| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | Chi-squared test single variance hypothesis |
| Difficulty | Standard +0.3 This is a straightforward application of chi-squared methods for variance: part (a) requires basic calculation of mean and variance from data, part (b) is a standard hypothesis test for variance using chi-squared, and parts (c)-(d) involve constructing a confidence interval with critical values provided. All techniques are routine S4 procedures with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks |
|---|---|
| \(\bar{x} = \frac{123+118+127+120+134+120+118+135+121}{9} = \frac{1116}{9} = 124\) | B1 |
| \(s^2 = \frac{1}{8}\left[\sum x^2 - \frac{(\sum x)^2}{9}\right] = \frac{1}{8}\left[139124 - \frac{1116^2}{9}\right] = \frac{1}{8}(139124-138384) = \frac{740}{8} = 42.5\) (awrt) | B1 |
| Answer | Marks |
|---|---|
| Test statistic: \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{8 \times 42.5}{25} = 13.6\) | M1 A1 |
| Critical value: \(\chi^2_8\) at 5% one-tail = \(15.507\) | B1 |
| \(13.6 < 15.507\), do not reject \(H_0\). Insufficient evidence that variance exceeds 25. The company's claim is supported. | A1 |
| Answer | Marks |
|---|---|
| \(\left(\frac{11 \times 8.17}{21.920},\ \frac{11 \times 8.17}{3.816}\right)\) | M1 |
| \(= (4.10, 23.5)\) (awrt) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Since 25 lies outside the 95% confidence interval, this supports Gurdip's claim that \(\sigma^2 < 25\) | B1 | Must reference CI and conclusion |
# Question 5:
## Part (a):
| $\bar{x} = \frac{123+118+127+120+134+120+118+135+121}{9} = \frac{1116}{9} = 124$ | B1 | |
|---|---|---|
| $s^2 = \frac{1}{8}\left[\sum x^2 - \frac{(\sum x)^2}{9}\right] = \frac{1}{8}\left[139124 - \frac{1116^2}{9}\right] = \frac{1}{8}(139124-138384) = \frac{740}{8} = 42.5$ (awrt) | B1 | |
## Part (b):
| Test statistic: $\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{8 \times 42.5}{25} = 13.6$ | M1 A1 | |
|---|---|---|
| Critical value: $\chi^2_8$ at 5% one-tail = $15.507$ | B1 | |
| $13.6 < 15.507$, do not reject $H_0$. Insufficient evidence that variance exceeds 25. The company's claim is supported. | A1 | |
## Part (c):
| $\left(\frac{11 \times 8.17}{21.920},\ \frac{11 \times 8.17}{3.816}\right)$ | M1 | |
|---|---|---|
| $= (4.10, 23.5)$ (awrt) | A1 | |
## Part (d):
| Since 25 lies outside the 95% confidence interval, this supports Gurdip's claim that $\sigma^2 < 25$ | B1 | Must reference CI and conclusion |
|---|---|---|\begin{enumerate}
\item A researcher is investigating the accuracy of IQ tests. One company offers IQ tests that it claims will give any individual's IQ with a standard deviation of 5
\end{enumerate}
The researcher takes these tests 9 times with the following results
$$123 , \quad 118 , \quad 127 , \quad 120 , \quad 134 , \quad 120 , \quad 118 , \quad 135 , \quad 121$$
(a) Find the sample mean, $\bar { x }$, and the sample variance, $s ^ { 2 }$, of these scores.\\
(2)
Given that any individual's IQ scores on these tests are independent and have a normal distribution,\\
(b) use the hypotheses
$$\mathrm { H } _ { 0 } : \sigma ^ { 2 } = 25 \text { against } \mathrm { H } _ { 1 } : \sigma ^ { 2 } > 25$$
to test the company's claim at the $5 \%$ significance level.\\
(4)
Gurdip works for the company and has taken these IQ tests 12 times. Gurdip claims that the sample variance of these 12 scores is $s ^ { 2 } = 8.17$\\
(c) Use this value of $s ^ { 2 }$ to calculate a $95 \%$ confidence interval for the variance of Gurdip's IQ test scores.\\[0pt]
[You may use $\mathrm { P } \left( \chi _ { 11 } ^ { 2 } > 3.816 \right) = 0.975$ and $\mathrm { P } \left( \chi _ { 11 } ^ { 2 } > 21.920 \right) = 0.025$ ]\\
(d) Assuming that $\sigma ^ { 2 } = 25$, comment on Gurdip's claim.
\hfill \mbox{\textit{Edexcel S4 2015 Q5 [9]}}