Consider the following hypothesis test:
$H_0:μ≤12$
$H_a : μ > 12$
A sample of 25 provided a sample mean $\bar x = 14$ and a sample standard deviation $S = 4.32$. What is the rejection rule using the critical value? What is your conclusion? ($α = 0.05$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Reject $H_0$ if $t>1.711$; conclude there is sufficient evidence that $μ>12$.
Given $n = 25, \bar x = 14, S = 4.32, μ_0= 12, α = 0.05$
$t =\frac{\bar x-μ_0}{S/\sqrt{n}}=\frac{14-12}{4.32/\sqrt{25}}$
$=\frac{10}{4.32}= 2.31$
$∴ t = 2.31$
and degrees of freedom $= 25-1 = 24$
Reject $H_0$, if $t≥ t_α$
$t_α = t_{0.05}$
From the table, $t_{0.05} = 1.711$ with $df = 24$
$∵2.31 > 1.711$
∴ Reject $H_0$.