A seed company claims that its plants flower in 42 days on average. A random sample of 9 plants has a mean flowering time of 46 days and a sample standard deviation of 6 days (calculated with divisor n − 1). The value of the t-statistic for testing the claim is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
The one-sample t-statistic is
\( \displaystyle t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)
Here \(\bar{x} = 46\), \(\mu = 42\), s = 6 and n = 9, so \(\sqrt{n} = 3\):
\( \displaystyle t = \frac{46 - 42}{6/3} \)
\( \displaystyle = \frac{4}{2} = 2 \)
Why the other options are incorrect
Option (1) 6 — Incorrect. This is 4 ÷ (6/9), dividing s by n instead of by \(\sqrt{n}\).
Option (2) \(\displaystyle\frac{2}{3}\) — Incorrect. This is 4 ÷ 6, dividing by s instead of by the standard error \(s/\sqrt{n} = 2\).
Option (3) 12 — Incorrect. This is \(4 \times \sqrt{9}\), which leaves out the standard deviation s.