Hypothesis Testing: Sample mean; assumptions — VJC 2025 H2 Math Prelim Paper 2
What this question tests
Question
A hospital introduces a new medication regimen to help patients recover from a viral infection more quickly. Previously, the average recovery time was \(7.2\) days.
After adopting the new regimen, a random sample of \(40\) patients is observed, and their recovery times, \(x\), in days, are recorded. The results are summarised below. \[ \sum(x - 7.2) = -42 \qquad \text{and} \qquad \sum(x - 7.2)^2 = 648 \]
- (a) Explain whether the hospital should carry out a one-tailed test or a two-tailed test.
- (b) Calculate unbiased estimates of the population mean and variance of the recovery times under the new regimen.
- (c) Carry out an appropriate test at the \(5\%\) significance level. You should state clearly the hypotheses for your test and define any parameters that you use.
- (d) Explain, in the context of the question, the meaning of `at the \(5\%\) significance level'.
- (e) Explain whether it would have been sufficient for the hospital to take a random sample of \(10\) patients and record their recovery times under the new regimen in order to carry out the hypothesis test.
- (f) A test at the \(1\%\) significance level concludes that the average recovery time did not improve from \(7.2\) days. Find the set of values of \(k\).
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(a) The hospital should carry out a one-tailed test since they want to test whether the new regimen has reduced the mean recovery time (a directional claim).
(b) Let \(y = x - 7.2\). Unbiased estimate of the population mean: \[ \bar{x} = \bar{y} + 7.2 = \dfrac{-42}{40} + 7.2 = 6.15 \]
Unbiased estimate of the population variance: \[\begin{aligned} s_x^2 = s_y^2 &= \dfrac{1}{n - 1}\!\left[\sum(x - 7.2)^2 - \dfrac{\left(\sum(x - 7.2)\right)^2}{n}\right]\\ &= \dfrac{1}{39}\!\left[648 - \dfrac{(-42)^2}{40}\right]\\ &= 15.4846\\ &\approx 15.5 \end{aligned}\]
(c) Let \(X\) days be the recovery time of a patient and \(\mu\) days be the population mean of \(X\). \[\begin{aligned} \mathrm{H}_0&: \mu = 7.2\\ \mathrm{H}_1&: \mu < 7.2 \end{aligned}\] Level of significance: \(5\%\).
Test statistic: Since \(n = 40\) is large, by the Central Limit Theorem, \(\bar{X}\) is approximately normally distributed. Under \(\mathrm{H}_0\), \[ Z = \dfrac{\bar{X} - 7.2}{S/\sqrt{n}} \sim \mathrm{N}(0,\, 1)\ \text{approximately.} \]
Computation: With \(\bar{x} = 6.15\), \(s_x^2 = 15.4846\), \(n = 40\): \[ p\text{-value} = 0.0457 \quad (\text{or}\ z = -1.6876) \]
Conclusion: Since the \(p\)-value \(0.0457 < 0.05\), \(\mathrm{H}_0\) is rejected at the \(5\%\) significance level. There is sufficient evidence to conclude that the mean recovery time has reduced under the new regimen.
(d) A \(5\%\) significance level means that there is a probability of \(0.05\) that the test concludes that the mean recovery time has reduced under the new regimen when in fact it is still \(7.2\) days.
(e) No. Since the distribution of recovery time under the new regimen (\(X\)) is unknown, the sample size would have to be sufficiently large for the Central Limit Theorem to be applied so that the sample mean recovery time is approximately normally distributed. A sample of \(10\) is not large enough.
(f) \[\begin{aligned} \mathrm{H}_0&: \mu = 7.2\\ \mathrm{H}_1&: \mu < 7.2 \end{aligned}\] Level of significance: \(1\%\).
Test statistic: Since \(n = 30\) is large, by the Central Limit Theorem, \(\bar{X}\) is approximately normal. Under \(\mathrm{H}_0\), \(Z = \dfrac{\bar{X} - 7.2}{S/\sqrt{n}} \sim \mathrm{N}(0,\, 1)\) approximately.
Computation: Unbiased estimate of population variance from sample standard deviation: \[ s^2 = \dfrac{n}{n - 1}\!\left(1.7\right)^2 = \dfrac{30}{29}\!\left(2.89\right) = 2.9897 \]
Rejection region: \(z < -2.3263\).
Since \(\mathrm{H}_0\) is not rejected: \[ \dfrac{k - 7.2}{\sqrt{2.9897/30}} \ge -2.3263 \quad \Rightarrow \quad k \ge 6.4656 \]
Hence \(\{k \in \mathbb{R} : k \ge 6.47\}\).