean $μ_0 = 295$ lies in the confidence interval $(294.3, 300.9)$, so $H_0$ cannot be rejected.