OCR MEI Further Statistics A AS 2019 June — Question 5 13 marks

Exam BoardOCR MEI
ModuleFurther Statistics A AS (Further Statistics A AS)
Year2019
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared goodness of fit
TypeChi-squared goodness of fit: Binomial
DifficultyStandard +0.3 This is a straightforward completion of a chi-squared goodness of fit test where the model and parameter are already specified (B(7,0.35)). Students need to compare a test statistic (likely calculated in earlier parts) with critical value and state a conclusion, then make a basic comment using the given graph. Requires only standard procedure application with minimal problem-solving.
Spec5.08f Hypothesis test: Spearman rank5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression5.09e Use regression: for estimation in context
5 A researcher is investigating births of females and males in a particular species of animal which very often produces litters of 7 offspring.
The table shows some data about the number of females per litter in 200 litters of 7 offspring. The researcher thinks that a binomial distribution \(\mathrm { B } ( 7 , p )\) may be an appropriate model for these data. (c) Complete the test at the \(5 \%\) significance level. Fig. 5 shows the probability distribution \(\mathrm { B } ( 7,0.35 )\) together with the relative frequencies of the observed data (the numbers of litters each divided by 200). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fd496303-10f1-450e-bbeb-421ab6f4de21-5_659_1285_342_319} \captionsetup{labelformat=empty} \caption{Fig. 5}
\end{figure} (d) Comment on the result of the test completed in part (c) by considering Fig. 5.
Question 5:
AnswerMarks Guidance
5(a) Mean number of females per litter =
200451...27 490
= 2.45
200 200
np = 2.45 with n = 7
AnswerMarks
(p = 0.35)B1
B1
AnswerMarks
[2]1.1
1.1For 7p = 2.45 leading to given answer
AG
SC There are 490 females out of 1400 offspring so
p = 490/1400 = 0.35 gets SC2
p = 490/1400 with no further explanation gets SC1
AnswerMarks
(b)prob = 0.18477… ×200 or 0.29848…×200
For 1, exp freq = 36.9553 4d.p.
For 2, exp freq = 59.6970 4 d.p.
(3753.5742)2
For 3 Contribution =
53.5742
AnswerMarks
= 5.1275M1
A1
M1
A1
AnswerMarks
[4]3.4
2.2a
1.1a
AnswerMarks
1.1For method for either
Allow second value to be found by subtraction.
Allow 36.96 and 59.70
Allow 5.13
AnswerMarks
(c)H : binomial model is suitable
0
H : binomial model is not suitable
1
X2 = 30.98
Refer to 2
4
Critical value at 5% level = 9.488
30.98 > 9.488 Significant result
There is sufficient evidence to suggest that the
AnswerMarks
binomial model is not suitableB1
B1
M1
A1
M1
A1
AnswerMarks
[6]3.3
1.1
3.4
1.1
1.1
AnswerMarks
2.2bAllow ‘the binomial model fits/does not fit the data’
For 4 degrees of freedom. No further marks form here if
wrong.
For comparison leading to a conclusion.
FT their test statistic
For non-assertive correct conclusion in terms of H .
1
AnswerMarks
(d)e.g.
More litters than expected have very few (0 or 1)
females.
The observed data shows a more positive skew
than the binomial model.
The diagram shows that the observed data has
more variation than the binomial distribution
AnswerMarks Guidance
would suggest.B1
[1]2.2b For any one of these comments
Accept suitable alternatives which highlight specific
differences between probabilities and relative frequencies.
Question 5:
5 | (a) | Mean number of females per litter =
200451...27 490
= 2.45
200 200
np = 2.45 with n = 7
(p = 0.35) | B1
B1
[2] | 1.1
1.1 | For 7p = 2.45 leading to given answer
AG
SC There are 490 females out of 1400 offspring so
p = 490/1400 = 0.35 gets SC2
p = 490/1400 with no further explanation gets SC1
(b) | prob = 0.18477… ×200 or 0.29848…×200
For 1, exp freq = 36.9553 4d.p.
For 2, exp freq = 59.6970 4 d.p.
(3753.5742)2
For 3 Contribution =
53.5742
= 5.1275 | M1
A1
M1
A1
[4] | 3.4
2.2a
1.1a
1.1 | For method for either
Allow second value to be found by subtraction.
Allow 36.96 and 59.70
Allow 5.13
(c) | H : binomial model is suitable
0
H : binomial model is not suitable
1
X2 = 30.98
Refer to 2
4
Critical value at 5% level = 9.488
30.98 > 9.488 Significant result
There is sufficient evidence to suggest that the
binomial model is not suitable | B1
B1
M1
A1
M1
A1
[6] | 3.3
1.1
3.4
1.1
1.1
2.2b | Allow ‘the binomial model fits/does not fit the data’
For 4 degrees of freedom. No further marks form here if
wrong.
For comparison leading to a conclusion.
FT their test statistic
For non-assertive correct conclusion in terms of H .
1
(d) | e.g.
More litters than expected have very few (0 or 1)
females.
The observed data shows a more positive skew
than the binomial model.
The diagram shows that the observed data has
more variation than the binomial distribution
would suggest. | B1
[1] | 2.2b | For any one of these comments
Accept suitable alternatives which highlight specific
differences between probabilities and relative frequencies.
5 A researcher is investigating births of females and males in a particular species of animal which very often produces litters of 7 offspring.\\
The table shows some data about the number of females per litter in 200 litters of 7 offspring. The researcher thinks that a binomial distribution $\mathrm { B } ( 7 , p )$ may be an appropriate model for these data.

(c) Complete the test at the $5 \%$ significance level.

Fig. 5 shows the probability distribution $\mathrm { B } ( 7,0.35 )$ together with the relative frequencies of the observed data (the numbers of litters each divided by 200).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fd496303-10f1-450e-bbeb-421ab6f4de21-5_659_1285_342_319}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}

(d) Comment on the result of the test completed in part (c) by considering Fig. 5.

\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2019 Q5 [13]}}
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