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 $S = 4.5$. Compute the value of the test statistic.

-1.54

The correct answer is Option (3) → -1.54

Given $μ_0 = 18, n = 48, \bar x = 17, S = 4.5$

$t =\frac{\bar x-μ_0}{S/\sqrt{n}}=\frac{17-18}{4.5/\sqrt{48}}$

$=\frac{-1×\sqrt{48}}{4.5}= -1.54$

$∴ t = -1.54$

and degrees of freedom $= 48-1 = 47$