OCR MEI S3 2008 June — Question 3 18 marks

Exam BoardOCR MEI
ModuleS3 (Statistics 3)
Year2008
SessionJune
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypePaired sample t-test
DifficultyStandard +0.3 This is a standard S3 question testing paired t-test and Wilcoxon signed-rank test with straightforward data. Part (a) requires stating hypotheses and assumptions (routine), calculating differences, finding mean/SD, and performing a t-test (mechanical). Part (b) involves ranking differences and applying Wilcoxon test (standard procedure). Both are textbook applications with no conceptual challenges or novel insights required, making this slightly easier than average.
Spec5.05c Hypothesis test: normal distribution for population mean5.07b Sign test: and Wilcoxon signed-rank

  1. A tea grower is testing two types of plant for the weight of tea they produce. A trial is set up in which each type of plant is grown at each of 8 sites. The total weight, in grams, of tea leaves harvested from each plant is measured and shown below.
    SiteABCDEFGH
    Type I225.2268.9303.6244.1230.6202.7242.1247.5
    Type II215.2242.1260.9241.7245.5204.7225.8236.0
    1. The grower intends to perform a \(t\) test to examine whether there is any difference in the mean yield of the two types of plant. State the hypotheses he should use and also any necessary assumption. [3]
    2. Carry out the test using a 5\% significance level. [7]
  2. The tea grower deals with many types of tea and employs tasters to rate them. The tasters do this by giving each tea a score out of 100. The tea grower wishes to compare the scores given by two of the tasters. Their scores for a random selection of 10 teas are as follows.
    TeaQRSTUVWXYZ
    Taster 169798563816585868977
    Taster 274759966756496949686
    Use a Wilcoxon test to examine, at the 5\% level of significance, whether it appears that, on the whole, the scores given to teas by these two tasters differ. [8]

(a)(i)
\(H_0: \mu_0 = 0\) (or \(\mu = \mu_I\))
\(H_1: \mu_0 \neq 0\) (or \(\mu_0 \neq \mu_I\))
where \(\mu_0\) is "mean for II – mean for I"
AnswerMarks Guidance
Normality of differences is required.B1, B1, B1 Both. Hypotheses in words only must include "population". For adequate verbal definition. Allow absence of "population" if correct notation \(\mu\) is used, but do NOT allow "\(\bar{X} = \bar{X}_B\)" or similar unless \(\bar{X}\) is clearly and explicitly stated to be a population mean.
(a)(ii)
MUST be PAIRED COMPARISON \(t\) test.
Differences are: 10.0, 26.8, 42.7, 2.4, –14.9, –2.0, 16.3, 11.5
\(\bar{d} = 11.6\), \(s_{n-1} = 17.707\)
Test statistic is \(\frac{11.6 - 0}{17.707/\sqrt{8}} = 1.852(92)\).
Refer to \(t_7\).
Double-tailed 5% point is 2.365.
Not significant.
AnswerMarks Guidance
Seems there is no difference between the mean yields of the two types of plant.B1, M1, A1, M1, A1, A1, A1, A1 \(s_n = 16.563\) but do NOT allow this here or in construction of test statistic, but FT from there. Allow \(c's\) \(\bar{d}\) and/or \(s_{n-1}\). Allow alternative: \(0 + (c's\) 2.365) \(\times \frac{17.707}{\sqrt{8}}\) (= 14.806) for subsequent comparison with \(\bar{d}\). (Or \(\bar{d} - (c's\) 2.365) \(\times \frac{17.707}{\sqrt{8}}\) (=-3.206) for comparison with 0.) c.a.o. but ft from here in any case if wrong. Use of \(0 - \bar{d}\) scores M1A0, but ft. No ft from here if wrong. No ft from here if wrong. ft only c's test statistic. ft only c's test statistic. Special case: (\(t_6\) and 2.306) can score 1 of these last 2 marks if either form of conclusion is given.
Question 3 (continued):
(b)
AnswerMarks Guidance
DiffRank of \ diff\
\(W_+ = 1 + 3 + 5 = 9\) (or \(W_- = 2 + 4 + 6 + 7 + 8 + 9 + 10 = 46\))
Refer to tables of Wilcoxon paired (single sample) statistic for \(n = 10\).
Lower (or upper if 46 used) double-tailed 5% point is 8 (or 47 if 46 used).
Result is not significant.
AnswerMarks Guidance
No evidence to suggest the tasters differ on the whole.M1, M1, A1, B1, M1, A1, A1, A1 For differences. ZERO in this section if differences not used. For ranks. FT from here if ranks wrong. (Condone if not combined as fully as shown above, but require top two cells combined as a minimum.) No ft from here if wrong. i.e. a 2-tail test. No ft from here if wrong. ft only c's test statistic. ft only c's test statistic.
## (a)(i)
$H_0: \mu_0 = 0$ (or $\mu = \mu_I$)

$H_1: \mu_0 \neq 0$ (or $\mu_0 \neq \mu_I$)

where $\mu_0$ is "mean for II – mean for I"

Normality of differences is required. | B1, B1, B1 | Both. Hypotheses in words only must include "population". For adequate verbal definition. Allow absence of "population" if correct notation $\mu$ is used, but do NOT allow "$\bar{X} = \bar{X}_B$" or similar unless $\bar{X}$ is clearly and explicitly stated to be a population mean.

## (a)(ii)
MUST be PAIRED COMPARISON $t$ test.

Differences are: 10.0, 26.8, 42.7, 2.4, –14.9, –2.0, 16.3, 11.5

$\bar{d} = 11.6$, $s_{n-1} = 17.707$

Test statistic is $\frac{11.6 - 0}{17.707/\sqrt{8}} = 1.852(92)$.

Refer to $t_7$.

Double-tailed 5% point is 2.365.

Not significant.

Seems there is no difference between the mean yields of the two types of plant. | B1, M1, A1, M1, A1, A1, A1, A1 | $s_n = 16.563$ but do NOT allow this here or in construction of test statistic, but FT from there. Allow $c's$ $\bar{d}$ and/or $s_{n-1}$. Allow alternative: $0 + (c's$ 2.365) $\times \frac{17.707}{\sqrt{8}}$ (= 14.806) for subsequent comparison with $\bar{d}$. (Or $\bar{d} - (c's$ 2.365) $\times \frac{17.707}{\sqrt{8}}$ (=-3.206) for comparison with 0.) c.a.o. but ft from here in any case if wrong. Use of $0 - \bar{d}$ scores M1A0, but ft. No ft from here if wrong. No ft from here if wrong. ft only c's test statistic. ft only c's test statistic. Special case: ($t_6$ and 2.306) can score 1 of these last 2 marks if either form of conclusion is given.

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# Question 3 (continued):

## (b)
| Diff | Rank of \|diff\| | –5, 4, –14, –3, 6, 1, –11, –8, –7, –9 | 4, 3, 10, 2, 5, 1, 9, 7, 6, 8 |

$W_+ = 1 + 3 + 5 = 9$ (or $W_- = 2 + 4 + 6 + 7 + 8 + 9 + 10 = 46$)

Refer to tables of Wilcoxon paired (single sample) statistic for $n = 10$.

Lower (or upper if 46 used) double-tailed 5% point is 8 (or 47 if 46 used).

Result is not significant.

No evidence to suggest the tasters differ on the whole. | M1, M1, A1, B1, M1, A1, A1, A1 | For differences. ZERO in this section if differences not used. For ranks. FT from here if ranks wrong. (Condone if not combined as fully as shown above, but require top two cells combined as a minimum.) No ft from here if wrong. i.e. a 2-tail test. No ft from here if wrong. ft only c's test statistic. ft only c's test statistic.

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\begin{enumerate}[label=(\alph*)]
\item A tea grower is testing two types of plant for the weight of tea they produce. A trial is set up in which each type of plant is grown at each of 8 sites. The total weight, in grams, of tea leaves harvested from each plant is measured and shown below.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Site & A & B & C & D & E & F & G & H \\
\hline
Type I & 225.2 & 268.9 & 303.6 & 244.1 & 230.6 & 202.7 & 242.1 & 247.5 \\
\hline
Type II & 215.2 & 242.1 & 260.9 & 241.7 & 245.5 & 204.7 & 225.8 & 236.0 \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}[label=(\roman*)]
\item The grower intends to perform a $t$ test to examine whether there is any difference in the mean yield of the two types of plant. State the hypotheses he should use and also any necessary assumption. [3]

\item Carry out the test using a 5\% significance level. [7]
\end{enumerate}

\item The tea grower deals with many types of tea and employs tasters to rate them. The tasters do this by giving each tea a score out of 100. The tea grower wishes to compare the scores given by two of the tasters. Their scores for a random selection of 10 teas are as follows.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Tea & Q & R & S & T & U & V & W & X & Y & Z \\
\hline
Taster 1 & 69 & 79 & 85 & 63 & 81 & 65 & 85 & 86 & 89 & 77 \\
\hline
Taster 2 & 74 & 75 & 99 & 66 & 75 & 64 & 96 & 94 & 96 & 86 \\
\hline
\end{tabular}
\end{center}

Use a Wilcoxon test to examine, at the 5\% level of significance, whether it appears that, on the whole, the scores given to teas by these two tasters differ. [8]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S3 2008 Q3 [18]}}