CAIE S2 2019 June — Question 5 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeCI from raw data list
DifficultyChallenging +1.2 This question requires understanding of hypothesis testing for a normal distribution with known variance, calculating a test statistic, and working backwards from a non-rejection decision to find the critical significance level. While it involves multiple steps (calculating sample mean, finding z-statistic, determining critical values), the concepts are standard S2 material and the reverse-engineering aspect adds modest problem-solving demand beyond routine hypothesis test questions.
Spec5.05c Hypothesis test: normal distribution for population mean
5 The amount of money, in dollars, spent by a customer on one visit to a certain shop is modelled by the distribution \(\mathrm { N } ( \mu , 1.94 )\). In the past, the value of \(\mu\) has been found to be 20.00 , but following a rearrangement in the shop, the manager suspects that the value of \(\mu\) has changed. He takes a random sample of 6 customers and notes how much they each spend, in dollars. The results are as follows.
15.50
17.60
17.30
22.00
23.50
31.00 The manager carries out a hypothesis test using a significance level of \(\alpha \%\). The test does not support his suspicion. Find the largest possible value of \(\alpha\).
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0\): Pop mean \(= 20\); \(H_1\): Pop mean \(\neq 20\)B1 Accept \(\mu\)
\(\frac{\sum x}{6} \left(= \frac{126.9}{6} = 21.15\right)\)M1 Attempted or 126.9 and 11.64 attempted
\(\frac{\text{"21.15"} - 20}{\sqrt{\frac{1.94}{6}}}\)M1 Must have \(\sqrt{6}\); or \(\frac{120 - 126.9}{\sqrt{11.64}}\) no mixed method
\(= 2.022\)A1
\(2(1 - \phi(\text{"2.022"})) \ ) \ 2(1 - \text{"0.9784"}) = 0.0432\)M1 FT \(2 \times (1 - \text{"}.9784\text{"})\)
\(\alpha = 4.32\) (3 sf)A1 FT Allow 4.3 or 4, if correct working seen, or clearly implied, as far as 0.0216; FT their \(z\), no error seen; One-tail test scores maximum 3/6
Total: 6 
## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean $= 20$; $H_1$: Pop mean $\neq 20$ | B1 | Accept $\mu$ |
| $\frac{\sum x}{6} \left(= \frac{126.9}{6} = 21.15\right)$ | M1 | Attempted or 126.9 and 11.64 attempted |
| $\frac{\text{"21.15"} - 20}{\sqrt{\frac{1.94}{6}}}$ | M1 | Must have $\sqrt{6}$; or $\frac{120 - 126.9}{\sqrt{11.64}}$ no mixed method |
| $= 2.022$ | A1 | |
| $2(1 - \phi(\text{"2.022"})) \ ) \ 2(1 - \text{"0.9784"}) = 0.0432$ | M1 | **FT** $2 \times (1 - \text{"}.9784\text{"})$ |
| $\alpha = 4.32$ (3 sf) | A1 | **FT** Allow 4.3 or 4, if correct working seen, or clearly implied, as far as 0.0216; **FT** their $z$, no error seen; One-tail test scores maximum 3/6 |
| **Total: 6** | | |

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5 The amount of money, in dollars, spent by a customer on one visit to a certain shop is modelled by the distribution $\mathrm { N } ( \mu , 1.94 )$. In the past, the value of $\mu$ has been found to be 20.00 , but following a rearrangement in the shop, the manager suspects that the value of $\mu$ has changed. He takes a random sample of 6 customers and notes how much they each spend, in dollars. The results are as follows.\\
15.50\\
17.60\\
17.30\\
22.00\\
23.50\\
31.00

The manager carries out a hypothesis test using a significance level of $\alpha \%$. The test does not support his suspicion. Find the largest possible value of $\alpha$.\\

\hfill \mbox{\textit{CAIE S2 2019 Q5 [6]}}
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