Hypothesis testing

FS1-4 Hypothesis testing — coverage pack

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

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

For a Poisson-mean test, state $H_0$ and $H_1$ in terms of the population parameter $\lambda$ or $\mu$ , not the observed count.
Under $H_0$ , combine independent observation periods into one Poisson total and use the direction of $H_1$ to choose the tail.
A critical region is the most extreme attainable tail whose probability under $H_0$ does not exceed the significance level; its probability is the actual size.
A common error is to say that $H_0$ is proved when it is not rejected; the conclusion should state that there is insufficient evidence for $H_1$ .

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 marks]
Answer
- Let $\lambda$ be the population mean number of alerts per shift
- $H_0:\lambda=4.2$ ; $H_1:\lambda>4.2$
Method: The claim supplies the null value. The word 'increased' makes the alternative upper-tailed, so use $H_1:\lambda>4.2$ .

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 its actual size.[7 marks]
Answer
- $H_0:\lambda=2.5$ ; $H_1:\lambda>2.5$
- Critical region $X\geq29$
- Size $=0.03433$
- Do not reject $H_0$ ; there is insufficient evidence that the mean rate has increased
Method: 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 $H_0$ .

Tier 3 · Hard

1. A Poisson model has mean $1.8$ events per batch. Twelve independent batches produce $14$ events in total. Test at a $5\%$ significance level whether the mean per batch has decreased. Give the critical region, its size and the $p$ -value.[8 marks]
Answer
- $H_0:\lambda=1.8$ ; $H_1:\lambda<1.8$
- Critical region $X\leq13$
- Size $=0.03329$
- $p$ -value $=0.05626$
- Do not reject $H_0$ ; there is insufficient evidence of a decrease
Method: Under $H_0$ , the total $X\sim\operatorname{Po}(12\times1.8)=\operatorname{Po}(21.6)$ . The lower-tail probabilities are $P(X\leq13)=0.0332864\ldots$ and $P(X\leq14)=0.0562576\ldots$ , so the $5\%$ critical region is $X\leq13$ with size $0.03329$ . The observed lower-tail $p$ -value is $P(X\leq14)=0.05626>0.05$ , so do not reject $H_0$ .

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

For a geometric-parameter test, hypotheses are stated in terms of the success probability $p$ .
A larger $p$ tends to produce an earlier first success, so small trial numbers support $H_1:p>p_0$ and large trial numbers support $H_1:p<p_0$ .
The sum of $r$ independent geometric variables with the same $p$ counts the trial of the $r$ th success and has a negative binomial distribution.
A common error is to reverse the tail: long waiting times are evidence for a smaller, not larger, success probability.

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 marks]
Answer
- $H_0:p=0.35$ ; $H_1:p<0.35$
- Large waiting times form the critical tail
Method: 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.

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 marks]
Answer
- Critical region $S\leq10$
- Size $=0.04735$
- Reject $H_0$ ; there is evidence that $p>0.3$
Method: 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 $H_0$ .

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. Find the $p$ -value and conclusion when $S=19$ .[8 marks]
Answer
- Critical region $S\geq22$
- Size $=0.03696$
- $p$ -value $=0.09417$
- Do not reject $H_0$ ; there is insufficient evidence that $p<0.4$
Method: Under $H_0$ , $S$ is negative binomial with $r=5$ and $p=0.4$ . The upper-tail probabilities are $P(S\geq22)=0.0369556\ldots$ and $P(S\geq21)=0.0509520\ldots$ , so the critical region is $S\geq22$ with size $0.03696$ . For $S=19$ , the $p$ -value is $P(S\geq19)=0.0941686\ldots>0.05$ , so do not reject $H_0$ .