| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| 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-sample t-test with straightforward application of procedures. Students must test for equal variances (F-test) then perform a pooled t-test. While it requires multiple steps and careful hypothesis statements, these are routine S4 procedures with no novel problem-solving or conceptual challenges beyond textbook exercises. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The two populations have equal variances | B1 | Must state assumption clearly |
| \(H_0: \sigma_1^2 = \sigma_2^2\), \(H_1: \sigma_1^2 \neq \sigma_2^2\) | B1 | Both hypotheses required |
| For Group 1: \(\bar{x}_1 = \frac{30+29+35+27+23+33+33+35+28}{9} = \frac{273}{9} = 30.33...\) | M1 | Attempt to find mean and \(s^2\) for Group 1 |
| \(s_1^2 = \frac{1}{8}\left(\sum x^2 - \frac{(\sum x)^2}{9}\right)\) | M1 | Correct method for \(s_1^2\) |
| \(\sum x^2 = 900+841+1225+729+529+1089+1089+1225+784 = 8411\) | ||
| \(s_1^2 = \frac{1}{8}\left(8411 - \frac{273^2}{9}\right) = \frac{1}{8}(8411 - 8281) = \frac{130}{8} = 16.25\) | A1 | Correct value |
| \(F = \frac{16.25}{12.9} = 1.260\) (or \(\frac{12.9}{16.25} = 0.794\)) | M1 | Correct ratio, larger/smaller |
| Critical value \(F_{8,6}(10\%) = 3.58\) (or \(F_{6,8} = 3.58\)) | B1 | Correct critical value at 10% |
| Since \(1.260 < 3.58\), do not reject \(H_0\); assumption of equal variances is reasonable | A1 | Correct conclusion in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_1 = \mu_2\), \(H_1: \mu_1 \neq \mu_2\) | B1 | Both hypotheses, must use \(\mu\) notation |
| Pooled variance: \(s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}\) | M1 | Correct pooled variance formula |
| \(s_p^2 = \frac{8 \times 16.25 + 6 \times 12.9}{14} = \frac{130 + 77.4}{14} = \frac{207.4}{14} = 14.814...\) | A1 | Correct value |
| \(t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} = \frac{30.33 - 31.29}{\sqrt{14.814\left(\frac{1}{9}+\frac{1}{7}\right)}}\) | M1 | Correct structure of test statistic |
| \(= \frac{-0.956...}{\sqrt{14.814 \times 0.2540...}} = \frac{-0.956...}{\sqrt{3.763...}} = \frac{-0.956...}{1.9399...} = -0.493\) | A1 | Correct value (allow \(\pm\)) |
| Degrees of freedom \(= 14\), critical value \(t_{14}(5\%) = 2.145\) (two-tailed) | B1 | Correct cv with correct df |
| Since \( | {-0.493} | < 2.145\), do not reject \(H_0\); no significant evidence of a difference between mean marks of the two groups |
# Question 7:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| The two populations have equal variances | B1 | Must state assumption clearly |
| $H_0: \sigma_1^2 = \sigma_2^2$, $H_1: \sigma_1^2 \neq \sigma_2^2$ | B1 | Both hypotheses required |
| For Group 1: $\bar{x}_1 = \frac{30+29+35+27+23+33+33+35+28}{9} = \frac{273}{9} = 30.33...$ | M1 | Attempt to find mean and $s^2$ for Group 1 |
| $s_1^2 = \frac{1}{8}\left(\sum x^2 - \frac{(\sum x)^2}{9}\right)$ | M1 | Correct method for $s_1^2$ |
| $\sum x^2 = 900+841+1225+729+529+1089+1089+1225+784 = 8411$ | | |
| $s_1^2 = \frac{1}{8}\left(8411 - \frac{273^2}{9}\right) = \frac{1}{8}(8411 - 8281) = \frac{130}{8} = 16.25$ | A1 | Correct value |
| $F = \frac{16.25}{12.9} = 1.260$ (or $\frac{12.9}{16.25} = 0.794$) | M1 | Correct ratio, larger/smaller |
| Critical value $F_{8,6}(10\%) = 3.58$ (or $F_{6,8} = 3.58$) | B1 | Correct critical value at 10% |
| Since $1.260 < 3.58$, do not reject $H_0$; assumption of equal variances is reasonable | A1 | Correct conclusion in context |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_1 = \mu_2$, $H_1: \mu_1 \neq \mu_2$ | B1 | Both hypotheses, must use $\mu$ notation |
| Pooled variance: $s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$ | M1 | Correct pooled variance formula |
| $s_p^2 = \frac{8 \times 16.25 + 6 \times 12.9}{14} = \frac{130 + 77.4}{14} = \frac{207.4}{14} = 14.814...$ | A1 | Correct value |
| $t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} = \frac{30.33 - 31.29}{\sqrt{14.814\left(\frac{1}{9}+\frac{1}{7}\right)}}$ | M1 | Correct structure of test statistic |
| $= \frac{-0.956...}{\sqrt{14.814 \times 0.2540...}} = \frac{-0.956...}{\sqrt{3.763...}} = \frac{-0.956...}{1.9399...} = -0.493$ | A1 | Correct value (allow $\pm$) |
| Degrees of freedom $= 14$, critical value $t_{14}(5\%) = 2.145$ (two-tailed) | B1 | Correct cv with correct df |
| Since $|{-0.493}| < 2.145$, do not reject $H_0$; no significant evidence of a difference between mean marks of the two groups | A1 | Correct conclusion in context |7. Two groups of students take the same examination.
A random sample of students is taken from each of the groups.
The marks of the 9 students from Group 1 are as follows
$$\begin{array} { l l l l l l l l l }
30 & 29 & 35 & 27 & 23 & 33 & 33 & 35 & 28
\end{array}$$
The marks, $x$, of the 7 students from Group 2 gave the following statistics
$$\bar { x } = 31.29 \quad s ^ { 2 } = 12.9$$
A test is to be carried out to see whether or not there is a difference between the mean marks of the two groups of students.
You may assume that the samples are taken from normally distributed populations and that they are independent.
\begin{enumerate}[label=(\alph*)]
\item State one other assumption that must be made in order to apply this test and show that this assumption is reasonable by testing it at a $10 \%$ level of significance. State your hypotheses clearly.
\item Stating your hypotheses clearly, test, using a significance level of $5 \%$, whether or not there is a difference between the mean marks of the two groups of students.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2014 Q7 [14]}}