Consider the following hypothesis test:
\(H_0: \mu = 18\)
\(H_1: \mu \ne 18\),
If a sample of 48 provided a sample mean \(\bar{x} = 17\) and a sample standard deviation \(\sigma = 4.5\), then the value of the t-test statistic is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -1.54
The population standard deviation is not known, so the one-sample t statistic is used. The given 4.5 is the sample standard deviation s (printed as σ), worked out with divisor n − 1, so the standard error of the mean is \(\frac{s}{\sqrt{n}}\):
\( \displaystyle t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{17 - 18}{4.5/\sqrt{48}} \)
\(\sqrt{48} = 6.928\)
\(4.5 \div 6.928 = 0.6495\)
\( \displaystyle t = \frac{-1}{0.6495} = -1.54 \)
The same test can also be written with the standard deviation s worked out with divisor n; then the standard error is \(\frac{s}{\sqrt{n - 1}}\), and \(t = \frac{-1}{4.5/\sqrt{47}} = -1.52\). Option (2) is the nearest value either way.