Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 6: Probability & Statistics 2 · official syllabus
HYPOTHESIS TESTS LAB
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Four short missions on binomial and Poisson rejection regions, a normal mean test and Type II error. Keep the null model distinct from a specified true alternative.
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Mission 1 of 4
Set a binomial rejection region
A coin is tossed independently 10 times. Test H0: p=0.5 against H1: p>0.5 for the heads probability, at the 5% level. Reject when X>=c. Find the smallest integer c whose null upper-tail probability is at most 0.05.
Which probability is compared with 0.05 for this upper-tail test?
Teal bars are the candidate upper rejection tail under the null B(10,0.5). Use the sum of these probabilities, not just the first bar.
Your cutoff is 6. Choose the smallest cutoff with null tail no more than 0.05.
Study notes
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State hypotheses about a population parameter 6.5
The null hypothesis H0 gives the model or parameter value being tested. The alternative H1 states the proposed direction or difference. For a coin with heads probability p, H0:p=0.5 and H1:p>0.5 is an upper-tail test; H1:p<0.5 is lower-tail, and H1:p not equal to 0.5 is two-tailed.
Write hypotheses in terms of the unknown population parameter, not the observed sample proportion or sample mean. Choose the tail from the question before examining how extreme the sample result is. A test statistic is a quantity calculated from the observed data whose null distribution is known or approximated.
The significance level is a bound on the probability of rejecting H0 when it is true. A rejection or critical region contains the statistic values leading to rejection. Its complement is often called the acceptance region, but membership means do not reject, not that the null has been proved true.
Compare a tail probability with the test level 6.5
For an upper-tail test, calculate the null probability of a result at least as large as observed; for a lower-tail test, use at least as small. Use the whole relevant tail, not only the probability of the exact observed value. State the null model, the probability, the comparison and the conclusion in context.
For a two-tailed test using equal tail allocation, construct each rejection tail with probability at most half the stated level. Discrete tails may not attain that probability exactly; follow any rejection-region or tail convention specified in the question rather than assuming every two-sided discrete p-value definition is identical.
A small tail probability is evidence against H0 under the model. It is not the probability that H0 is true, and a non-rejection does not prove H0. Connect the conclusion to the original population claim, such as evidence that the defect rate has increased.
Construct a binomial rejection region by direct calculation 6.5
With independent tosses X~B(10,0.5) under H0 and H1:p>0.5, seek the smallest c such that P(X>=c)<=0.05. P(X>=8)=(45+10+1)/1024=56/1024=0.0546875 exceeds 0.05, while P(X>=9)=11/1024, approximately 0.0107 to 4 d.p., does not.
Reject for X>=9. For an observed 9 heads, there is sufficient evidence at the 5% level that p exceeds 0.5. The actual Type I probability is 11/1024, approximately 1.07%, rather than exactly 5%; discrete rejection regions often have actual size below the nominal level.
At a specified true p=0.7, a Type II error for this test is X<=8. Its probability is 1-[10(0.7)^9(0.3)+(0.7)^10], approximately 0.8507 to 4 d.p. Use the alternative value 0.7 for this probability, not the null value 0.5.
Construct a Poisson rejection region by direct calculation 6.5
For a fixed production interval, suppose X~Poisson(2) under H0 and H1:lambda>2. In a 5% upper-tail test, P(X>=5)=1-exp(-2) sum from k=0 to 4 of 2^k/k!, approximately 0.0527. P(X>=6), calculated with the sum through 5, is approximately 0.0166. Reject for X>=6.
An observation of 5 defects does not reject H0 under that rule. There is insufficient evidence at 5% that the mean number of defects per stated interval has increased; this does not demonstrate that it remains exactly 2.
If the true mean is instead 4 per interval, Type II error is P_Poisson(4)(X<=5)=exp(-4) sum from k=0 to 5 of 4^k/k!, approximately 0.7851 to 4 d.p. The interval unit and length must match in hypotheses, model and conclusion.
Use normal approximations with the correct boundary 6.5
A binomial or Poisson null distribution can sometimes be approximated by a normal distribution. Check the approximation conditions and use its null mean and variance: np and np(1-p) for binomial, lambda and lambda for Poisson. Discrete counts require a continuity correction before standardising.
Example: under B(100,0.5), the upper tail X>=60 is approximated by Y>59.5 for Y~N(50,25). The standardised boundary is (59.5-50)/5=1.9, giving an approximate upper tail of 0.0287 to 4 d.p. At 5% this is evidence in the upper direction; the exact binomial tail is different.
For Poisson, the syllabus normal-approximation guideline is lambda>15, approximately. Under Poisson(20), X>=29 uses normal boundary 28.5 and z=8.5/sqrt(20), about 1.90 to 2 d.p. Do not move the boundary the wrong way or use lambda as the normal standard deviation.
Test a population mean with a normal sampling distribution 6.5
For independent observations from a normal population with known standard deviation sigma, test H0:mu=mu0 using Z=(sample mean-mu0)/(sigma/sqrt(n)), which is standard normal under H0. Large-sample cases can use a normal approximation, with estimated variance if necessary. Do not silently apply this small-sample normal formula to an unknown-variance case.
For H0:mu=100 g against H1:mu>100 g, sigma=10 g and n=25 give standard error 2 g. Using supplied rounded 5% critical z=1.645, reject for sample mean>103.29 g. A sample mean of 104 g gives z=2, so there is evidence that the population mean exceeds 100 g. No continuity correction applies to this continuous sampling mean.
At a two-sided 5% level the usual rounded critical values are -1.96 and 1.96; the two tails each carry approximately 2.5%. The selected tail and significance level determine the critical value, not the sign or magnitude of the eventual sample result.
Distinguish Type I error, Type II error and power 6.5
A Type I error rejects a true null hypothesis. Its probability is calculated under H0 and is bounded by the chosen significance level for the specified rejection-region test. A Type II error does not reject H0 when an alternative is true. Its probability beta depends on which alternative parameter value is specified.
For the preceding mean test, suppose the true mean is 104 g while sigma remains 10 g. Type II error is sample mean<=103.29 g. Its probability is Phi((103.29-104)/2)=Phi(-0.355), approximately 0.3613 to 4 d.p. Power at this alternative is 1-beta, approximately 0.6387. The rejection boundary used the rounded z=1.645.
Draw the fixed rejection boundary against the null distribution for Type I probability and against the specified alternative distribution for Type II probability. Lowering the significance level generally trades less Type I risk for more Type II risk at a fixed sample size and alternative; a larger independent sample can improve power for the same level.
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