CAIE S2 2018 November — Question 5 10 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2018
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicType I/II errors and power of test
TypeIdentify which error type was made
DifficultyStandard +0.3 This is a straightforward hypothesis testing question covering standard S2 content: calculating unbiased estimates (routine formulas), conducting a one-tailed z-test (standard procedure), and identifying which type of error is possible given the test outcome. All parts follow textbook methods with no novel insight required, making it slightly easier than average.
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

5 The numbers of basketball courts in a random sample of 70 schools in South Mowland are summarised in the table.
Number of basketball courts01234\(> 4\)
Number of schools228261040
  1. Calculate unbiased estimates for the population mean and variance of the number of basketball courts per school in South Mowland.
    The mean number of basketball courts per school in North Mowland is 1.9 .
  2. Test at the \(5 \%\) significance level whether the mean number of basketball courts per school in South Mowland is less than the mean for North Mowland.
  3. State, with a reason, which of the errors, Type I or Type II, might have been made in the test in part (ii).

Question 5(i):
AnswerMarks Guidance
\(\hat{\mu} = \frac{126}{70} = \frac{9}{5} = 1.8\)B1
\(\Sigma x^2 f = 286\)B1 Seen or implied
\(\text{Est}(\sigma^2) = \frac{70}{69}\left(\frac{\Sigma x^2 f}{70} - 1.8^2\right)\)M1 oe attempted
\(= 0.858\) or \(\frac{296}{345}\)A1 Note: Final answer for var 0.846 (biased) and no working implies B1 for 286
Question 5(ii):
AnswerMarks Guidance
\(H_0: \mu = 1.9\), \(H_1: \mu < 1.9\)B1 Or 'pop mean'; not just 'mean'
\(\frac{1.8 - 1.9}{\sqrt{\frac{0.858}{70}}}\)M1 Standardise with their values from (i). Must have \(\sqrt{70}\). No SD/Var mix
\(= -0.903\)A1 Accept \(\pm\)
\(0.903 < 1.645\)M1 comp 1.645; allow comp 1.96 if \(H_1: \mu \neq 1.9\); or comp \(1 - \phi(0.903) = 0.182\) or \(0.183\) with \(0.05\)
No evidence that mean no. courts in S is less than in NA1ft No contradictions. ft their 0.903, but not comp 1.96. Accept cv method: \(\text{cv} = 1.718\) M1A1 \(1.718 < 1.8\) M1 conclusion A1
Question 5(iii):
AnswerMarks Guidance
Type II because \(H_0\) was not rejectedB1ft ft their conclusion. If \(H_0\) rejected, 'Type I because \(H_0\) rejected' B1. No conclusion in (ii) scores B0
## Question 5(i):

| $\hat{\mu} = \frac{126}{70} = \frac{9}{5} = 1.8$ | B1 | |
|---|---|---|
| $\Sigma x^2 f = 286$ | B1 | Seen or implied |
| $\text{Est}(\sigma^2) = \frac{70}{69}\left(\frac{\Sigma x^2 f}{70} - 1.8^2\right)$ | M1 | oe attempted |
| $= 0.858$ or $\frac{296}{345}$ | A1 | Note: Final answer for var 0.846 (biased) and no working implies B1 for 286 |

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## Question 5(ii):

| $H_0: \mu = 1.9$, $H_1: \mu < 1.9$ | B1 | Or 'pop mean'; not just 'mean' |
|---|---|---|
| $\frac{1.8 - 1.9}{\sqrt{\frac{0.858}{70}}}$ | M1 | Standardise with their values from (i). Must have $\sqrt{70}$. No SD/Var mix |
| $= -0.903$ | A1 | Accept $\pm$ |
| $0.903 < 1.645$ | M1 | comp 1.645; allow comp 1.96 if $H_1: \mu \neq 1.9$; or comp $1 - \phi(0.903) = 0.182$ or $0.183$ with $0.05$ |
| No evidence that mean no. courts in S is less than in N | A1ft | No contradictions. ft their 0.903, but not comp 1.96. Accept cv method: $\text{cv} = 1.718$ M1A1 $1.718 < 1.8$ M1 conclusion A1 |

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## Question 5(iii):

| Type II because $H_0$ was not rejected | B1ft | ft their conclusion. If $H_0$ rejected, 'Type I because $H_0$ rejected' B1. No conclusion in (ii) scores B0 |
|---|---|---|

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5 The numbers of basketball courts in a random sample of 70 schools in South Mowland are summarised in the table.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of basketball courts & 0 & 1 & 2 & 3 & 4 & $> 4$ \\
\hline
Number of schools & 2 & 28 & 26 & 10 & 4 & 0 \\
\hline
\end{tabular}
\end{center}

(i) Calculate unbiased estimates for the population mean and variance of the number of basketball courts per school in South Mowland.\\

The mean number of basketball courts per school in North Mowland is 1.9 .\\
(ii) Test at the $5 \%$ significance level whether the mean number of basketball courts per school in South Mowland is less than the mean for North Mowland.\\

(iii) State, with a reason, which of the errors, Type I or Type II, might have been made in the test in part (ii).\\

\hfill \mbox{\textit{CAIE S2 2018 Q5 [10]}}