Moderate -0.8 This is a straightforward application of the standard confidence interval formula for a proportion with no complications. Students need only recall the formula p̂ ± 1.96√(p̂(1-p̂)/n), substitute the given values (33/120 = 0.275), and calculate. It requires no problem-solving or conceptual insight beyond basic formula recall, making it easier than average.
2 The manager of a video hire shop wishes to estimate the proportion of videos damaged by customers. He takes a random sample of 120 videos and finds that 33 of them are damaged. Find a \(95 \%\) confidence interval for the true proportion of videos that are being damaged when hired from this shop.
Calculation of correct form \(np \pm z\sqrt{npq}\) (accept just one side of interval); Division by 120 (BOTH sides); Use of 0.275; Correct answer
**EITHER**
$0.275 \pm 1.96 \times \frac{0.275 \times 0.725}{120}$
$0.195 < p < 0.355$ | M2, B1, A1 (4 marks) | Calculation of correct form $p \pm z\sqrt{\frac{pq}{n}}$ (SR M1 if only one side of interval seen); Use of $p = 0.275$; For correct answer
**OR**
$33 \pm 1.96 \sqrt{\frac{120 \times 0.275 \times 0.725}{120}}$
$23.413 < p < 42.586$
$0.195 < p < 0.355$ | M1, M1, B1, A1 (4 marks) | Calculation of correct form $np \pm z\sqrt{npq}$ (accept just one side of interval); Division by 120 (BOTH sides); Use of 0.275; Correct answer
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2 The manager of a video hire shop wishes to estimate the proportion of videos damaged by customers. He takes a random sample of 120 videos and finds that 33 of them are damaged. Find a $95 \%$ confidence interval for the true proportion of videos that are being damaged when hired from this shop.
\hfill \mbox{\textit{CAIE S2 2002 Q2 [4]}}