FS1-4 Hypothesis testing — revision question pack

2 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FS1-4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

FS1-4.1 · Extend ideas of hypothesis tests to test for the mean of a Poisson distribution.

Explanation

A Poisson-mean test states $H_0$ and $H_1$ in terms of the population parameter $\lambda$ or $\mu$ , not the observed count. Under $H_0$ , scale the mean to the complete observation period; independent periods may be combined into one Poisson total.
The direction of $H_1$ selects the tail.
A critical region is the most extreme attainable tail whose probability under $H_0$ does not exceed the significance level, and that probability is the size of the test.
A $p$ -value is the null probability of an outcome at least as extreme as observed.
Examiners expect a contextual conclusion: reject $H_0$ for evidence supporting $H_1$ , or do not reject it for insufficient evidence.

Worked example

A process has claimed mean $3$ faults per day. Over $5$ independent days, $24$ faults occur. At the $5\%$ level, test whether the daily mean has increased, given that $P(\operatorname{Po}(15)\geq23)=0.0327$ and $P(\operatorname{Po}(15)\geq22)=0.0531$ .

1.Let $\lambda$ be the daily population mean: $H_0:\lambda=3$ , $H_1:\lambda>3$ .
2.Under $H_0$ , the total $X\sim\operatorname{Po}(15)$ .
3.The critical region is $X\geq23$ , with size $0.0327$ .
4.Since $24$ is critical, reject $H_0$ .

Answer: There is sufficient evidence at the $5\%$ level that the daily mean fault rate has increased.

Common mistakes

Don't state hypotheses using the sample total rather than the population mean parameter.
Don't use the daily mean as the Poisson parameter for a multi-day total.
Don't say that failing to reject $H_0$ proves the null hypothesis.

Exam tip

A full test conclusion must name the significance level, evidence and contextual parameter direction.

Tier 1 · Easy

1.
A manager claims that the mean number of alerts per shift is $4.2$ . State hypotheses to test whether the mean has increased, defining your parameter.
(2)
(Total for Question 1 is 2 marks)
2.
The count $X$ has distribution $\operatorname{Po}(\lambda)$ . A test at the $5\%$ level of significance of whether the population mean has increased from $3$ uses the critical region $X\geq7$ . State the hypotheses, find and interpret the size of the test, and give the conclusion when $X=8$ is observed.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Under $H_0$ , defects occur at a mean rate of $2.5$ per hour. Eight independent hours give a total of $28$ defects. Test at a $5\%$ significance level whether the mean rate has increased. State the critical region and the size of the test.
(7)
(Total for Question 1 is 7 marks)
2.
A process is observed for $1.5$ , $2$ , $0.75$ and $1.25$ hours, producing counts $6,7,3,4$ . Let $\lambda$ be its population mean count per hour. Test $H_0:\lambda=2$ against $H_1:\lambda>2$ at the $5\%$ level of significance by comparing $P(X\geq\text{observed})$ with $0.05$ . State an assumption needed to pool the four periods.
(6)
(Total for Question 2 is 6 marks)
3.
The number of blemishes on a ceramic tile is modelled by a Poisson distribution. Let $\lambda$ denote an appropriate population parameter. Six independent tiles have blemish counts $1,0,1,0,0,1$ . Test whether the mean has decreased from $1.4$ at the $5\%$ level of significance using the supplied critical region $T\leq3$ . Define $\lambda$ , state $H_0$ and $H_1$ , show that the size of this test is less than $0.05$ , and state your conclusion in context.
(7)
(Total for Question 3 is 7 marks)

Tier 3 · Hard

1.
A Poisson model has mean $1.8$ events per batch. Twelve independent batches produce $14$ events in total. Carry out a test at the $5\%$ significance level of whether the mean per batch has decreased, giving the critical region, its size and your conclusion.
(8)
(Total for Question 1 is 8 marks)
2.
A total of $17$ events is recorded over a continuous period of $4.5$ hours. Test whether the population hourly mean $\lambda$ differs from $2.2$ at the $5\%$ level of significance by placing at most $2.5\%$ in each tail. Find the exact discrete critical region, the size of the test and the conclusion.
(8)
(Total for Question 2 is 8 marks)
3.
Pinholes are modelled by a Poisson process with a common constant rate along the sheet material. Counts on disjoint lengths are independent. Four lengths measure $1.2$ , $1.8$ , $2.5$ and $1.5$ metres, with respective counts $2,4,7,9$ . Let $\lambda$ denote an appropriate population parameter. At the $5\%$ level of significance, test whether the mean differs from $1.75$ using the given critical region $T\leq5$ or $T\geq21$ for the total count $T$ . Define $\lambda$ , state $H_0$ and $H_1$ , calculate the size of the test and state your conclusion in context.
(8)
(Total for Question 3 is 8 marks)
4.
Customer counts in independent one-hour periods are Poisson variables with population mean $\lambda$ per hour. Two periods give counts $3$ and $6$ . At significance level $\alpha$ , an upper-tail test of $H_0:\lambda=1.7$ against $H_1:\lambda>1.7$ uses the critical region $T\geq c$ , where $c$ is the smallest integer for which $P(T\geq c\mid H_0)\leq\alpha$ . Find the set of values of $\alpha$ for which the critical region is $T\geq8$ . Find the power when $\lambda=2.6$ and state the conclusion for the observed counts. Give probabilities to $4$ significant figures.
(8)
(Total for Question 4 is 8 marks)
5.
Genuine signals arrive as a Poisson process with population mean rate $\lambda$ per hour. Independently, false alerts arrive as a Poisson process with known mean rate $0.4$ per hour. During $6$ hours, $18$ alerts are recorded in total. At the $5\%$ level of significance, test $H_0:\lambda=1.5$ against $H_1:\lambda>1.5$ . Find the critical region and the size of the test, and state the conclusion in context. Give probabilities to $4$ significant figures.
(8)
(Total for Question 5 is 8 marks)

FS1-4.2 · Extend hypothesis testing to test for the parameter p of a geometric distribution.

Explanation

A geometric-parameter test states hypotheses in terms of the success probability $p$ . A larger $p$ tends to produce earlier success, so small waiting times support $H_1:p>p_0$ and large waiting times support $H_1:p<p_0$ .
The sum of $r$ independent geometric variables with common $p$ has a negative binomial distribution counting the trial of the $r$ th success.
Use this distribution to find a discrete critical region whose null probability does not exceed the significance level.
The size of the test may be below the nominal level.
Examiners expect the tail to be justified from the effect of changing $p$ and conclusions to refer to evidence about the population success probability.

Worked example

Four independent geometric waiting times have common parameter $p$ . Their sum is $S$ . Test $H_0:p=0.5$ against $H_1:p<0.5$ . Explain which tail is critical and interpret a result in that tail.

1.A smaller $p$ makes successes less frequent and waiting times larger.
2.Under $H_0$ , $S$ is negative binomial with $r=4$ , $p=0.5$ .
3.Therefore the critical region has form $S\geq c$ , where $c$ is the least integer making $P(S\geq c)\leq\alpha$ .

Answer: Use the upper tail; a critical result gives evidence that the success probability is below $0.5$ .

Common mistakes

Don't use the lower waiting-time tail for the alternative $p<p_0$ .
Don't model a sum of geometric waiting times as another geometric variable.
Don't state hypotheses using the observed sum $S$ rather than the parameter $p$ .

Exam tip

Before calculating a critical value, write one sentence linking larger or smaller $p$ to waiting-time direction.

Tier 1 · Easy

1.
A geometric model uses $p=0.35$ . State hypotheses for testing whether the probability of success has fallen, and state which tail of the waiting time is critical.
(3)
(Total for Question 1 is 3 marks)
2.
One geometric observation $X$ records the trial of the first success. Test $H_0:p=0.08$ against $H_1:p>0.08$ . The level of significance is $10\%$ . Determine the critical region and its size, then conclude when the observed stopping time is $X=1$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Six independent geometric observations count trials to first success. Test $H_0:p=0.3$ against $H_1:p>0.3$ at the $5\%$ level of significance using their sum $S$ . Find the critical region and its size, then state the conclusion when $S=9$ .
(7)
(Total for Question 1 is 7 marks)
2.
One geometric observation $X$ records the trial of the first success. Test $H_0:p=0.12$ against $H_1:p<0.12$ at the $5\%$ level of significance. Find the least integer $c$ for which $X\geq c$ is a critical region, state the size of the test exactly, and conclude when $X=27$ is observed.
(6)
(Total for Question 2 is 6 marks)
3.
Three independent scanners each repeat independent attempts until they read a label successfully. Their stopping times are $5$ , $6$ and $9$ , and each attempt has the same population success probability $p$ . At the $5\%$ level of significance, test whether $p$ is below $0.30$ using the supplied critical region $S\geq20$ for the sum $S$ . State $H_0$ and $H_1$ , find the size of the test, and state your conclusion in context.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Five independent geometric observations have common parameter $p$ . Use their sum $S$ to test $H_0:p=0.4$ against $H_1:p<0.4$ at the $5\%$ level. Determine the critical region and its size, then carry out the test and state your conclusion when $S=19$ .
(8)
(Total for Question 1 is 8 marks)
2.
Eight independent sensors have common response probability $p$ on each trial, and their geometric stopping times are $2,3,3,4,5,6,9,11$ . Use the sum $S$ as the test statistic for $H_0:p=0.25$ against $H_1:p\ne0.25$ at the $10\%$ level. Allocate no more than $5\%$ of the null probability to either tail; determine both rejection tails and the overall size, then use the data to state your conclusion in context.
(8)
(Total for Question 2 is 8 marks)
3.
Seven mutually independent calibration checks each repeat independent trials, with common pass probability $p$ on every trial, until the first pass. Their stopping times are $2,3,4,5,6,7,9$ . At the $10\%$ level of significance, test whether $p$ differs from $0.32$ using the given critical region $S\leq11$ or $S\geq35$ for the sum $S$ . Each rejection tail was allocated at most $5\%$ of the null probability. State $H_0$ and $H_1$ , find the size of the test, explain why the size of the test is below $10\%$ , state one assumption needed to model the seven stopping times as independent geometric variables with a common $p$ , and state your conclusion in context.
(8)
(Total for Question 3 is 8 marks)
4.
Each of $r$ independent inspection devices repeats independent attempts until its first successful recognition, with the same population success probability $p$ . Let $S$ be the sum of the $r$ stopping times. A test of $H_0:p=0.45$ against $H_1:p>0.45$ rejects $H_0$ when $S\leq r+1$ . At the $5\%$ level of significance, find the least possible value of $r$ and the power when $p=0.65$ , giving the power to $4$ decimal places. For the chosen value of $r$ , an observed sum is $7$ ; state the conclusion in context.
(8)
(Total for Question 4 is 8 marks)
5.
A recognition time $X$ is geometric with population success probability $p$ . At the $10\%$ level of significance, a test of $H_0:p=p_0$ against $H_1:p>p_0$ rejects $H_0$ when $X\leq2$ . The size of the test is $0.0975$ . Find $p_0$ . An observed recognition time is $2$ ; state the conclusion in context. Find the power of the test and the probability of a Type II error when $p=0.2$ .
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FS1-4.1 · Extend ideas of hypothesis tests to test for the mean of a Poisson distribution.

Tier 1 · Easy

Mark scheme for FS1-4.1 Tier 1 · Easy
Question	Scheme	Marks
1	Let $\lambda$ be the population mean number of alerts per shift $H_0:\lambda=4.2$ ; $H_1:\lambda>4.2$	2
(2 marks)		2
Notes
The claim supplies the null value. The word 'increased' makes the alternative upper-tailed, so use $H_1:\lambda>4.2$ .
2	$H_0:\lambda=3$ ; $H_1:\lambda>3$ The size of the test is $P(X\geq7)=0.0335$ to $4$ decimal places If $H_0$ is true, there is a $3.35\%$ chance of rejecting it Since $8$ is in the critical region, reject $H_0$ ; there is evidence that the mean has increased	4
(4 marks)		4
Notes
An increase gives $H_0:\lambda=3$ against $H_1:\lambda>3$ . Under $H_0$ , the supplied region has probability $P(X\geq7)=1-P(X\leq6)=0.0335085353\ldots$ , which is the chance of rejecting a true null hypothesis. The observation $8$ is critical, so reject $H_0$ in favour of an increased mean.

Tier 2 · Standard

Mark scheme for FS1-4.1 Tier 2 · Standard
Question	Scheme	Marks
1	$H_0:\lambda=2.5$ ; $H_1:\lambda>2.5$ Critical region $X\geq29$ Size $=0.03433$ Do not reject $H_0$ ; there is insufficient evidence that the mean rate has increased	7
(7 marks)		7
Notes
Under $H_0$ , the eight-hour total $X\sim\operatorname{Po}(20)$ . Calculator tails give $P(X\geq29)=0.0343335\ldots\leq0.05$ , while $P(X\geq28)=0.0524807\ldots>0.05$ . Thus the critical region is $X\geq29$ and its size is $0.03433$ . Since $28$ is not in the critical region, do not reject $H_0$ .
2	The total exposure is $5.5$ hours The observed total is $20$ Under $H_0$ , $X\sim\operatorname{Po}(11)$ Assume the counts in the four periods are independent and share the same constant hourly mean $P(X\geq20)=0.0093$ to $4$ decimal places, which is less than $0.05$ Reject $H_0$ ; there is evidence that the hourly mean exceeds $2$	6
(6 marks)		6
Notes
The exposure is $1.5+2+0.75+1.25=5.5$ hours and the observed total is $6+7+3+4=20$ . Assuming independent counts and a common constant hourly mean allows the four periods to be pooled. Under $H_0$ , the total $X\sim\operatorname{Po}(5.5\times2)=\operatorname{Po}(11)$ . The upper-tail probability is $P(X\geq20)=0.0092894579\ldots=0.0093$ to $4$ decimal places. Since $0.0093<0.05$ , reject $H_0$ ; there is evidence that the hourly mean exceeds $2$ .
3	$\lambda$ is the population mean number of blemishes per tile $H_0:\lambda=1.4$ ; $H_1:\lambda<1.4$ Under $H_0$ , $T\sim\operatorname{Po}(8.4)$ The size of the supplied critical region is $P(T\leq3)=0.0322604<0.05$ The observed total is $T=3$ Since $T\leq3$ , reject $H_0$ There is evidence at the $5\%$ level that the population mean number of blemishes per tile has decreased	7
(7 marks)		7
Notes
The alternative is lower-tailed because a decrease is being tested. Under $H_0$ , add the six independent tile counts to get $T\sim\operatorname{Po}(6\times1.4)=\operatorname{Po}(8.4)$ . The size of the supplied critical region is $P(T\leq3)=0.0322603658\ldots<0.05$ . The data give $T=3$ , which lies in the supplied critical region. Therefore reject $H_0$ ; there is evidence at the $5\%$ level that the mean blemish count per tile has decreased.

Tier 3 · Hard

Mark scheme for FS1-4.1 Tier 3 · Hard
Question	Scheme	Marks
1	$H_0:\lambda=1.8$ $H_1:\lambda<1.8$ Under $H_0$ , the total $X\sim\operatorname{Po}(21.6)$ A decrease requires a lower-tail critical region $P(X\leq13)=0.03329\leq0.05$ $P(X\leq14)=0.05626>0.05$ , so the critical region is $X\leq13$ and its size is $0.03329$ The observed total $14$ is not critical, so do not reject $H_0$ There is insufficient evidence that the mean per batch has decreased	8
(8 marks)		8
Notes
Under $H_0$ , the total $X\sim\operatorname{Po}(12\times1.8)=\operatorname{Po}(21.6)$ . A decrease requires the lower tail. The adjacent probabilities are $P(X\leq13)=0.0332864\ldots\leq0.05$ and $P(X\leq14)=0.0562576\ldots>0.05$ , so the $5\%$ critical region is $X\leq13$ with size $0.03329$ . The observed total $14$ is not critical, so do not reject $H_0$ ; there is insufficient evidence of a decrease.
2	$H_0:\lambda=2.2$ ; $H_1:\lambda\ne2.2$ Under $H_0$ , $X\sim\operatorname{Po}(9.9)$ The observed total is $17$ The lower critical tail is $X\leq3$ The upper critical tail is $X\geq17$ The size of the test is $0.0361$ to $4$ decimal places Reject $H_0$ There is evidence that the hourly mean differs from $2.2$	8
(8 marks)		8
Notes
The observation period is $4.5$ hours and the observed total is $17$ . Under $H_0$ , $X\sim\operatorname{Po}(4.5\times2.2)=\operatorname{Po}(9.9)$ . In the lower tail, $P(X\leq3)=0.0111197883\ldots$ but $P(X\leq4)=0.0312021214\ldots>0.025$ . In the upper tail, $P(X\geq17)=0.0249361000\ldots$ but $P(X\geq16)=0.0453548029\ldots>0.025$ . Thus the critical region is $X\leq3$ or $X\geq17$ , and the size of the test is $0.0360558883\ldots=0.0361$ . Since $17$ is critical, reject $H_0$ .
3	$\lambda$ is the population mean number of pinholes per metre, with $H_0:\lambda=1.75$ and $H_1:\lambda\ne1.75$ The total length is $7$ metres The observed total is $T=22$ Under $H_0$ , $T\sim\operatorname{Po}(12.25)$ $P(T\leq5)=0.01738$ to $4$ significant figures $P(T\geq21)=0.01423$ to $4$ significant figures The size of the test is $0.03161$ to $4$ significant figures Since $22$ is critical, reject $H_0$ ; there is evidence at the $5\%$ level that the population mean number of pinholes per metre differs from $1.75$	8
(8 marks)		8
Notes
The total exposure is $1.2+1.8+2.5+1.5=7$ metres and the observed total is $2+4+7+9=22$ . Under $H_0$ , independence and a common constant rate give $T\sim\operatorname{Po}(7\times1.75)=\operatorname{Po}(12.25)$ . The two probabilities in the given region are $P(T\leq5)=0.0173782531\ldots$ and $P(T\geq21)=0.0142277193\ldots$ , so the size is $0.0316059724\ldots$ . The observed total lies in the upper part of the critical region, so reject $H_0$ ; there is evidence that the mean pinhole rate differs from $1.75$ per metre.
4	The observed total is $T=9$ Under $H_0$ , $T\sim\operatorname{Po}(3.4)$ $P(T\geq8\mid H_0)=0.02307$ to $4$ significant figures $P(T\geq7\mid H_0)=0.05785$ to $4$ significant figures The critical region is $T\geq8$ when $0.02307\ldots\leq\alpha<0.05785\ldots$ When $\lambda=2.6$ , $T\sim\operatorname{Po}(5.2)$ and the power is $P(T\geq8)=0.1551$ to $4$ significant figures The observed total $9$ is critical, so reject $H_0$ There is evidence at significance level $\alpha$ in the stated set that the population mean customer count exceeds $1.7$ per hour	8
(8 marks)		8
Notes
The observed total is $3+6=9$ . Under $H_0$ , adding the two independent counts gives $T\sim\operatorname{Po}(3.4)$ . The adjacent upper-tail probabilities are $P(T\geq8)=0.0230739361\ldots$ and $P(T\geq7)=0.0578532207\ldots$ . Therefore $8$ is the smallest critical boundary precisely when $0.0230739361\ldots\leq\alpha<0.0578532207\ldots$ . If $\lambda=2.6$ , then $T\sim\operatorname{Po}(5.2)$ and the rejection probability is $P(T\geq8)=0.1550784100\ldots$ . Since the observed total is $9$ , it lies in the critical region; there is evidence that the population mean customer count exceeds $1.7$ per hour.
5	Under $H_0$ , the genuine-signal and false-alert counts have distributions $\operatorname{Po}(9)$ and $\operatorname{Po}(2.4)$ respectively Under $H_0$ , the total $T\sim\operatorname{Po}(11.4)$ $P(T\geq18)=0.04276$ to $4$ significant figures $P(T\geq17)=0.07196$ to $4$ significant figures The critical region is $T\geq18$ The size of the test is $0.04276$ to $4$ significant figures The observed total $18$ is critical, so reject $H_0$ There is evidence at the $5\%$ level that the population mean rate of genuine signals is greater than $1.5$ per hour	8
(8 marks)		8
Notes
Under $H_0$ , the independent genuine-signal and false-alert counts over $6$ hours have parameters $6(1.5)=9$ and $6(0.4)=2.4$ . Their total is therefore $T\sim\operatorname{Po}(11.4)$ . The upper-tail probabilities are $P(T\geq18)=0.0427583\ldots$ and $P(T\geq17)=0.0719565\ldots$ , so the largest rejection region with size at most $0.05$ is $T\geq18$ and its size is $0.04276$ to $4$ significant figures. Since the observed total is $18$ , reject $H_0$ ; there is evidence that the population mean genuine-signal rate exceeds $1.5$ per hour.

FS1-4.2 · Extend hypothesis testing to test for the parameter p of a geometric distribution.

Tier 1 · Easy

Mark scheme for FS1-4.2 Tier 1 · Easy
Question	Scheme	Marks
1	$H_0:p=0.35$ ; $H_1:p<0.35$ Large waiting times form the critical tail	3
(3 marks)		3
Notes
A fall gives the lower-tailed parameter alternative $p<0.35$ . Smaller success probabilities make the first success take longer, so evidence lies in the upper tail of the geometric variable.
2	The critical region is $X=1$ Its size is $0.08$ Since $X=1$ is critical, reject $H_0$ ; there is evidence that the success probability exceeds $0.08$	3
(3 marks)		3
Notes
A larger success probability shortens the wait, so use the lower tail. Under $H_0$ , $P(X\leq1)=0.08\leq0.10$ , whereas $P(X\leq2)=1-0.92^2=0.1536>0.10$ . Hence the exact critical region is $X=1$ and its size is $0.08$ . The observed value is critical, so reject $H_0$ in favour of $p>0.08$ .

Tier 2 · Standard

Mark scheme for FS1-4.2 Tier 2 · Standard
Question	Scheme	Marks
1	Critical region $S\leq10$ Size $=0.04735$ Reject $H_0$ ; there is evidence that $p>0.3$	7
(7 marks)		7
Notes
Under $H_0$ , $S$ is negative binomial with $r=6$ and $p=0.3$ . Since larger $p$ gives smaller sums, use the lower tail. Calculator probabilities give $P(S\leq10)=0.0473490\ldots$ and $P(S\leq11)=0.0782248\ldots$ , so the critical region is $S\leq10$ with size $0.04735$ . Since $9\leq10$ , reject $H_0$ .
2	Smaller $p$ makes large values of $X$ critical, and $P(X\geq c)=0.88^{c-1}$ under $H_0$ Solve $0.88^{c-1}\leq0.05$ $0.88^{23}=0.0528569\ldots>0.05$ but $0.88^{24}=0.0465140\ldots\leq0.05$ , so $c=25$ The critical region is $X\geq25$ The size of the test is exactly $0.88^{24}$ Since $27$ is critical, reject $H_0$ ; there is evidence that $p<0.12$	6
(6 marks)		6
Notes
A smaller success probability increases the waiting time, so use the upper tail. Under $H_0$ , $P(X\geq c)=(1-0.12)^{c-1}=0.88^{c-1}$ . Since $0.88^{23}=0.0528568721\ldots>0.05$ but $0.88^{24}=0.0465140475\ldots\leq0.05$ , the least possible threshold is $c=25$ . Thus the critical region is $X\geq25$ and the exact size of the test is $0.88^{24}$ . The observation $27$ is critical, so reject $H_0$ in favour of $p<0.12$ .
3	$H_0:p=0.30$ ; $H_1:p<0.30$ Under $H_0$ , $S$ is negative binomial with $r=3$ and $p=0.30$ $P(S\geq20)=0.04622$ to $4$ significant figures The observed sum is $S=5+6+9=20$ Since $20$ is in the supplied critical region, reject $H_0$ There is evidence at the $5\%$ level that the scanners' population success probability is below $0.30$	6
(6 marks)		6
Notes
A smaller $p$ produces longer waits, so the upper tail supports $H_1$ . Under $H_0$ , the sum of three independent geometric stopping times is negative binomial with $r=3$ and $p=0.30$ . Hence the size of the supplied region is $P(S\geq20)=\sum_{s=20}^{\infty}\binom{s-1}{2}(0.30)^3(0.70)^{s-3}=0.0462236831\ldots$ . The observed sum is $5+6+9=20$ , which is critical. Therefore reject $H_0$ ; there is evidence at the $5\%$ level that the scanners' population success probability is below $0.30$ .

Tier 3 · Hard

Mark scheme for FS1-4.2 Tier 3 · Hard
Question	Scheme	Marks
1	$H_0:p=0.4$ $H_1:p<0.4$ Under $H_0$ , $S$ is negative binomial with $r=5$ and $p=0.4$ A smaller $p$ requires an upper-tail critical region $P(S\geq22)=0.03696\leq0.05$ $P(S\geq21)=0.05095>0.05$ , so the critical region is $S\geq22$ and its size is $0.03696$ The observed sum $19$ is not critical, so do not reject $H_0$ There is insufficient evidence that $p<0.4$	8
(8 marks)		8
Notes
Under $H_0$ , $S$ is negative binomial with $r=5$ and $p=0.4$ . A smaller $p$ makes larger sums more likely, so use the upper tail. The adjacent probabilities are $P(S\geq22)=0.0369556\ldots\leq0.05$ and $P(S\geq21)=0.0509520\ldots>0.05$ , so the critical region is $S\geq22$ with size $0.03696$ . Since the observed sum $19$ is not critical, do not reject $H_0$ ; there is insufficient evidence that $p<0.4$ .
2	$H_0:p=0.25$ ; $H_1:p\ne0.25$ Under $H_0$ , $S$ is negative binomial with $r=8$ and $p=0.25$ The lower critical tail is $S\leq17$ , with probability $0.04024$ The upper critical tail is $S\geq51$ , with probability $0.04526$ The critical region is $S\leq17$ or $S\geq51$ The combined size of the test is $0.08549$ The observed sum is $43$ Do not reject $H_0$ ; there is insufficient evidence that the sensors' response probability differs from $0.25$	8
(8 marks)		8
Notes
The sum of eight independent geometric variables is negative binomial with $r=8$ . Under $H_0$ , $P(S\leq17)=0.0402367802\ldots\leq0.05$ , while $P(S\leq18)=0.0569479807\ldots>0.05$ . Also $P(S\geq51)=0.0452558465\ldots\leq0.05$ , while $P(S\geq50)=0.0526704725\ldots>0.05$ . Hence the two tails are $S\leq17$ and $S\geq51$ , and the combined size of the test is $0.0402367802\ldots+0.0452558465\ldots=0.0854926267\ldots$ . The observed sum is $43$ , which is not in either critical tail, so do not reject $H_0$ ; there is insufficient evidence that the response probability differs from $0.25$ .
3	$H_0:p=0.32$ ; $H_1:p\ne0.32$ The observed sum is $S=36$ Under $H_0$ , $S$ is negative binomial with $r=7$ and $p=0.32$ The size of the test is $0.07954$ to $4$ significant figures The size is below $10\%$ because discreteness prevents the two rejection tails from containing exactly $5\%$ each without exceeding their limits The seven stopping times are independent and every trial has the same constant pass probability $p$ The observed sum lies in the upper rejection tail, so reject $H_0$ There is evidence at the $10\%$ level that the calibration checks' population pass probability differs from $0.32$	8
(8 marks)		8
Notes
The sum is $S=2+3+4+5+6+7+9=36$ . Under $H_0$ , the sum of seven independent geometric stopping times is negative binomial with $r=7$ and $p=0.32$ . The verified tails have probabilities $P(S\leq11)=0.0309307662\ldots$ and $P(S\geq35)=0.0486101547\ldots$ , so the size is $0.0795409209\ldots<0.10$ . Because $S$ is discrete, adding the next attainable value to either rejection tail would take that tail over its allocated $5\%$ , so an exact overall size of $10\%$ cannot be attained with these tails. The model assumes the seven stopping times are independent and that every trial has the same constant pass probability $p$ . Since $36\geq35$ , reject $H_0$ ; there is evidence that the population pass probability for the calibration checks differs from $0.32$ .
4	$S\leq r+1$ means at most one failed attempt before the $r$ th success Under $H_0$ , $S$ is negative binomial with parameters $r$ and $0.45$ $P(S\leq r+1)=(0.45)^r[1+0.55r]$ For $r=5$ , the size is $0.06920>0.05$ For $r=6$ , the size is $0.03571\leq0.05$ The least possible value is $r=6$ When $p=0.65$ , the power is $(0.65)^6[1+6(0.35)]=0.2338$ to $4$ decimal places With $r=6$ , the observed sum $S=7$ lies in $S\leq7$ , so reject $H_0$ ; there is evidence at the $5\%$ level that the population recognition probability exceeds $0.45$	8
(8 marks)		8
Notes
The event $S\leq r+1$ means that there are at most one failed attempt before the $r$ th success. Under $H_0$ , its probability is $(0.45)^r+r(0.55)(0.45)^r=(0.45)^r[1+0.55r]$ . This is $0.0691980\ldots$ at $r=5$ and $0.0357062\ldots$ at $r=6$ , so $6$ is least. At $p=0.65$ , the rejection probability is $(0.65)^6[1+6(0.35)]=0.2337986\ldots$ . For $r=6$ the critical region is $S\leq7$ , so the observed sum $S=7$ gives evidence that the population recognition probability exceeds $0.45$ .
5	Under $H_0$ , $P(X>2)=(1-p_0)^2$ , so $P(X\leq2)=1-(1-p_0)^2$ $(1-p_0)^2=0.9025$ $p_0=0.05$ The observed value $2$ is critical, so reject $H_0$ There is evidence at the $10\%$ level that the population recognition probability exceeds $0.05$ When $p=0.2$ , the power is $P(X\leq2)=1-(0.8)^2=0.36$ The probability of a Type II error is $1-0.36=0.64$	7
(7 marks)		7
Notes
For a geometric variable, $P(X\leq2)=1-P(X>2)=1-(1-p)^2$ . Under $H_0$ , the given size therefore gives $1-(1-p_0)^2=0.0975$ , so $(1-p_0)^2=0.9025$ and the valid solution is $p_0=0.05$ . Since $2$ lies in the rejection region, there is evidence that the population recognition probability exceeds $0.05$ . When $p=0.2$ , the rejection probability, and hence the power, is $1-0.8^2=0.36$ . The complementary Type II error probability is $0.64$ .