| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a standard S3 hypothesis testing question requiring routine application of one-sample t-test procedures and confidence interval formulas. While it involves multiple parts and some calculation, all techniques are textbook exercises with no novel problem-solving required. The final part requires simple algebraic manipulation of the confidence interval width formula. Slightly easier than average due to straightforward application of standard procedures. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 1\), \(H_1: \mu < 1\) | B1 | Both hypotheses. In words only must include "population" |
| where \(\mu\) is the mean weight of the cakes | B1 | For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) used, but do NOT allow "\(\bar{X} = ...\)" or similar unless \(\bar{X}\) is clearly stated to be a population mean |
| \(\bar{x} = 0.957375\), \(s_{n-1} = 0.07314(55)\) | B1 | \(s_n = 0.06842\) but do NOT allow this here or in construction of test statistic, but FT from there |
| Test statistic is \(\frac{0.957375 - 1}{\frac{0.07314}{\sqrt{8}}}\) | M1 | Allow c's \(\bar{x}\) and/or \(s_{n-1}\). Allow alternative: \(1 + (c\text{'s} - 1.895) \times \frac{0.07314}{\sqrt{8}}\) (= 0.950997) for subsequent comparison with \(\bar{x}\). Or \(\bar{x} - (c\text{'s} - 1.895) \times \frac{0.07314}{\sqrt{8}}\) (= 1.006377) for comparison with 1 |
| \(= -1.648(24)\) | A1 | c.a.o. but ft from here in any case if wrong. Use of \(1 - \bar{x}\) scores M1A0, but ft |
| Refer to \(t_7\) | M1 | No ft from here if wrong. \(P(t < -1.648(24)) = 0.0716\) |
| Single-tailed 5% point is \(-1.895\) | A1 | Must be minus 1.895 unless absolute values are being compared. No ft from here if wrong |
| Not significant | A1 | ft only c's test statistic |
| Insufficient evidence to suggest that the cakes are underweight on average | A1 | ft only c's test statistic |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CI is given by \(0.957375 \pm\) | M1 | |
| \(2.365\) | B1 | |
| \(\times \frac{0.07314}{\sqrt{8}}\) | M1 | |
| \(= 0.957375 \pm 0.061156 = (0.896(2), 1.018(5))\) | A1 | c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to \(t_7\) is OK |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} \pm 1.96 \times \sqrt{\frac{0.006}{n}}\) | M1 | Structure correct, incl. use of Normal |
| B1 | Normal | |
| A1 | 1.96 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 \times 1.96 \times \sqrt{\frac{0.006}{n}} < 0.025\) | M1 | Set up appropriate inequation. Condone an equation |
| \(n > \left(\frac{2 \times 1.96}{0.025}\right)^2 \times 0.006 = 147.517\) | M1 | Attempt to rearrange and solve |
| So take \(n = 148\) | A1 | c.a.o. (expressed as an integer). S.C. Allow max M1A1(c.a.o.) when the factor "2" is missing. (\(n > 36.879\)) |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 1$, $H_1: \mu < 1$ | B1 | Both hypotheses. In words only must include "population" |
| where $\mu$ is the mean weight of the cakes | B1 | For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ used, but do NOT allow "$\bar{X} = ...$" or similar unless $\bar{X}$ is clearly stated to be a population mean |
| $\bar{x} = 0.957375$, $s_{n-1} = 0.07314(55)$ | B1 | $s_n = 0.06842$ but do NOT allow this here or in construction of test statistic, but FT from there |
| Test statistic is $\frac{0.957375 - 1}{\frac{0.07314}{\sqrt{8}}}$ | M1 | Allow c's $\bar{x}$ and/or $s_{n-1}$. Allow alternative: $1 + (c\text{'s} - 1.895) \times \frac{0.07314}{\sqrt{8}}$ (= 0.950997) for subsequent comparison with $\bar{x}$. Or $\bar{x} - (c\text{'s} - 1.895) \times \frac{0.07314}{\sqrt{8}}$ (= 1.006377) for comparison with 1 |
| $= -1.648(24)$ | A1 | c.a.o. but ft from here in any case if wrong. Use of $1 - \bar{x}$ scores M1A0, but ft |
| Refer to $t_7$ | M1 | No ft from here if wrong. $P(t < -1.648(24)) = 0.0716$ |
| Single-tailed 5% point is $-1.895$ | A1 | Must be minus 1.895 unless absolute values are being compared. No ft from here if wrong |
| Not significant | A1 | ft only c's test statistic |
| Insufficient evidence to suggest that the cakes are underweight on average | A1 | ft only c's test statistic |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| CI is given by $0.957375 \pm$ | M1 | |
| $2.365$ | B1 | |
| $\times \frac{0.07314}{\sqrt{8}}$ | M1 | |
| $= 0.957375 \pm 0.061156 = (0.896(2), 1.018(5))$ | A1 | c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to $t_7$ is OK |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} \pm 1.96 \times \sqrt{\frac{0.006}{n}}$ | M1 | Structure correct, incl. use of Normal |
| | B1 | Normal |
| | A1 | 1.96 |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 \times 1.96 \times \sqrt{\frac{0.006}{n}} < 0.025$ | M1 | Set up appropriate inequation. Condone an equation |
| $n > \left(\frac{2 \times 1.96}{0.025}\right)^2 \times 0.006 = 147.517$ | M1 | Attempt to rearrange and solve |
| So take $n = 148$ | A1 | c.a.o. (expressed as an integer). S.C. Allow max M1A1(c.a.o.) when the factor "2" is missing. ($n > 36.879$) |
---2 Pat makes and sells fruit cakes at a local market. On her stall a sign states that the average weight of the cakes is 1 kg . A trading standards officer carries out a routine check of a random sample of 8 of Pat's cakes to ensure that they are not underweight, on average. The weights, in kg , that he records are as follows.
$$\begin{array} { l l l l l l l l }
0.957 & 1.055 & 0.983 & 0.917 & 1.015 & 0.865 & 1.013 & 0.854
\end{array}$$
(i) On behalf of the trading standards officer, carry out a suitable test at a $5 \%$ level of significance, stating your hypotheses clearly. Assume that the weights of Pat's fruit cakes are Normally distributed.\\
(ii) Find a 95\% confidence interval for the true mean weight of Pat's fruit cakes.
Pat's husband, Tony, is the owner of a factory which makes and supplies fruit cakes to a large supermarket chain. A large random sample of $n$ of these cakes has mean weight $\bar { x } \mathrm {~kg}$ and variance $0.006 \mathrm {~kg} ^ { 2 }$.\\
(iii) Write down, in terms of $n$ and $\bar { x }$, a $95 \%$ confidence interval for the true mean weight of cakes produced in Tony's factory.\\
(iv) What is the size of the smallest sample that should be taken if the width of the confidence interval in part (iii) is to be 0.025 kg at most?
\hfill \mbox{\textit{OCR MEI S3 2009 Q2 [19]}}