A company has been producing steel tubes of mean inner diameter of 2 cm. A sample of 10 tubes gives an inner diameter of 2.01 cm and a variance of 0.0004 cm2. The value of test statistic is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.5
The sample is small (n = 10) and the population standard deviation is not known, so the t-test for a single mean is used.
\(\mu = 2,\quad \bar{x} = 2.01,\quad n = 10\)
\(s^2 = 0.0004,\quad s = 0.02\)
The given variance is that of the sample itself (divisor n), so the test statistic is
\( \displaystyle t = \frac{\bar{x} - \mu}{s/\sqrt{n-1}} \)
\( \displaystyle = \frac{2.01 - 2}{0.02/\sqrt{9}} \)
\( \displaystyle = \frac{0.01 \times 3}{0.02} = 1.5 \)
Tip: When the sample variance is given, the denominator is \(s/\sqrt{n-1}\), not \(s/\sqrt{n}\). Using \(\sqrt{10}\) here gives about 1.58, which is not among the options.