AQA S3 2011 June — Question 5 8 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2011
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeConfidence interval for difference of proportions
DifficultyStandard +0.3 This is a standard confidence interval question for difference of proportions with straightforward calculations. Students need to state assumptions (random sampling, independence), calculate sample proportions (72/160 and 102/250), apply the normal approximation formula with z=2.576, and compute the interval. While it requires multiple steps and careful arithmetic, it follows a well-practiced procedure with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec5.05d Confidence intervals: using normal distribution
5
An examination of 160 e-mails received by Gopal showed that 72 had attachments. An examination of 250 e-mails received by Haley showed that 102 had attachments.
Stating two necessary assumptions about the selection of e-mails, construct an approximate \(99 \%\) confidence interval for the difference between the proportion of e-mails received by Gopal that have attachments and the proportion of e-mails received by Haley that have attachments.
(8 marks)
Question 5:
AnswerMarks Guidance
AnswerMark Guidance
Assumption 1: The e-mails are selected independently (from each other / between the two people)B1 Must be stated clearly
Assumption 2: Each e-mail is equally likely to be selected / random sampleB1 Must be stated clearly
\(\hat{p}_G = \frac{72}{160} = 0.45\), \(\hat{p}_H = \frac{102}{250} = 0.408\)B1 Both values correct
\(\hat{p}_G - \hat{p}_H = 0.45 - 0.408 = 0.042\)M1 Difference of proportions
\(SE = \sqrt{\frac{0.45 \times 0.55}{160} + \frac{0.408 \times 0.592}{250}}\)M1 Correct form of standard error using sample proportions
\(= \sqrt{0.001547 + 0.000966} = \sqrt{0.002513} = 0.05013\)A1 Correct value
\(99\%\) CI uses \(z = 2.576\)B1 Correct \(z\)-value
\(0.042 \pm 2.576 \times 0.05013\)M1 Correct structure of CI
\(= 0.042 \pm 0.1291\)
\((-0.0871,\ 0.1711)\) or \((-0.087,\ 0.171)\)A1 Accept equivalent forms, correct to 3 s.f. 
# Question 5:

| Answer | Mark | Guidance |
|--------|------|----------|
| Assumption 1: The e-mails are selected independently (from each other / between the two people) | B1 | Must be stated clearly |
| Assumption 2: Each e-mail is equally likely to be selected / random sample | B1 | Must be stated clearly |
| $\hat{p}_G = \frac{72}{160} = 0.45$, $\hat{p}_H = \frac{102}{250} = 0.408$ | B1 | Both values correct |
| $\hat{p}_G - \hat{p}_H = 0.45 - 0.408 = 0.042$ | M1 | Difference of proportions |
| $SE = \sqrt{\frac{0.45 \times 0.55}{160} + \frac{0.408 \times 0.592}{250}}$ | M1 | Correct form of standard error using sample proportions |
| $= \sqrt{0.001547 + 0.000966} = \sqrt{0.002513} = 0.05013$ | A1 | Correct value |
| $99\%$ CI uses $z = 2.576$ | B1 | Correct $z$-value |
| $0.042 \pm 2.576 \times 0.05013$ | M1 | Correct structure of CI |
| $= 0.042 \pm 0.1291$ | | |
| $(-0.0871,\ 0.1711)$ or $(-0.087,\ 0.171)$ | A1 | Accept equivalent forms, correct to 3 s.f. |

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5 \\
An examination of 160 e-mails received by Gopal showed that 72 had attachments. An examination of 250 e-mails received by Haley showed that 102 had attachments. \\
Stating two necessary assumptions about the selection of e-mails, construct an approximate $99 \%$ confidence interval for the difference between the proportion of e-mails received by Gopal that have attachments and the proportion of e-mails received by Haley that have attachments. \\
(8 marks) \\

\hfill \mbox{\textit{AQA S3 2011 Q5 [8]}}
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Q1 7 Q2 7 Q3 13 Q4 5 Q5 8 Q6 13 Q7 9 Q8 13