| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Calculate Type I error probability |
| Difficulty | Standard +0.3 This is a straightforward S4 hypothesis testing question requiring calculation of Type I error (significance level) using binomial probabilities. Part (a) and (c) involve direct application of P(X > k | Hâ‚€) using tables or formulas. Part (b) is algebraic manipulation of binomial expressions. The question is methodical but requires only standard techniques with no novel insight, making it slightly easier than average for S4 material. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| \(p\) | 0.10 | 0.15 | 0.20 | 0.25 |
| Power | 0.07 | \(s\) | 0.32 | 0.47 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(5, p)\); Size \(= P(\text{reject } H_0 \mid p = 0.05) = P(X > 1 \mid p = 0.05)\) | M1 | M1 for finding \(P(X>1)\) |
| \(= 1 - 0.9774 = 0.0226\) | A1 | A1 awrt 0.0226 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Power \(= 1 - P(0) - P(1)\) | M1 | |
| \(= 1-(1-p)^5 - 5(1-p)^4 p\) | M1 | M1 for \(1-(1-p)^5 - 5(1-p)^4p\) |
| \(= 1-(1-p)^4(1-p+5p) = 1-(1-p)^4(1+4p)\) | A1cso | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y \sim B(10, p)\); \(P(\text{Type I error}) = P(Y > 2 \mid p = 0.05)\) | M1 | M1 for finding \(P(Y>2)\) |
| \(= 1 - 0.9885 = 0.0115\) | A1 | A1 awrt 0.0115 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = 0.18\) | B1 | B1 0.18 cao (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Graph of power function for deputy's test plotted (blue curve, below black curve for small \(p\), crossing near \(p \approx 0.12\)) | B1ft | B1ft their value of \(s\) (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| i Intersection at \(p \approx 0.12\)–\(0.13\) ("their graphs intersection") | B1ft | |
| ii If \(p > 0.12\) the deputy's test is more powerful | B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| More powerful for \(p < 0.12\) and \(p\) unlikely to be above 0.12; it would cost more / take longer / more to sample | B1 | If give first statement, must suggest \(p\) unlikely to be above 0.12 (1) |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(5, p)$; Size $= P(\text{reject } H_0 \mid p = 0.05) = P(X > 1 \mid p = 0.05)$ | M1 | M1 for finding $P(X>1)$ |
| $= 1 - 0.9774 = 0.0226$ | A1 | A1 awrt 0.0226 (2) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Power $= 1 - P(0) - P(1)$ | M1 | |
| $= 1-(1-p)^5 - 5(1-p)^4 p$ | M1 | M1 for $1-(1-p)^5 - 5(1-p)^4p$ |
| $= 1-(1-p)^4(1-p+5p) = 1-(1-p)^4(1+4p)$ | A1cso | (3) |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y \sim B(10, p)$; $P(\text{Type I error}) = P(Y > 2 \mid p = 0.05)$ | M1 | M1 for finding $P(Y>2)$ |
| $= 1 - 0.9885 = 0.0115$ | A1 | A1 awrt 0.0115 (2) |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = 0.18$ | B1 | B1 0.18 cao (1) |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Graph of power function for deputy's test plotted (blue curve, below black curve for small $p$, crossing near $p \approx 0.12$) | B1ft | B1ft their value of $s$ (1) |
## Part (f)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **i** Intersection at $p \approx 0.12$–$0.13$ ("their graphs intersection") | B1ft | |
| **ii** If $p > 0.12$ the deputy's test is more powerful | B1 | (2) |
## Part (g)
| Answer/Working | Mark | Guidance |
|---|---|---|
| More powerful for $p < 0.12$ and $p$ unlikely to be above 0.12; it would cost more / take longer / more to sample | B1 | If give first statement, must suggest $p$ unlikely to be above 0.12 (1) |\begin{enumerate}
\item A manager in a sweet factory believes that the machines are working incorrectly and the proportion $p$ of underweight bags of sweets is more than $5 \%$. He decides to test this by randomly selecting a sample of 5 bags and recording the number $X$ that are underweight. The manager sets up the hypotheses $\mathrm { H } _ { 0 } : p = 0.05$ and $\mathrm { H } _ { 1 } : p > 0.05$ and rejects the null hypothesis if $x > 1$.\\
(a) Find the size of the test.\\
(b) Show that the power function of the test is
\end{enumerate}
$$1 - ( 1 - p ) ^ { 4 } ( 1 + 4 p )$$
The manager goes on holiday and his deputy checks the production by randomly selecting a sample of 10 bags of sweets. He rejects the hypothesis that $p = 0.05$ if more than 2 underweight bags are found in the sample.\\
(c) Find the probability of a Type I error using the deputy's test.
\section*{Question 3 continues on page 12}
The table below gives some values, to 2 decimal places, of the power function for the deputy's test.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$p$ & 0.10 & 0.15 & 0.20 & 0.25 \\
\hline
Power & 0.07 & $s$ & 0.32 & 0.47 \\
\hline
\end{tabular}
\end{center}
(d) Find the value of $s$.
The graph of the power function for the manager's test is shown in Figure 1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0bc6c296-9cbe-498b-89d9-c034b1b246e4-08_1157_1436_847_260}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
(e) On the same axes, draw the graph of the power function for the deputy's test.\\
(f) (i) State the value of $p$ where these graphs intersect.\\
(ii) Compare the effectiveness of the two tests if $p$ is greater than this value.
The deputy suggests that they should use his sampling method rather than the manager's.\\
(g) Give a reason why the manager might not agree to this change.
\hfill \mbox{\textit{Edexcel S4 2010 Q3 [12]}}