| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | One-tailed hypothesis test (upper tail, H₁: p > p₀) |
| Difficulty | Standard +0.3 This is a straightforward one-tailed binomial hypothesis test with standard structure: state hypotheses, find critical value or p-value, make conclusion. Part (b) is a simple calculation of mean and standard deviation for a binomial distribution. Both parts require only routine application of S2 formulas with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(H_0: p = 0.35\), \(H_1: p > 0.35\) | B1 | Both hypotheses correct (may use \(p\) or \(\pi\)); must have \(H_0\) and \(H_1\) |
| Probability route: \(P(X \geq 11) = 1 - P(X \leq 10) = 1 - 0.9468 = 0.0532\) CR route: \(P(X \leq 11) = 0.9804\), \(P(X \geq 12) = 0.0196\), CR: \(X \geq 12\) | M1 | For writing or using \(1-P(X \leq 10)\), or if leading to CR allow \(P(X \leq 11) =\) awrt 0.980; condone 0.98 or \(P(X \geq 12) =\) awrt 0.0196; NB M1 A1 for \(0.9468 < 0.95\) |
| \(= 0.0532\) / CR: \(X \geq 12\) | A1 | For 0.0532 or CR: \(X \geq 12\) oe |
| Do not reject \(H_0\) / not significant / 11 does not lie in the CR | dM1 | Dependent on 1st M1; correct statement; follow through their probability/CR and \(H_1\); must have critical region not just critical value; do not allow non-contextual conflicting statements |
| Hadi's belief is not supported / the proportion of customers paying by credit card is not greater than 35% | A1cso | Fully correct solution and correct contextual statement; underlined words required |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(X \sim \text{B}(20, 0.35)\); \([E(X)=] 7\) | B1 | \(E(X) = 7\) |
| \(\text{S.D.} = \sqrt{20 \times 0.35 \times 0.65} = \sqrt{4.55} = 2.133...\) | B1 | For correct expression for standard deviation |
| \("7" + 2 \times "2.133..."\) or \(11 - "7" < 2 \times "2.133..."\) or \(\frac{11-"7"}{\sqrt{"4.55"}}\) or \(11 - 2 \times "2.133..."\) | M1 | For using 'mean' \(+ 2 \times\) 'standard deviation' or \(11 -\) "their mean" \(< 2 \times\) 'standard deviation' or \(\frac{11 - \text{"their } E(X)\text{"}}{\text{"their sd"}}\); if \(E(X)\) and sd not given then allow \(5 < E(X) < 10\) and \(0 < \text{sd} < 5\) |
| \(11 < 11.266\) / \(4 < 4.266\) / \(1.875 < 2\) / \(6.734 < 7\) | A1cso | For comparison with 11 or 4 or 2; allow awrt (11.3 or 4.27 or 1.88 or 6.43) oe and no errors seen |
# Question 4:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $H_0: p = 0.35$, $H_1: p > 0.35$ | B1 | Both hypotheses correct (may use $p$ or $\pi$); must have $H_0$ and $H_1$ |
| **Probability route:** $P(X \geq 11) = 1 - P(X \leq 10) = 1 - 0.9468 = 0.0532$ **CR route:** $P(X \leq 11) = 0.9804$, $P(X \geq 12) = 0.0196$, CR: $X \geq 12$ | M1 | For writing or using $1-P(X \leq 10)$, or if leading to CR allow $P(X \leq 11) =$ awrt 0.980; condone 0.98 or $P(X \geq 12) =$ awrt 0.0196; NB M1 A1 for $0.9468 < 0.95$ |
| $= 0.0532$ / CR: $X \geq 12$ | A1 | For 0.0532 or CR: $X \geq 12$ oe |
| Do not reject $H_0$ / not significant / 11 does not lie in the CR | dM1 | Dependent on 1st M1; correct statement; follow through their probability/CR and $H_1$; must have critical region not just critical value; do not allow non-contextual conflicting statements |
| Hadi's belief is **not** supported / the proportion of customers paying by credit card is not greater than 35% | A1cso | Fully correct solution and correct contextual statement; underlined words required |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $X \sim \text{B}(20, 0.35)$; $[E(X)=] 7$ | B1 | $E(X) = 7$ |
| $\text{S.D.} = \sqrt{20 \times 0.35 \times 0.65} = \sqrt{4.55} = 2.133...$ | B1 | For correct expression for standard deviation |
| $"7" + 2 \times "2.133..."$ or $11 - "7" < 2 \times "2.133..."$ or $\frac{11-"7"}{\sqrt{"4.55"}}$ or $11 - 2 \times "2.133..."$ | M1 | For using 'mean' $+ 2 \times$ 'standard deviation' or $11 -$ "their mean" $< 2 \times$ 'standard deviation' or $\frac{11 - \text{"their } E(X)\text{"}}{\text{"their sd"}}$; if $E(X)$ and sd not given then allow $5 < E(X) < 10$ and $0 < \text{sd} < 5$ |
| $11 < 11.266$ / $4 < 4.266$ / $1.875 < 2$ / $6.734 < 7$ | A1cso | For comparison with 11 or 4 or 2; allow awrt (11.3 or 4.27 or 1.88 or 6.43) oe and no errors seen |
---\begin{enumerate}
\item At a shop, past figures show that $35 \%$ of customers pay by credit card. Following the shop's decision to no longer charge a fee for using a credit card, a random sample of 20 customers is taken and 11 are found to have paid by credit card.
\end{enumerate}
Hadi believes that the proportion of customers paying by credit card is now greater than 35\%\\
(a) Test Hadi's belief at the $5 \%$ level of significance. State your hypotheses clearly.
For a random sample of 20 customers,\\
(b) show that 11 lies less than 2 standard deviations above the mean number of customers paying by credit card.\\
You may assume that $35 \%$ is the true proportion of customers who pay by credit card.\\
\hfill \mbox{\textit{Edexcel S2 2019 Q4 [9]}}