OCR MEI S3 2007 January — Question 4 18 marks

Exam BoardOCR MEI
ModuleS3 (Statistics 3)
Year2007
SessionJanuary
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared goodness of fit
TypeChi-squared goodness of fit: Normal
DifficultyStandard +0.3 Part (a) is a standard chi-squared goodness of fit test with given expected frequencies requiring only calculation of test statistic, degrees of freedom adjustment, and comparison to critical value. Part (b) is a routine Wilcoxon signed-rank test on paired data. Both are textbook applications with no novel insight required, though slightly above average due to the two-part structure and need to handle the degrees of freedom correctly in part (a).
Spec5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous5.07b Sign test: and Wilcoxon signed-rank5.07d Paired vs two-sample: selection

4
  1. An amateur weather forecaster has been keeping records of air pressure, measured in atmospheres. She takes the measurement at the same time every day using a barometer situated in her garden. A random sample of 100 of her observations is summarised in the table below. The corresponding expected frequencies for a Normal distribution, with its two parameters estimated by sample statistics, are also shown in the table.
    Pressure ( \(a\) atmospheres)Observed frequencyFrequency as given by Normal model
    \(a \leqslant 0.98\)41.45
    \(0.98 < a \leqslant 0.99\)65.23
    \(0.99 < a \leqslant 1.00\)913.98
    \(1.00 < a \leqslant 1.01\)1523.91
    \(1.01 < a \leqslant 1.02\)3726.15
    \(1.02 < a \leqslant 1.03\)2118.29
    \(1.03 < a\)810.99
    Carry out a test at the \(5 \%\) level of significance of the goodness of fit of the Normal model. State your conclusion carefully and comment on your findings.
  2. The forecaster buys a new digital barometer that can be linked to her computer for easier recording of observations. She decides that she wishes to compare the readings of the new barometer with those of the old one. For a random sample of 10 days, the readings (in atmospheres) of the two barometers are shown below.
    DayABCDEFGHIJ
    Old0.9921.0051.0011.0111.0260.9801.0201.0251.0421.009
    New0.9851.0031.0021.0141.0220.9881.0301.0161.0471.025
    Use an appropriate Wilcoxon test to examine at the \(10 \%\) level of significance whether there is any reason to suppose that, on the whole, readings on the old and new barometers do not agree.

Question 4:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Combine first two rows: Obs = 10, Exp = 6.68M1 Combine first two rows
\(X^2 = \dfrac{(10-6.68)^2}{6.68} + \ldots = 1.6501+1.7740+3.3203+4.5018+0.4015+0.8135\)M1
\(= 12.46(12)\)A1
d.o.f. \(= 6-3 = 3\) Require d.o.f. \(=\) No. cells used \(- 3\)
Refer to \(\chi^2_3\). Upper 5% point is 7.815M1, A1 No ft if wrong
\(12.46 > 7.815\) \(\therefore\) Result is significantE1 ft only c's test statistic
Seems the Normal model does not fit the data at the 5% levelE1 ft only c's test statistic
Biggest discrepancy is in class \(1.01 < a \leq 1.02\)E1
Model overestimates in classes ..., but underestimates in classes ...E1 Any two suitable comments
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Differences (Old \(-\) New): \(0.007,\ 0.002,\ -0.001,\ -0.003,\ 0.004,\ -0.008,\ -0.010,\ 0.009,\ -0.005,\ -0.016\)M1 For differences. ZERO if differences not used
Ranks of \(\lvert\text{diff}\rvert\): \(6,\ 2,\ 1,\ 3,\ 4,\ 7,\ 9,\ 8,\ 5,\ 10\)M1, A1 For ranks of \(\lvert\text{difference}\rvert\); all correct
\(W_+ = 6+2+4+8 = 20\)B1 Or \(W_- = 1+3+7+9+5+10 = 35\)
Refer to Wilcoxon single sample (paired) tables for \(n=10\)M1 No ft if wrong
Lower two-tail 10% point is \(\ldots 10\)M1, A1 Or if 35 used, upper point is 45
\(20 > 10\) \(\therefore\) Result is not significantE1 ft only c's test statistic
Seems there is no reason to suppose the barometers differE1 ft only c's test statistic
# Question 4:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Combine first two rows: Obs = 10, Exp = 6.68 | M1 | Combine first two rows |
| $X^2 = \dfrac{(10-6.68)^2}{6.68} + \ldots = 1.6501+1.7740+3.3203+4.5018+0.4015+0.8135$ | M1 | |
| $= 12.46(12)$ | A1 | |
| d.o.f. $= 6-3 = 3$ | | Require d.o.f. $=$ No. cells used $- 3$ |
| Refer to $\chi^2_3$. Upper 5% point is 7.815 | M1, A1 | No ft if wrong |
| $12.46 > 7.815$ $\therefore$ Result is significant | E1 | ft only c's test statistic |
| Seems the Normal model does not fit the data at the 5% level | E1 | ft only c's test statistic |
| Biggest discrepancy is in class $1.01 < a \leq 1.02$ | E1 | |
| Model overestimates in classes ..., but underestimates in classes ... | E1 | Any two suitable comments | [9 marks] |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differences (Old $-$ New): $0.007,\ 0.002,\ -0.001,\ -0.003,\ 0.004,\ -0.008,\ -0.010,\ 0.009,\ -0.005,\ -0.016$ | M1 | For differences. ZERO if differences not used |
| Ranks of $\lvert\text{diff}\rvert$: $6,\ 2,\ 1,\ 3,\ 4,\ 7,\ 9,\ 8,\ 5,\ 10$ | M1, A1 | For ranks of $\lvert\text{difference}\rvert$; all correct |
| $W_+ = 6+2+4+8 = 20$ | B1 | Or $W_- = 1+3+7+9+5+10 = 35$ |
| Refer to Wilcoxon single sample (paired) tables for $n=10$ | M1 | No ft if wrong |
| Lower two-tail 10% point is $\ldots 10$ | M1, A1 | Or if 35 used, upper point is 45 |
| $20 > 10$ $\therefore$ Result is not significant | E1 | ft only c's test statistic |
| Seems there is no reason to suppose the barometers differ | E1 | ft only c's test statistic | [9 marks] |
4
\begin{enumerate}[label=(\alph*)]
\item An amateur weather forecaster has been keeping records of air pressure, measured in atmospheres. She takes the measurement at the same time every day using a barometer situated in her garden. A random sample of 100 of her observations is summarised in the table below. The corresponding expected frequencies for a Normal distribution, with its two parameters estimated by sample statistics, are also shown in the table.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Pressure ( $a$ atmospheres) & Observed frequency & Frequency as given by Normal model \\
\hline
$a \leqslant 0.98$ & 4 & 1.45 \\
\hline
$0.98 < a \leqslant 0.99$ & 6 & 5.23 \\
\hline
$0.99 < a \leqslant 1.00$ & 9 & 13.98 \\
\hline
$1.00 < a \leqslant 1.01$ & 15 & 23.91 \\
\hline
$1.01 < a \leqslant 1.02$ & 37 & 26.15 \\
\hline
$1.02 < a \leqslant 1.03$ & 21 & 18.29 \\
\hline
$1.03 < a$ & 8 & 10.99 \\
\hline
\end{tabular}
\end{center}

Carry out a test at the $5 \%$ level of significance of the goodness of fit of the Normal model. State your conclusion carefully and comment on your findings.
\item The forecaster buys a new digital barometer that can be linked to her computer for easier recording of observations. She decides that she wishes to compare the readings of the new barometer with those of the old one. For a random sample of 10 days, the readings (in atmospheres) of the two barometers are shown below.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
Day & A & B & C & D & E & F & G & H & I & J \\
\hline
Old & 0.992 & 1.005 & 1.001 & 1.011 & 1.026 & 0.980 & 1.020 & 1.025 & 1.042 & 1.009 \\
\hline
New & 0.985 & 1.003 & 1.002 & 1.014 & 1.022 & 0.988 & 1.030 & 1.016 & 1.047 & 1.025 \\
\hline
\end{tabular}
\end{center}

Use an appropriate Wilcoxon test to examine at the $10 \%$ level of significance whether there is any reason to suppose that, on the whole, readings on the old and new barometers do not agree.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S3 2007 Q4 [18]}}