Moderate -0.3 This is a standard S2 hypothesis testing question following a routine template: state conditions, write hypotheses, apply continuity correction, and compare to critical value. While it requires multiple steps (np/nq check, standardization, z-table lookup), each step is procedural with no novel problem-solving or conceptual insight needed. Slightly easier than average due to its formulaic nature.
5. (a) State the conditions under which the normal distribution may be used as an approximation to the binomial distribution.
A company sells seeds and claims that 55\% of its pea seeds germinate.
(b) Write down a reason why the company should not justify their claim by testing all the pea seeds they produce.
To test the company's claim, a random sample of 220 pea seeds was planted.
(c) State the hypotheses for a two-tailed test of the company's claim.
Given that 135 of the 220 pea seeds germinated,
(d) use a normal approximation to test, at the \(5 \%\) level of significance, whether or not the company's claim is justified.
M1: attempting continuity correction (Method 1: \(135/85 \pm 0.5\); Method 2: \(x \pm 0.5\)). M1: standardising using their mean and SD, using 134.5, 135, 135.5, 85, 85.5 or 84.5. A1: correct \(z\) value awrt \(\pm 1.83\) or \(\pm\frac{134.5-121}{\sqrt{54.45}}\) (or equivalent)
\(= P(Z \geq 1.8295\ldots)\)
\(= 1 - 0.9664\)
\(= 0.0336/0.0337\) \(\quad x = 135.96\)
A1
awrt 0.0336/0.0337 or awrt 136 (allow 126 if one-tail test in (c)), or comparison of awrt 1.83 with 1.96 (1.6449)
Accept \(H_0\), not in CR, not significant
M1
A correct statement. Accept \(H_0\) oe if 2-tailed test in (c); reject \(H_0\) oe if 1-tailed test. Do not allow contradictions of non-contextual statements
The company's claim is justified or \(55\%\) of its pea seeds germinate
A1cso
A correct contextual statement including bold/underlined words for a 2-tailed test. This is not a follow-through mark
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $n$ is large and $p$ close to $0.5$ | B1B1 | B1: accept $n > 50$ (or any number bigger than 50). B1: $p$ close to 0.5. NB do not accept $np>5$, $nq>5$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| There would be no pea seeds left | B1 | Must have the idea of no peas left. Must mention either **pea** or **seeds** |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: p = 0.55 \quad H_1: p \neq 0.55$ | B1 | Both hypotheses correct. Must use $p$ or $\pi$ and 0.55 oe. Accept the hypotheses in part (d) |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim N(121, 54.45)$ | B1 | Correct mean and variance. May be seen in standardisation formula as 121 and $\sqrt{54.45}$ or 7.38 to 2dp or implied by correct answer |
| $P(X \geq 134.5) = P\!\left(Z \geq \frac{134.5-121}{\sqrt{54.45}}\right)$ or $\pm\frac{x-0.5-121}{\sqrt{54.45}} = 1.96$ | M1M1A1 | M1: attempting continuity correction (Method 1: $135/85 \pm 0.5$; Method 2: $x \pm 0.5$). M1: standardising using their mean and SD, using 134.5, 135, 135.5, 85, 85.5 or 84.5. A1: correct $z$ value awrt $\pm 1.83$ **or** $\pm\frac{134.5-121}{\sqrt{54.45}}$ (or equivalent) |
| $= P(Z \geq 1.8295\ldots)$ | | |
| $= 1 - 0.9664$ | | |
| $= 0.0336/0.0337$ $\quad x = 135.96$ | A1 | awrt 0.0336/0.0337 **or** awrt 136 (allow 126 if one-tail test in (c)), or comparison of awrt 1.83 with 1.96 (1.6449) |
| Accept $H_0$, not in CR, not significant | M1 | A correct statement. Accept $H_0$ oe if 2-tailed test in (c); reject $H_0$ oe if 1-tailed test. Do not allow contradictions of non-contextual statements |
| The **company's claim** is justified **or** $55\%$ of its pea **seeds germinate** | A1cso | A correct contextual statement including bold/underlined words for a 2-tailed test. This is not a follow-through mark |
5. (a) State the conditions under which the normal distribution may be used as an approximation to the binomial distribution.
A company sells seeds and claims that 55\% of its pea seeds germinate.\\
(b) Write down a reason why the company should not justify their claim by testing all the pea seeds they produce.
To test the company's claim, a random sample of 220 pea seeds was planted.\\
(c) State the hypotheses for a two-tailed test of the company's claim.
Given that 135 of the 220 pea seeds germinated,\\
(d) use a normal approximation to test, at the $5 \%$ level of significance, whether or not the company's claim is justified.\\
\hfill \mbox{\textit{Edexcel S2 2014 Q5 [11]}}