| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard two-part hypothesis testing question requiring an F-test for equal variances followed by a two-sample t-test. While it involves multiple steps (calculating sample variances from summary statistics, conducting two hypothesis tests with different significance levels), these are routine S4 procedures with no novel problem-solving required. The question is slightly above average difficulty due to the need to compute variances from Σx² and the two-test structure, but remains a textbook exercise. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| No. of butterflies | Sample mean \(\bar { x }\) | \(\sum x ^ { 2 }\) | |
| Females | 7 | 50.6 | 17956.5 |
| Males | 10 | 53.2 | 28335.1 |
| Answer | Marks |
|---|---|
| \(H_0 : \sigma_F^2 = \sigma_M^2\) | B1 |
| \(H_1 : \sigma_F^2 \neq \sigma_M^2\) | |
| \(s_F^2 = \frac{1}{6}(17956.5 - 7 \times 50.6^2) = \frac{33.98}{6} = 5.66333...\) | B1 |
| \(s_M^2 = \frac{1}{9}(28335.1 - 10 \times 53.2^2) = \frac{32.7}{9} = 3.63333...\) | B1 |
| \(\frac{s_F^2}{s_M^2} = 1.5587...\) (Reciprocal 0.6415) | M1 A1 |
| \(F_{6,9} = 3.37\) (or 0.24) | B1 |
| Not in critical region. Variances of the two distributions are the same | A1 |
Total: 7 marks
| Answer | Marks |
|---|---|
| \(H_0 : \mu_F = \mu_M\) | B1 |
| \(H_1 : \mu_F < \mu_M\) | |
| Pooled estimate \(s^2 = \frac{6 \times 5.66333... + 9 \times 3.63333}{15} = 4.44533\) | M1 |
| \(s = 2.11\) | |
| \(t = \frac{50.6 - 53.2}{2.11\sqrt{\frac{1}{7} + \frac{1}{10}}} = \pm 2.50\) | M1 A1 |
| C.V. \(t_{15}(5\%) = \pm 1.753\) | B1 |
| Significant. The mean length of the females forewing is less than the length of the males forewing | A1 |
Total: 6 marks
**Part a:**
$H_0 : \sigma_F^2 = \sigma_M^2$ | B1 |
$H_1 : \sigma_F^2 \neq \sigma_M^2$ | |
$s_F^2 = \frac{1}{6}(17956.5 - 7 \times 50.6^2) = \frac{33.98}{6} = 5.66333...$ | B1 |
$s_M^2 = \frac{1}{9}(28335.1 - 10 \times 53.2^2) = \frac{32.7}{9} = 3.63333...$ | B1 |
$\frac{s_F^2}{s_M^2} = 1.5587...$ (Reciprocal 0.6415) | M1 A1 |
$F_{6,9} = 3.37$ (or 0.24) | B1 |
Not in critical region. Variances of the two distributions are the same | A1 |
Total: 7 marks
**Part b:**
$H_0 : \mu_F = \mu_M$ | B1 |
$H_1 : \mu_F < \mu_M$ | |
Pooled estimate $s^2 = \frac{6 \times 5.66333... + 9 \times 3.63333}{15} = 4.44533$ | M1 |
$s = 2.11$ | |
$t = \frac{50.6 - 53.2}{2.11\sqrt{\frac{1}{7} + \frac{1}{10}}} = \pm 2.50$ | M1 A1 |
C.V. $t_{15}(5\%) = \pm 1.753$ | B1 |
Significant. The mean length of the females forewing is less than the length of the males forewing | A1 |
Notes:
- (a) need to have variance and the same o.e
- (b) need female and forewing(wing)
Total: 6 marks
---3. The lengths, $x \mathrm {~mm}$, of the forewings of a random sample of male and female adult butterflies are measured. The following statistics are obtained from the data.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& No. of butterflies & Sample mean $\bar { x }$ & $\sum x ^ { 2 }$ \\
\hline
Females & 7 & 50.6 & 17956.5 \\
\hline
Males & 10 & 53.2 & 28335.1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Assuming the lengths of the forewings are normally distributed test, at the $10 \%$ level of significance, whether or not the variances of the two distributions are the same. State your hypotheses clearly.
\item Stating your hypotheses clearly test, at the $5 \%$ level of significance, whether the mean length of the forewings of the female butterflies is less than the mean length of the forewings of the male butterflies.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2007 Q3 [13]}}