| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2002 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward two-part hypothesis testing question requiring a one-sample t-test and a chi-squared test for variance. Both are standard S4 procedures with clearly stated hypotheses and routine calculations from given summary statistics. The question requires no novel insight—just careful application of learned techniques. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 150.9\) [accept \(\geq 150.9\)], \(H_1: \mu < 150.9\) | B1 | both |
| \(s^2 = \frac{1}{29}\left(646904.1 - \frac{(4400.7)^2}{30}\right) = \frac{1365.727}{29} = 47.1\) | M1 | |
| Test statistic \(t = \frac{30}{s/\sqrt{30}} = -3.36\) | M1 A1 | |
| Critical value \(t_{29}(5\%) = (-)1.669\) | B1 | |
| Significant, evidence to confirm doctor's statement | A1 ft | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \sigma^2 = 36\), \(H_1: \sigma^2 \neq 36\) | B1 | both |
| Test statistic \(\frac{(n-1)s^2}{\sigma^2} = \frac{1365.727}{36} = 37.9\) | M1 A1 | |
| Critical values \(\chi_{29}^2(5\%)\) upper tail \(= 45.722\), \(\chi_{29}^2(5\%)\) lower tail \(= 16.047\) | B1 B1 | not significant |
| Insufficient evidence that variance of the heights of female Indians is different from that of females in the UK | A1 ft | (6) |
## Part (a)
$H_0: \mu = 150.9$ [accept $\geq 150.9$], $H_1: \mu < 150.9$ | B1 | both
$s^2 = \frac{1}{29}\left(646904.1 - \frac{(4400.7)^2}{30}\right) = \frac{1365.727}{29} = 47.1$ | M1 |
Test statistic $t = \frac{30}{s/\sqrt{30}} = -3.36$ | M1 A1 |
Critical value $t_{29}(5\%) = (-)1.669$ | B1 |
Significant, evidence to confirm doctor's statement | A1 ft | (6)
## Part (b)
$H_0: \sigma^2 = 36$, $H_1: \sigma^2 \neq 36$ | B1 | both
Test statistic $\frac{(n-1)s^2}{\sigma^2} = \frac{1365.727}{36} = 37.9$ | M1 A1 |
Critical values $\chi_{29}^2(5\%)$ upper tail $= 45.722$, $\chi_{29}^2(5\%)$ lower tail $= 16.047$ | B1 B1 | not significant
Insufficient evidence that variance of the heights of female Indians is different from that of females in the UK | A1 ft | (6)
**Total: (12 marks)**
---A recent census in the U.K. revealed that the heights of females in the U.K. have a mean of 160.9 cm. A doctor is studying the heights of female Indians in a remote region of South America. The doctor measured the height, $x$ cm, of each of a random sample of 30 female Indians and obtained the following statistics.
$$\Sigma x = 4400.7, \quad \Sigma x^2 = 646904.41.$$
The heights of female Indians may be assumed to follow a normal distribution.
The doctor presented the results of the study in a medical journal and wrote 'the female Indians in this region are more than 10 cm shorter than females in the U.K.'
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly and using a 5% level of significance, test the doctor's statement.
[6]
\end{enumerate}
The census also revealed that the standard deviation of the heights of U.K. females was 6.0 cm.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Stating your hypotheses clearly test, at the 5% level of significance, whether or not there is evidence that the variance of the heights of female Indians is different from that of females in the U.K.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2002 Q4 [12]}}