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.

Confidence interval: (294.3, 300.9). Since 295 lies inside this interval, fail to reject $H_0$.

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.