CAIE S2 2012 June — Question 6 11 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2012
SessionJune
Marks11
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Mark schemeDownload PDF ↗
TopicType I/II errors and power of test
TypeState meaning of Type II error
DifficultyStandard +0.3 This is a straightforward hypothesis testing question covering standard S2 content: calculating sample statistics from frequency data, performing a two-tailed z-test for a mean, and explaining Type I/II errors. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average for A-level statistics.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean

6 A survey taken last year showed that the mean number of computers per household in Branley was 1.66 . This year a random sample of 50 households in Branley answered a questionnaire with the following results.
Number of computers01234\(> 4\)
Number of households512181050
  1. Calculate unbiased estimates for the population mean and variance of the number of computers per household in Branley this year.
  2. Test at the \(5 \%\) significance level whether the mean number of computers per household has changed since last year.
  3. Explain whether it is possible that a Type I error may have been made in the test in part (ii).
  4. State what is meant by a Type II error in the context of the test in part (ii), and give the set of values of the test statistic that could lead to a Type II error being made.

(i)
B1 \(\bar{x} = 1.96\)
M1 \((\Sigma x^2 f = 254)\); \(S^2 = \frac{x - 1.96^2}{49}\left(\frac{254}{50}\right)\)
A1 \(= \frac{1548}{1225}\) or \(1.2637\)
[3] \(S^2\) correct sub in formula
(ii)
B1 Ho: Pop mean \(= 1.66\)
M1 H1: Pop mean \(\neq 1.66\)
A1 \(\frac{1.96-1.66}{\sqrt{1.2637/50}} = 1.887\)
M1 \(z = 1.96\); \(1.887 < 1.96\)
A1ft No evidence that mean has changed
[5] If H1: Pop mean \(> 1.66\) B0; \(z = 1.645\) M1; In context Evidence mean has changed A1ft
(iii)
B1f No because Ho not rejected
[1] If Ho rejected in (ii): Yes because Ho rejected
(iv)
B1 State mean not changed when it has
B1 In context State mean not increased when it has
[2] \(-1.96 < \) test stat \(< 1.96\); test stat \(< 1.645\)
**(i)**

**B1** $\bar{x} = 1.96$

**M1** $(\Sigma x^2 f = 254)$; $S^2 = \frac{x - 1.96^2}{49}\left(\frac{254}{50}\right)$

**A1** $= \frac{1548}{1225}$ or $1.2637$

**[3]** $S^2$ correct sub in formula

**(ii)**

**B1** Ho: Pop mean $= 1.66$

**M1** H1: Pop mean $\neq 1.66$

**A1** $\frac{1.96-1.66}{\sqrt{1.2637/50}} = 1.887$

**M1** $z = 1.96$; $1.887 < 1.96$

**A1ft** No evidence that mean has changed

**[5]** If H1: Pop mean $> 1.66$ B0; $z = 1.645$ M1; In context Evidence mean has changed A1ft

**(iii)**

**B1f** No because Ho not rejected

**[1]** If Ho rejected in (ii): Yes because Ho rejected

**(iv)**

**B1** State mean not changed when it has

**B1** In context State mean not increased when it has

**[2]** $-1.96 < $ test stat $< 1.96$; test stat $< 1.645$

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6 A survey taken last year showed that the mean number of computers per household in Branley was 1.66 . This year a random sample of 50 households in Branley answered a questionnaire with the following results.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of computers & 0 & 1 & 2 & 3 & 4 & $> 4$ \\
\hline
Number of households & 5 & 12 & 18 & 10 & 5 & 0 \\
\hline
\end{tabular}
\end{center}

(i) Calculate unbiased estimates for the population mean and variance of the number of computers per household in Branley this year.\\
(ii) Test at the $5 \%$ significance level whether the mean number of computers per household has changed since last year.\\
(iii) Explain whether it is possible that a Type I error may have been made in the test in part (ii).\\
(iv) State what is meant by a Type II error in the context of the test in part (ii), and give the set of values of the test statistic that could lead to a Type II error being made.

\hfill \mbox{\textit{CAIE S2 2012 Q6 [11]}}