CAIE S2 2015 June — Question 4 8 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2015
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicType I/II errors and power of test
TypeState meaning of Type I error
DifficultyStandard +0.3 This is a straightforward hypothesis testing question requiring standard procedures: defining Type I error in context, performing a two-tailed z-test with given values, and identifying which error type is possible. All steps are routine applications of textbook methods with no novel problem-solving required, making it slightly easier than average.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean

4 In the past, the flight time, in hours, for a particular flight has had mean 6.20 and standard deviation 0.80 . Some new regulations are introduced. In order to test whether these new regulations have had any effect upon flight times, the mean flight time for a random sample of 40 of these flights is found.
  1. State what is meant by a Type I error in this context.
  2. The mean time for the sample of 40 flights is found to be 5.98 hours. Assuming that the standard deviation of flight times is still 0.80 hours, test at the \(5 \%\) significance level whether the population mean flight time has changed.
  3. State, with a reason, which of the errors, Type I or Type II, might have been made in your answer to part (ii).

AnswerMarks
(i) Conclude flight times affected when in fact they have not been.B1
(ii) \(H_0\): Pop mean (or \(\mu\)) = 6.2B1
\(H_1\): Pop mean (or \(\mu\)) \(\neq\) 6.2B1
Total: 2
AnswerMarks
(iii) \(\frac{5.98 - 6.2}{\frac{0.8}{\sqrt{40}}} = -1.739\) (±) Accept (±)1.74M1 A1
comp \(z = 1.96\)B1
No evidence that flight times affected; \(H_0\) was not rejected or equivalentB1
Type IIB1*
B1*dep
Or accept pop mean changed from 6.2 although pop mean has not changed from 6.2
Allow with 40 instead of \(\sqrt{40}\)
Allow SD/Var mix
(CV method 5.952 or 6.2279 M1 A1)
For valid comparison
or \(P(z < -1.739) = 0.041 > 0.025\) or \(5.98 > 5.952\) or \(6.2 < 6.228\)
and correct conclusion
If in (ii) \(H_0\) was rejected, then: \(H_0\) rejected B1; Type I B1dep
Total: 8
(i) Conclude flight times affected when in fact they have not been. | B1

(ii) $H_0$: Pop mean (or $\mu$) = 6.2 | B1

$H_1$: Pop mean (or $\mu$) $\neq$ 6.2 | B1

**Total: 2**

(iii) $\frac{5.98 - 6.2}{\frac{0.8}{\sqrt{40}}} = -1.739$ (±) Accept (±)1.74 | M1 A1

comp $z = 1.96$ | B1

No evidence that flight times affected; $H_0$ was not rejected or equivalent | B1

Type II | B1*

| B1*dep

Or accept pop mean changed from 6.2 although pop mean has not changed from 6.2

Allow with 40 instead of $\sqrt{40}$

Allow SD/Var mix

(CV method 5.952 or 6.2279 M1 A1)

For valid comparison

or $P(z < -1.739) = 0.041 > 0.025$ or $5.98 > 5.952$ or $6.2 < 6.228$

and correct conclusion

If in (ii) $H_0$ was rejected, then: $H_0$ rejected B1; Type I B1dep

**Total: 8**

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4 In the past, the flight time, in hours, for a particular flight has had mean 6.20 and standard deviation 0.80 . Some new regulations are introduced. In order to test whether these new regulations have had any effect upon flight times, the mean flight time for a random sample of 40 of these flights is found.\\
(i) State what is meant by a Type I error in this context.\\
(ii) The mean time for the sample of 40 flights is found to be 5.98 hours. Assuming that the standard deviation of flight times is still 0.80 hours, test at the $5 \%$ significance level whether the population mean flight time has changed.\\
(iii) State, with a reason, which of the errors, Type I or Type II, might have been made in your answer to part (ii).

\hfill \mbox{\textit{CAIE S2 2015 Q4 [8]}}