OCR MEI Further Statistics B AS 2018 June — Question 5 10 marks

Exam BoardOCR MEI
ModuleFurther Statistics B AS (Further Statistics B AS)
Year2018
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeTwo-tail z-test
DifficultyModerate -0.3 This is a straightforward one-sample z-test with all information provided (known variance, sample mean given, standard hypotheses). Part (i) requires routine application of the z-test procedure with clear hypotheses, test statistic calculation, and conclusion. Parts (ii) and (iii) test basic knowledge of alternative tests (sign test and t-test). Slightly easier than average due to the sample mean being pre-calculated and the test being a standard textbook application with no complications.
Spec5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

5 The flight time between two airports is known to be Normally distributed with mean 3.75 hours and standard deviation 0.21 hours. A new airline starts flying the same route. The flight times for a random sample of 12 flights with the new airline are shown in the spreadsheet (Fig. 5), together with the sample mean. \begin{table}[h]
ABCDEFGHIJKL
13.5953.7233.5843.6433.6693.6973.5503.6743.9243.5633.3303.706
2
3Mean3.638
\captionsetup{labelformat=empty} \caption{Fig. 5}
\end{table} \section*{(i) In this question you must show detailed reasoning.} You should assume that:
  • the flight times for the new airline are Normally distributed,
  • the standard deviation of the flight times is still 0.21 hours.
Carry out a test at the \(5 \%\) significance level to investigate whether the mean flight time for the new airline is less than 3.75 hours.
(ii) If both of the assumptions in part (i) were false, name an alternative test that you could carry out to investigate average flight times, stating any assumption necessary for this test.
(iii) If instead the flight times were still Normally distributed but the standard deviation was not known to be 0.21 hours, name another test that you could carry out.

Question 5:
AnswerMarks Guidance
5(i) H : μ = 3.75 H : μ < 3.75
0 1
Where μ is the population mean flight time with the
new airline in hours
Test statistic is
= -1.848
Critical value (1-tailed) at 5% level is (-)1.645
-1.848 < -1.645 so significant ( reject H )
0
Sufficient evidence to suggest that the average
flight time with the new airline is less than 3.75
AnswerMarks
hoursB1
B1
M1
A1
B1
M1
A1
AnswerMarks
[7]DR
Hypotheses in words only must
include “population”.
For definition in context.
Allow use of exact value for mean =
3.6382 leading to -1.844
AnswerMarks
No FT if not ±1.645For critical value of X method,
critical value
= 3.75 – 1.645×0.21/√12 gets M1 =
3.650 gets A1 with B1 for 1.645 than
sixth mark for comparison with
3.638 and significant
For p-value method fifth mark is for
p-value of 0.0323 than sixth for
comparison with 0.05 and significant
AnswerMarks
(ii)Wilcoxon signed rank test
(Underlying distribution) needs to be symmetricalB1
B1
AnswerMarks
[2]Allow Wilcoxon single sample test
Ignore ‘around the median’ or
AnswerMarks
‘around the mean’ following thisDo not allow just ‘Wilcoxon test’ nor
‘Wilcoxon rank test’
Do not allow ‘Median = mean’
AnswerMarks Guidance
(iii)Single sample t-test B1
[1]Condone ‘t-test’ without ‘single
sample’
Question 5:
5 | (i) | H : μ = 3.75 H : μ < 3.75
0 1
Where μ is the population mean flight time with the
new airline in hours
Test statistic is
= -1.848
Critical value (1-tailed) at 5% level is (-)1.645
-1.848 < -1.645 so significant ( reject H )
0
Sufficient evidence to suggest that the average
flight time with the new airline is less than 3.75
hours | B1
B1
M1
A1
B1
M1
A1
[7] | DR
Hypotheses in words only must
include “population”.
For definition in context.
Allow use of exact value for mean =
3.6382 leading to -1.844
No FT if not ±1.645 | For critical value of X method,
critical value
= 3.75 – 1.645×0.21/√12 gets M1 =
3.650 gets A1 with B1 for 1.645 than
sixth mark for comparison with
3.638 and significant
For p-value method fifth mark is for
p-value of 0.0323 than sixth for
comparison with 0.05 and significant
(ii) | Wilcoxon signed rank test
(Underlying distribution) needs to be symmetrical | B1
B1
[2] | Allow Wilcoxon single sample test
Ignore ‘around the median’ or
‘around the mean’ following this | Do not allow just ‘Wilcoxon test’ nor
‘Wilcoxon rank test’
Do not allow ‘Median = mean’
(iii) | Single sample t-test | B1
[1] | Condone ‘t-test’ without ‘single
sample’
5 The flight time between two airports is known to be Normally distributed with mean 3.75 hours and standard deviation 0.21 hours. A new airline starts flying the same route. The flight times for a random sample of 12 flights with the new airline are shown in the spreadsheet (Fig. 5), together with the sample mean.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline
 & A & B & C & D & E & F & G & H & I & J & K & L \\
\hline
1 & 3.595 & 3.723 & 3.584 & 3.643 & 3.669 & 3.697 & 3.550 & 3.674 & 3.924 & 3.563 & 3.330 & 3.706 \\
\hline
2 &  &  &  &  &  &  &  &  &  &  &  &  \\
\hline
3 & Mean & 3.638 &  &  &  &  &  &  &  &  &  &  \\
\hline
 &  &  &  &  &  &  &  &  &  &  &  &  \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{table}

\section*{(i) In this question you must show detailed reasoning.}
You should assume that:

\begin{itemize}
  \item the flight times for the new airline are Normally distributed,
  \item the standard deviation of the flight times is still 0.21 hours.
\end{itemize}

Carry out a test at the $5 \%$ significance level to investigate whether the mean flight time for the new airline is less than 3.75 hours.\\
(ii) If both of the assumptions in part (i) were false, name an alternative test that you could carry out to investigate average flight times, stating any assumption necessary for this test.\\
(iii) If instead the flight times were still Normally distributed but the standard deviation was not known to be 0.21 hours, name another test that you could carry out.

\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2018 Q5 [10]}}