OCR MEI S2 2011 June — Question 4 18 marks

Exam BoardOCR MEI
ModuleS2 (Statistics 2)
Year2011
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared test of independence
TypeExpected frequencies partially provided
DifficultyStandard +0.3 This is a standard chi-squared test of independence with all calculations scaffolded (contributions given, test statistic computed). Students only need to state hypotheses, verify one expected frequency calculation, and compare to critical value. Slightly above average difficulty due to being a hypothesis test rather than pure calculation, but the extensive scaffolding makes it routine for S2 level.
Spec5.06a Chi-squared: contingency tables

4
  1. In a survey on internet usage, a random sample of 200 people is selected. The people are asked how much they have spent on internet shopping during the last three months. The results, classified by amount spent and sex, are shown in the table.
    \multirow{2}{*}{}Sex\multirow{2}{*}{Row totals}
    MaleFemale
    \multirow{5}{*}{Amount spent}Nothing283462
    Less than £50172138
    £50 up to £200222648
    £200 up to £1000231639
    £1000 or more8513
    Column totals98102200
    1. Write down null and alternative hypotheses for a test to examine whether there is any association between amount spent and sex of person. The contributions to the test statistic for the usual \(\chi ^ { 2 }\) test are shown in the table below.
      \multirow{2}{*}{}Sex
      MaleFemale
      \multirow{5}{*}{Amount spent}Nothing0.18650.1791
      Less than £500.14090.1354
      £50 up to £2000.09820.0944
      £200 up to £10000.79180.7608
      £1000 or more0.41710.4007
      The sum of these contributions, correct to 3 decimal places, is 3.205.
    2. Calculate the expected frequency for females spending nothing. Verify the corresponding contribution, 0.1791 , to the test statistic.
    3. Carry out the test at the \(5 \%\) level of significance, stating your conclusion clearly.
  2. A bakery sells loaves specified as having a mean weight of 400 grams. It is known that the weights of these loaves are Normally distributed and that the standard deviation is 5.7 grams. An inspector suspects that the true mean weight may be less than 400 grams. In order to test this, the inspector takes a random sample of 6 loaves. Carry out a suitable test at the \(5 \%\) level, given that the weights, in grams, of the 6 loaves are as follows. \(\begin{array} { l l l l l l } 392.1 & 405.8 & 401.3 & 387.4 & 391.8 & 400.6 \end{array}\) RECOGNISING ACHIEVEMENT

Question 4:
Part 4(a)(i)
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0\): no association between amount spent and sex \(H_1\): some association between amount spent and sexB1 (both) Hypotheses must be right way round, in context, and must not mention 'correlation'
Total: 1
Part 4(a)(ii)
AnswerMarks Guidance
AnswerMarks Guidance
Expected frequency \(= 62 \times 102 \div 200 = 31.62\)B1 Do not allow 31.6
Contribution \(= (34 - 31.62)^2 / 31.62 = 0.1791\)M1 A1 For valid attempt at \(\frac{(O-E)^2}{E}\); NB answer given
Total: 3
Question 4(a)(iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Refer to \(X_4^2\); Critical value at 5% level = 9.488B1 For 4 degrees of freedom
Critical value = 9.488B1 CAO For cv
\(3.205 < 9.488\)M1
Result is not significantA1 For not significant; allow \(p = 0.524\), \(0.524 > 0.05\); conclusion must be stated to earn A1; allow 'do not reject \(H_0\)' and condone 'accept \(H_0\)' or 'reject \(H_1\)'; FT if cv consistent with their d.o.f.
There is insufficient evidence to suggest any association between amount spent and sex.E1 Dependent on previous A1 and final comment must be in context and not mention correlation; SC1 for correct final conclusion where previous A1 omitted but M1 awarded
5
Question 4(b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0: \mu = 400\)B1 Hypotheses in words must refer to population mean
\(H_1: \mu < 400\)B1
Where \(\mu\) denotes the population mean (weight of the loaves)B1 For definition of \(\mu\)
\(\bar{x} = 396.5\)B1 For sample mean
Test statistic \(= \dfrac{396.5 - 400}{5.7/\sqrt{6}} = \dfrac{-3.5}{2.327} = -1.504\)M1, A1FT M1 must include \(\sqrt{6}\); use their sample mean; condone numerator reversed for M1 but award A1 only if test statistic of 1.504 is compared with a positive \(z\)-value
5% level 1-tailed critical value of \(z = -1.645\)B1 For \(\pm 1.645\)
\(-1.504 > -1.645\) so not significantM1 For sensible comparison leading to a conclusion; dependent on previous M1
There is insufficient evidence to reject \(H_0\)
There is insufficient evidence to suggest that the true mean weight of the loaves is lower than the minimum specified value of 400 grams.A1 For conclusion in context; FT their sample mean only if hypotheses are correct
9
Additional Notes Q4(b):
AnswerMarks Guidance
MethodWorking Marks
\(\sigma\) estimatedSample mean 7.079... used in place of 5.7 leads to test statistic \(-1.212\)... M1A0, remaining marks available
Critical Value Method\(400 - 1.645 \times 5.7 \div \sqrt{6} = 396.17...\) M1B1...A1
\(400 + 1.645 \times 5.7 \div \sqrt{6}\)M1B1A0
\(396.5 > 396.2\)M1 for sensible comparison (B1 for 396.5)
90% CI MethodCI centred on 396.5; \(\pm 1.645 \times 5.7 \div \sqrt{6}\) M1 B1
\(= (392.67, 400.33)\); contains 400A1, M1
Probability Method\(P(\bar{x} < 396.5) = 0.0663\) M1 A1 (and B1 for 396.5)
\(0.0663 > 0.05\)M1 for sensible comparison
Two-tailed testMax B1 B0 B1 B1 M1 A1 B1 (for \(cv = -1.96\)) M1 A0
# Question 4:

## Part 4(a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: no association between amount spent and sex $H_1$: some association between amount spent and sex | B1 (both) | Hypotheses must be right way round, in context, and must not mention 'correlation' |
| **Total: 1** | | |

## Part 4(a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected frequency $= 62 \times 102 \div 200 = 31.62$ | B1 | Do not allow 31.6 |
| Contribution $= (34 - 31.62)^2 / 31.62 = 0.1791$ | M1 A1 | For valid attempt at $\frac{(O-E)^2}{E}$; **NB answer given** |
| **Total: 3** | | |

# Question 4(a)(iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Refer to $X_4^2$; Critical value at 5% level = 9.488 | B1 | For 4 degrees of freedom |
| Critical value = 9.488 | B1 CAO | For cv |
| $3.205 < 9.488$ | M1 | |
| Result is not significant | A1 | For not significant; allow $p = 0.524$, $0.524 > 0.05$; conclusion must be stated to earn A1; allow 'do not reject $H_0$' and condone 'accept $H_0$' or 'reject $H_1$'; FT if cv consistent with their d.o.f. |
| There is insufficient evidence to suggest any association between amount spent and sex. | E1 | Dependent on previous A1 and final comment must be in context and not mention correlation; SC1 for correct final conclusion where previous A1 omitted but M1 awarded |
| | **5** | |

---

# Question 4(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 400$ | B1 | Hypotheses in words must refer to population mean |
| $H_1: \mu < 400$ | B1 | |
| Where $\mu$ denotes the population mean (weight of the loaves) | B1 | For definition of $\mu$ |
| $\bar{x} = 396.5$ | B1 | For sample mean |
| Test statistic $= \dfrac{396.5 - 400}{5.7/\sqrt{6}} = \dfrac{-3.5}{2.327} = -1.504$ | M1, A1FT | M1 must include $\sqrt{6}$; use their sample mean; condone numerator reversed for M1 but award A1 only if test statistic of 1.504 is compared with a positive $z$-value |
| 5% level 1-tailed critical value of $z = -1.645$ | B1 | For $\pm 1.645$ |
| $-1.504 > -1.645$ so not significant | M1 | For sensible comparison leading to a conclusion; dependent on previous M1 |
| There is insufficient evidence to reject $H_0$ | | |
| There is insufficient evidence to suggest that the true mean weight of the loaves is lower than the minimum specified value of 400 grams. | A1 | For conclusion in context; FT their sample mean only if hypotheses are correct |
| | **9** | |

---

# Additional Notes Q4(b):

| Method | Working | Marks |
|---|---|---|
| $\sigma$ estimated | Sample mean 7.079... used in place of 5.7 leads to test statistic $-1.212$... | M1A0, remaining marks available |
| Critical Value Method | $400 - 1.645 \times 5.7 \div \sqrt{6} = 396.17...$ | M1B1...A1 |
| | $400 + 1.645 \times 5.7 \div \sqrt{6}$ | M1B1A0 |
| | $396.5 > 396.2$ | M1 for sensible comparison (B1 for 396.5) |
| 90% CI Method | CI centred on 396.5; $\pm 1.645 \times 5.7 \div \sqrt{6}$ | M1 B1 |
| | $= (392.67, 400.33)$; contains 400 | A1, M1 |
| Probability Method | $P(\bar{x} < 396.5) = 0.0663$ | M1 A1 (and B1 for 396.5) |
| | $0.0663 > 0.05$ | M1 for sensible comparison |
| Two-tailed test | Max B1 B0 B1 B1 M1 A1 B1 (for $cv = -1.96$) M1 A0 | |
4
\begin{enumerate}[label=(\alph*)]
\item In a survey on internet usage, a random sample of 200 people is selected. The people are asked how much they have spent on internet shopping during the last three months. The results, classified by amount spent and sex, are shown in the table.

\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{2}{|c|}{Sex} & \multirow{2}{*}{Row totals} \\
\hline
 &  & Male & Female &  \\
\hline
\multirow{5}{*}{Amount spent} & Nothing & 28 & 34 & 62 \\
\hline
 & Less than £50 & 17 & 21 & 38 \\
\hline
 & £50 up to £200 & 22 & 26 & 48 \\
\hline
 & £200 up to £1000 & 23 & 16 & 39 \\
\hline
 & £1000 or more & 8 & 5 & 13 \\
\hline
\multicolumn{2}{|c|}{Column totals} & 98 & 102 & 200 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Write down null and alternative hypotheses for a test to examine whether there is any association between amount spent and sex of person.

The contributions to the test statistic for the usual $\chi ^ { 2 }$ test are shown in the table below.

\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{2}{|c|}{Sex} \\
\hline
 &  & Male & Female \\
\hline
\multirow{5}{*}{Amount spent} & Nothing & 0.1865 & 0.1791 \\
\hline
 & Less than £50 & 0.1409 & 0.1354 \\
\hline
 & £50 up to £200 & 0.0982 & 0.0944 \\
\hline
 & £200 up to £1000 & 0.7918 & 0.7608 \\
\hline
 & £1000 or more & 0.4171 & 0.4007 \\
\hline
\end{tabular}
\end{center}

The sum of these contributions, correct to 3 decimal places, is 3.205.
\item Calculate the expected frequency for females spending nothing. Verify the corresponding contribution, 0.1791 , to the test statistic.
\item Carry out the test at the $5 \%$ level of significance, stating your conclusion clearly.
\end{enumerate}\item A bakery sells loaves specified as having a mean weight of 400 grams. It is known that the weights of these loaves are Normally distributed and that the standard deviation is 5.7 grams. An inspector suspects that the true mean weight may be less than 400 grams. In order to test this, the inspector takes a random sample of 6 loaves. Carry out a suitable test at the $5 \%$ level, given that the weights, in grams, of the 6 loaves are as follows.\\
$\begin{array} { l l l l l l } 392.1 & 405.8 & 401.3 & 387.4 & 391.8 & 400.6 \end{array}$

RECOGNISING ACHIEVEMENT
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S2 2011 Q4 [18]}}