OCR MEI S1 — Question 1 18 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of binomial distributions
TypeExpected value and most likely value
DifficultyStandard +0.3 This is a straightforward S1 binomial hypothesis testing question covering standard bookwork: calculating probabilities, expectation, mode, and finding a critical region. All parts follow routine procedures with no novel problem-solving required, though the multi-part structure and critical region calculation place it slightly above average difficulty.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail

1 A multinational accountancy firm receives a large number of job applications from graduates each year. On average \(20 \%\) of applicants are successful. A researcher in the human resources department of the firm selects a random sample of 17 graduate applicants.
  1. Find the probability that at least 4 of the 17 applicants are successful.
  2. Find the expected number of successful applicants in the sample.
  3. Find the most likely number of successful applicants in the sample, justifying your answer. It is suggested that mathematics graduates are more likely to be successful than those from other fields. In order to test this suggestion, the researcher decides to select a new random sample of 17 mathematics graduate applicants. The researcher then carries out a hypothesis test at the \(5 \%\) significance level.
  4. (A) Write down suitable null and alternative hypotheses for the test.
    (B) Give a reason for your choice of the alternative hypothesis.
  5. Find the critical region for the test at the \(5 \%\) level, showing all of your calculations.
  6. Explain why the critical region found in part (v) would be unaltered if a \(10 \%\) significance level were used.

Question 1
(i)
\(X \sim B(17, 0.2)\)
\(P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.5489 = 0.4511\)
B1 for 0.5489
M1 for \(1 -\) their 0.5489
A1 CAO
(ii)
\(E(X) = np = 17 \times 0.2 = 3.4\)
M1 for product
A1 CAO
(iii)
\(P(X = 2) = 0.3096 - 0.1182 = 0.1914\)
\(P(X = 3) = 0.5489 - 0.3096 = 0.2393\)
\(P(X = 4) = 0.7582 - 0.5489 = 0.2093\)
So 3 applicants is most likely
B1 for 0.2393
B1 for 0.2093
A1 CAO dep on both B1s
(iv)
(A) Let \(p\) = probability of a randomly selected maths graduate applicant being successful (for population)
\(H_0: p = 0.2\)
\(H_1: p > 0.2\)
(B) \(H_1\) has this form as the suggestion is that mathematics graduates are more likely to be successful.
B1 for definition of \(p\) in context
B1 for \(H_0\)
B1 for \(H_1\)
E1
(v)
Let \(X \sim B(17, 0.2)\)
\(P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.8943 = 0.1057 > 5\%\)
\(P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.9623 = 0.0377 < 5\%\)
So critical region is \(\{7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17\}\)
B1 for 0.1057
B1 for 0.0377
M1 for at least one comparison with \(5\%\)
A1 CAO for critical region dep on M1 and at least one B1
(vi)
Because \(P(X \geq 6) = 0.1057 > 10\%\)
Either: comment that 6 is still outside the critical region
Or comparison \(P(X \geq 7) = 0.0377 < 10\%\)
E1
E1
# Question 1

## (i)
$X \sim B(17, 0.2)$

$P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.5489 = 0.4511$

B1 for 0.5489

M1 for $1 -$ their 0.5489

A1 CAO

## (ii)
$E(X) = np = 17 \times 0.2 = 3.4$

M1 for product

A1 CAO

## (iii)
$P(X = 2) = 0.3096 - 0.1182 = 0.1914$

$P(X = 3) = 0.5489 - 0.3096 = 0.2393$

$P(X = 4) = 0.7582 - 0.5489 = 0.2093$

So 3 applicants is most likely

B1 for 0.2393

B1 for 0.2093

A1 CAO dep on both B1s

## (iv)
(A) Let $p$ = probability of a randomly selected maths graduate applicant being successful (for population)

$H_0: p = 0.2$

$H_1: p > 0.2$

(B) $H_1$ has this form as the suggestion is that mathematics graduates are more likely to be successful.

B1 for definition of $p$ in context

B1 for $H_0$

B1 for $H_1$

E1

## (v)
Let $X \sim B(17, 0.2)$

$P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.8943 = 0.1057 > 5\%$

$P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.9623 = 0.0377 < 5\%$

So critical region is $\{7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17\}$

B1 for 0.1057

B1 for 0.0377

M1 for at least one comparison with $5\%$

A1 CAO for critical region dep on M1 and at least one B1

## (vi)
Because $P(X \geq 6) = 0.1057 > 10\%$

Either: comment that 6 is still outside the critical region

Or comparison $P(X \geq 7) = 0.0377 < 10\%$

E1

E1
1 A multinational accountancy firm receives a large number of job applications from graduates each year. On average $20 \%$ of applicants are successful.

A researcher in the human resources department of the firm selects a random sample of 17 graduate applicants.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that at least 4 of the 17 applicants are successful.
\item Find the expected number of successful applicants in the sample.
\item Find the most likely number of successful applicants in the sample, justifying your answer.

It is suggested that mathematics graduates are more likely to be successful than those from other fields. In order to test this suggestion, the researcher decides to select a new random sample of 17 mathematics graduate applicants. The researcher then carries out a hypothesis test at the $5 \%$ significance level.
\item (A) Write down suitable null and alternative hypotheses for the test.\\
(B) Give a reason for your choice of the alternative hypothesis.
\item Find the critical region for the test at the $5 \%$ level, showing all of your calculations.
\item Explain why the critical region found in part (v) would be unaltered if a $10 \%$ significance level were used.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI S1  Q1 [18]}}