FS1-8 Quality of tests — revision question pack

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

Checked against Edexcel 9FM0 section FS1-8. 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

FS1-8.1 · Type I and Type II errors. Size and Power of Test. The power function.

Explanation

A Type I error rejects $H_0$ when it is true; a Type II error does not reject $H_0$ when a specified alternative is true. The size is $P(\text{reject }H_0\mid H_0\text{ true})$ , the attainable null probability of the critical region, which may differ from a nominal significance level in a discrete test.
The power function is $\pi(\theta)=P_\theta(\text{reject }H_0)$ .
At a specified alternative, power equals one minus the Type II error probability.
A more effective test has higher power against relevant alternatives while controlling size.
Examiners expect errors to be described in context and power to be calculated from the critical region at the alternative parameter.

A typical upper-tailed power function, with the test size shown at the null parameter.

Worked example

For $X\sim\operatorname{Bin}(10,p)$ , a test rejects $H_0:p=0.3$ when $X\geq6$ . Find the size, write the power function and express the Type II error probability at $p=0.6$ .

1.Size $=P_{0.3}(X\geq6)=\sum_{x=6}^{10}\binom{10}{x}0.3^x0.7^{10-x}$ .
2. $\pi(p)=\sum_{x=6}^{10}\binom{10}{x}p^x(1-p)^{10-x}$ .
3.At $p=0.6$ , Type II probability $=1-\pi(0.6)=P_{0.6}(X\leq5)$ .

Answer: Size $=0.04735$ to $5$ significant figures; power is the stated upper-tail sum, and Type II probability at $p=0.6$ is $1-\pi(0.6)$ .

Common mistakes

Don't call the nominal $5\%$ level the size without calculating the discrete critical-region probability.
Don't calculate power at an alternative parameter using the null distribution instead.
Don't add power and Type I error probability rather than power and Type II error probability.

Exam tip

Use the same critical-region event for size and power; only the parameter value in its probability changes.

Tier 1 · Easy

1.
In a test of $H_0:p=0.4$ against $H_1:p>0.4$ , describe Type I and Type II errors in terms of $p$ , and define the size of the test.
(4)
(Total for Question 1 is 4 marks)
2.
Under $H_0$ , $X\sim\operatorname{Bin}(8,0.5)$ . A test has critical region $X\geq7$ . Find the size of the test, giving an exact value.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Let $X\sim\operatorname{Bin}(12,p)$ . A test of $H_0:p=0.3$ against $H_1:p>0.3$ rejects $H_0$ when $X\geq7$ . Find the size, the power at $p=0.5$ , and the probability of a Type II error at $p=0.5$ .
(6)
(Total for Question 1 is 6 marks)
2.
For one observation $X\sim\operatorname{Po}(\lambda)$ , test $H_0:\lambda=3$ against $H_1:\lambda>3$ . Find the critical region, of the form $X\geq c$ , for a test at the $5\%$ level of significance. Write down the size of this test and calculate the power at $\lambda=5$ .
(5)
(Total for Question 2 is 5 marks)
3.
For one observation $X\sim\operatorname{Po}(\lambda)$ , a test of $H_0:\lambda=2.5$ against $H_1:\lambda>2.5$ has the critical region $X\geq6$ . Find the size, write down the power function, and calculate the power and Type II error probability when the actual value is $\lambda=4$ . State whether the size is below $5\%$ . Give each decimal probability to $4$ decimal places.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
For $X\sim\operatorname{Bin}(12,p)$ , a test of $H_0:p=0.4$ against $H_1:p>0.4$ uses critical region $X\geq8$ . Write its power function. Calculate its size, its power and Type II error probability when $p=0.65$ , and the expected number of rejections in $40$ independent repetitions at $p=0.65$ . Comment on whether it is a $5\%$ test.
(9)
(Total for Question 1 is 9 marks)
2.
Let $X\sim\operatorname{Bin}(12,p)$ . For $H_0:p=0.5$ , $H_1:p\neq0.5$ , compare $C_A=\{0,1,11,12\}$ with $C_B=\{0,10,11,12\}$ . Show that both sizes are at most $5\%$ and calculate each test's power at $p=0.25$ and $p=0.75$ . State, with a reason, which critical region you would use if departures above $0.5$ are of most interest.
(7)
(Total for Question 2 is 7 marks)
3.
For one observation $X\sim B(9,p)$ , test $H_0:p=1/3$ against $H_1:p>1/3$ using a critical region $X\geq c$ , where $c$ is an integer. When the actual value under the alternative is $p=2/3$ , find $c$ so that the Type I and Type II error probabilities are equal. Give their common exact value, write down the size and the power at $p=2/3$ , and state whether the size is below $5\%$ .
(9)
(Total for Question 3 is 9 marks)
4.
A quality inspector records the number $Y$ of components inspected up to and including the first defective component. The model is $Y\sim\operatorname{Geo}(p)$ , where $p$ is the probability that a component is defective. A test at the $5\%$ level of significance uses $H_0:p=0.35$ , $H_1:p<0.35$ and critical region $Y\geq8$ . Find the size of the test. Show that the power function is $\pi(p)=(1-p)^7$ . Find the power and the Type II error probability when $p=0.25$ , giving each decimal probability to $4$ decimal places. State the conclusion of the test if $Y=9$ is observed.
(7)
(Total for Question 4 is 7 marks)
5.
Two tests of $H_0:p=0.3$ against $H_1:p>0.3$ are considered at the $5\%$ level of significance. Test A uses $X\sim B(8,p)$ and rejects $H_0$ when $X\geq6$ . Test B uses $Y\sim B(12,p)$ and rejects $H_0$ when $Y\geq8$ . Find both sizes. Show that the power function for Test A is $p^6(21p^2-48p+28)$ and that the power function for Test B is $p^8(330p^4-1440p^3+2376p^2-1760p+495)$ . Find both powers exactly when $p=0.5$ and recommend a test for detecting this alternative.
(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.

FS1-8.1 · Type I and Type II errors. Size and Power of Test. The power function.

Tier 1 · Easy

Mark scheme for FS1-8.1 Tier 1 · Easy
Question	Scheme	Marks
1	Type I: conclude that $p>0.4$ when $p=0.4$ Type II: fail to conclude that $p>0.4$ when a specified value $p>0.4$ is true Size: $P(\text{reject }H_0\mid p=0.4)$	4
(4 marks)		4
Notes
Translate the two decision errors using the null value and the direction of the alternative. The size is the probability of the Type I error evaluated at the null parameter.
2	$P(X\geq7)=\{\binom87+\binom88\}/2^8$ The size is $9/256$ (that is, $0.03515625$ )	2
(2 marks)		2
Notes
The size is the null probability of the critical region. Since $p=0.5$ , $P(X\geq7)=[\binom87+\binom88](0.5)^8=(8+1)/256=9/256=0.03515625$ .

Tier 2 · Standard

Mark scheme for FS1-8.1 Tier 2 · Standard
Question	Scheme	Marks
1	Size $=0.03860$ Power at $p=0.5$ is $0.38721$ Type II probability $=0.61279$	6
(6 marks)		6
Notes
The size is $P_{0.3}(X\geq7)=\sum_{x=7}^{12}\binom{12}{x}(0.3)^x(0.7)^{12-x}=0.0386008\ldots$ . At $p=0.5$ , power is $P_{0.5}(X\geq7)=0.3872070\ldots$ . The Type II probability is its complement, $1-0.3872070=0.6127930\ldots$ .
2	$P_3(X\geq6)=0.0839179420\ldots>0.05$ while $P_3(X\geq7)=0.0335085353\ldots<0.05$ The critical region is $X\geq7$ The size of the test is $0.03351$ to $4$ significant figures $\pi(5)=P_5(X\geq7)$ The power at $\lambda=5$ is $0.2378$ to $4$ decimal places	5
(5 marks)		5
Notes
For an upper-tail rule, test adjacent thresholds. Under $H_0$ , $P_3(X\geq6)=0.0839179420\ldots$ , which exceeds $0.05$ , while $P_3(X\geq7)=0.0335085353\ldots$ , so the critical region is $X\geq7$ . Its size is $0.0335085353\ldots=0.03351$ to $4$ significant figures. At $\lambda=5$ , the power is $P_5(X\geq7)=0.2378165370\ldots=0.2378$ to $4$ decimal places.
3	The size is $P_{2.5}(X\geq6)$ $P_{2.5}(X\geq6)=0.0420$ to $4$ decimal places $\pi(\lambda)=P_{\lambda}(X\geq6)=1-e^{-\lambda}\displaystyle\sum_{x=0}^{5}\lambda^x/x!$ $\pi(4)=P_4(X\geq6)=0.2149$ to $4$ decimal places At $\lambda=4$ , the Type II error probability is $1-\pi(4)=0.7851$ to $4$ decimal places The size is below $0.05$	6
(6 marks)		6
Notes
The size is the null probability of the given critical region: $P_{2.5}(X\geq6)=1-e^{-2.5}\sum_{x=0}^{5}2.5^x/x!=0.0420210382\ldots=0.0420$ to $4$ decimal places. At a general parameter value the same rejection event gives $\pi(\lambda)=1-e^{-\lambda}\sum_{x=0}^{5}\lambda^x/x!$ . Hence $\pi(4)=1-e^{-4}\sum_{x=0}^{5}4^x/x!=0.2148696130\ldots=0.2149$ to $4$ decimal places. The Type II probability at $\lambda=4$ is $1-\pi(4)=0.7851303870\ldots=0.7851$ to $4$ decimal places. Since $0.042021\ldots<0.05$ , the test has size below $5\%$ .

Tier 3 · Hard

Mark scheme for FS1-8.1 Tier 3 · Hard
Question	Scheme	Marks
1	$\pi(p)=\sum_{x=8}^{12}\binom{12}{x}p^x(1-p)^{12-x}$ Size $=0.05731$ Power at $p=0.65$ is $0.5833$ to $4$ decimal places Type II probability $=0.4167$ to $4$ decimal places Expected rejections $=23.3$ It is not a $5\%$ test because its size exceeds $0.05$	9
(9 marks)		9
Notes
The power function is the critical-region probability at a general $p$ . At the null value, $\pi(0.4)=0.0573099\ldots$ , which is the size. At $p=0.65$ , the power is $0.5833$ to $4$ decimal places, so the Type II probability is $0.4167$ to $4$ decimal places. Using the unrounded power, the expected number rejected across $40$ repetitions is $23.3338\ldots$ . Since $0.05731>0.05$ , the test does not have size at most $5\%$ .
2	$\operatorname{size}(A)=26/4096=0.00634765625$ $\operatorname{size}(B)=80/4096=0.01953125$ Both sizes are below $0.05$ $\pi_A(0.25)=\pi_A(0.75)=0.1584$ to $4$ decimal places $\pi_B(0.25)=0.03171$ to $4$ significant figures $\pi_B(0.75)=0.3907$ to $4$ decimal places Choose $C_B$ when increases above $0.5$ are the priority, while noting its lower power at $p=0.25$	7
(7 marks)		7
Notes
At $p=0.5$ , $P(C_A)=[1+12+12+1]/4096=26/4096=0.00634765625$ and $P(C_B)=[1+66+12+1]/4096=80/4096=0.01953125$ , so both control size below $0.05$ . Direct binomial sums give $\pi_A(0.25)=0.1583839655\ldots$ and $\pi_B(0.25)=0.0317139626\ldots$ ; by symmetry $\pi_A(0.75)=0.1583839655\ldots$ , while $\pi_B(0.75)=0.3906750679\ldots$ . The rounded values are respectively $0.1584$ , $0.03171$ , $0.1584$ and $0.3907$ . Region $C_B$ is preferable for upper departures because its power at $p=0.75$ is much higher, although its asymmetry sacrifices lower-tail power.
3	$\alpha=P_{1/3}(X\geq c)$ $\beta=P_{2/3}(X<c)$ at the stated actual value When $X\sim B(9,2/3)$ , set $Z=9-X$ , so $Z\sim B(9,1/3)$ $\beta=P(Z\geq10-c)$ for $Z\sim B(9,1/3)$ Equality requires $c=10-c$ , so $c=5$ $\alpha=\displaystyle\sum_{x=5}^{9}\binom9x(1/3)^x(2/3)^{9-x}=2851/19683$ $\beta=2851/19683$ The size is $2851/19683=0.1448458\ldots>0.05$ The power at $p=2/3$ is $1-\beta=16832/19683$	9
(9 marks)		9
Notes
The Type I probability is $\alpha=P_{1/3}(X\geq c)$ . At the stated actual value $p=2/3$ , a Type II error is $X<c$ . If $X\sim B(9,2/3)$ , set $Z=9-X$ ; then $Z\sim B(9,1/3)$ and $\beta=P(Z\geq10-c)$ . Comparing upper tails from the same $B(9,1/3)$ distribution, which are strictly decreasing in their integer threshold, equality occurs when $c=10-c$ , giving $c=5$ . Direct summation gives $\alpha=P_{1/3}(X\geq5)=2851/19683$ . Symmetry gives the same exact value for $\beta$ . Thus the size is $2851/19683=0.1448458060\ldots$ , which exceeds $0.05$ , and the power at $p=2/3$ is $1-\beta=16832/19683=0.8551541940\ldots$ .
4	$P_p(Y\geq8)=P_p(\text{the first seven trials are failures})$ $P_p(Y\geq8)=(1-p)^7$ The size is $(1-0.35)^7=0.0490$ to $4$ decimal places Therefore $\pi(p)=(1-p)^7$ $\pi(0.25)=0.75^7=0.1335$ to $4$ decimal places At $p=0.25$ , the Type II error probability is $1-0.1335=0.8665$ to $4$ decimal places Since $9\geq8$ , reject $H_0$ ; there is sufficient evidence that the probability that a component is defective is below $0.35$	7
(7 marks)		7
Notes
The rejection event $Y\geq8$ occurs exactly when the first seven inspected components are not defective. At a general value of $p$ its probability is therefore $(1-p)^7$ , which proves that $\pi(p)=(1-p)^7$ . Under $H_0$ , the size is $0.65^7=0.0490222789\ldots=0.0490$ . At the stated alternative value, the power is $0.75^7=2187/16384=0.1334838867\ldots=0.1335$ , so the Type II error probability is $1-0.1334838867\ldots=0.8665161133\ldots=0.8665$ . An observed value of $9$ lies in the critical region, so reject $H_0$ ; in context, there is sufficient evidence that the probability that a component is defective is below $0.35$ .
5	$\operatorname{size}(A)=P_{0.3}(X\geq6)=0.01129$ to $4$ significant figures $\operatorname{size}(B)=P_{0.3}(Y\geq8)=0.009489$ to $4$ significant figures $\pi_A(p)=\displaystyle\sum_{x=6}^{8}\binom8x p^x(1-p)^{8-x}=p^6(21p^2-48p+28)$ $\pi_B(p)=\displaystyle\sum_{y=8}^{12}\binom{12}{y}p^y(1-p)^{12-y}=p^8(330p^4-1440p^3+2376p^2-1760p+495)$ $\pi_A(0.5)=(\binom86+\binom87+\binom88)/2^8=37/256$ $\pi_B(0.5)=(\binom{12}{8}+\binom{12}{9}+\binom{12}{10}+\binom{12}{11}+\binom{12}{12})/2^{12}=397/2048$ Recommend Test B at $p=0.5$ because $397/2048>37/256$ , while both tests have size below $5\%$	7
(7 marks)		7
Notes
Direct binomial tails under $p=0.3$ give sizes $0.01129221\ldots$ and $0.0094893711\ldots$ , both below $5\%$ . Expanding the rejection probabilities gives $\pi_A(p)=p^6[28(1-p)^2+8p(1-p)+p^2]=p^6(21p^2-48p+28)$ and $\pi_B(p)=p^8\sum_{j=0}^{4}\binom{12}{8+j}p^j(1-p)^{4-j}=p^8(330p^4-1440p^3+2376p^2-1760p+495)$ . At $p=0.5$ , the binomial coefficients give $\pi_A=37/256$ and $\pi_B=794/4096=397/2048$ . Since $397/2048>37/256$ , Test B has the greater probability of detecting this stated alternative.