Consider the following hypothesis test:
$Н_0: μ = 18$
$H_a: μ≠18$
A sample of 48 provided a sample mean $\bar x = 17$ and a sample standard deviation $S = 4.5$. What is the rejection rule using the critical value? What is your conclusion? $(α = 0.05)$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Reject $H_0$ if $t>-2.012$. Since $t=−1.54$, fail to reject $H_0$.
Given $μ_0 = 18, n = 48, \bar x = 17, S = 4.5, α = 0.05$
$t =\frac{\bar x-μ_0}{S/\sqrt{n}}=\frac{17-18}{4.5/\sqrt{48}}$
$=\frac{-1×\sqrt{48}}{4.5}= -1.54$
$∴ t = -1.54$
and degrees of freedom $= 48-1 = 47$
Reject $H_0$ if $t≤-t_{α/2}$ or $t≥t_{α/2}$.
Here, $t = -1.54$ and $t_{α/2} = 0.025$
From the table, $t_{0.025} = 2.012$ with $df = 47$
$∵ -1.54 > -2.012$
So, do not reject $H_0$.