OCR MEI S1 2006 January — Question 3 8 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2006
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeOne-tailed hypothesis test (lower tail, H₁: p < p₀)
DifficultyModerate -0.3 This is a straightforward application of binomial hypothesis testing with clear structure: part (i) is routine probability calculation using binomial tables, and part (ii) is a standard one-tailed test where the method is prescribed and the critical value comparison is straightforward. The question requires no novel insight—just following the standard hypothesis testing procedure taught in S1.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion

3 Over a long period of time, 20\% of all bowls made by a particular manufacturer are imperfect and cannot be sold.
  1. Find the probability that fewer than 4 bowls from a random sample of 10 made by the manufacturer are imperfect. The manufacturer introduces a new process for producing bowls. To test whether there has been an improvement, each of a random sample of 20 bowls made by the new process is examined. From this sample, 2 bowls are found to be imperfect.
  2. Show that this does not provide evidence, at the \(5 \%\) level of significance, of a reduction in the proportion of imperfect bowls. You should show your hypotheses and calculations clearly.

Question 3:
(i)
\(X \sim B(10, 0.2)\)
AnswerMarks Guidance
\(P(X < 4) = P(X \leq 3)\)M1
\(= 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791\)A1 Accept 0.879
(ii)
AnswerMarks Guidance
\(H_0: p = 0.2\), \(H_1: p < 0.2\)B1 Both hypotheses correct
\(X \sim B(20, 0.2)\) under \(H_0\)
AnswerMarks Guidance
\(P(X \leq 2) = 0.0692 + 0.2182 \times ...\)M1 Correct method
\(P(X \leq 2) = 0.2061\)A1
\(0.2061 > 0.05\), so not significantM1 Correct comparison
No evidence at 5% level of a reduction in proportion of imperfect bowlsA1 A1 Conclusion in context
# Question 3:

**(i)**
$X \sim B(10, 0.2)$
$P(X < 4) = P(X \leq 3)$ | M1 | 
$= 0.1074 + 0.2684 + 0.3020 + 0.2013 = 0.8791$ | A1 | Accept 0.879

**(ii)**
$H_0: p = 0.2$, $H_1: p < 0.2$ | B1 | Both hypotheses correct
$X \sim B(20, 0.2)$ under $H_0$
$P(X \leq 2) = 0.0692 + 0.2182 \times ...$ | M1 | Correct method
$P(X \leq 2) = 0.2061$ | A1 | 
$0.2061 > 0.05$, so not significant | M1 | Correct comparison
No evidence at 5% level of a reduction in proportion of imperfect bowls | A1 A1 | Conclusion in context

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3 Over a long period of time, 20\% of all bowls made by a particular manufacturer are imperfect and cannot be sold.\\
(i) Find the probability that fewer than 4 bowls from a random sample of 10 made by the manufacturer are imperfect.

The manufacturer introduces a new process for producing bowls. To test whether there has been an improvement, each of a random sample of 20 bowls made by the new process is examined. From this sample, 2 bowls are found to be imperfect.\\
(ii) Show that this does not provide evidence, at the $5 \%$ level of significance, of a reduction in the proportion of imperfect bowls. You should show your hypotheses and calculations clearly.

\hfill \mbox{\textit{OCR MEI S1 2006 Q3 [8]}}