Population, census, sample, sampling unit and sampling frame; advantages and disadvantages of a census and of a sample

Statistics and their sampling distributions, found by listing every possible sample in small cases

Null and alternative hypotheses, test statistic and significance level

Critical region, critical value and actual significance level

One-tailed and two-tailed tests

Tests for the proportion p of a binomial distribution and the mean λ of a Poisson distribution, using tables

Tests using a Normal approximation with a continuity correction, and conclusions in context

1. A hypothesis test assumes a value for a population parameter, works out how likely the observed data would be under that assumption, and rejects the assumption only if the data would be unlikely enough.

2. Census: every member of the population is observed. Sample: only some are, chosen from a sampling frame (a list of the sampling units).

3. A statistic is a quantity calculated from the sample alone (no unknown parameters); its probability distribution over all possible samples is its sampling distribution.

4. H₀ states a value of the parameter, e.g. p=0.15 or λ=4. H₁ says how it may differ: <, > (one-tailed) or ≠ (two-tailed).

5. The critical region is the set of values so unlikely under H₀ that we reject it. The actual significance level is the probability of landing in it when H₀ is true; for discrete distributions it is rarely exactly the stated level.

6. Conclude in two parts: a decision (reject H₀ or not) and a sentence in context with no certainty: "there is (in)sufficient evidence to suggest …".

Units missing from the frame can never be chosen, so a poor frame biases the sample, however carefully it is randomised.

WORKED EXAMPLE · QUESTION

A sports centre has 3200 registered members. The manager wants to know what members think of proposed new opening hours. She has an up-to-date list of the names of all registered members. (a) Identify the population and a suitable sampling frame. (b) State the sampling unit. (c) Give one advantage and one disadvantage of asking every member (a census) rather than a sample.

SOLUTION

(a) Population = everyone the question is about; frame = the list you draw from. Population: the 3200 registered members; sampling frame: the list of names of registered members

(b) One item that could be selected: a registered member

(c) Advantage: every member's opinion is included, so the result is exact for the centre. Disadvantage: contacting 3200 members and processing their replies takes a long time and costs a lot.

A statistic is a quantity calculated only from the observations in a sample: the sample mean, the median, the total, the largest value, the number of successes. It must not contain unknown parameters, so X̄-μ (with μ unknown) is not a statistic.

Different samples give different values, so a statistic is a random variable; its distribution over all possible samples is its sampling distribution. For a small population, list every possible sample, find its probability (multiply, and count the arrangements), calculate the statistic, and collect equal values.

WORKED EXAMPLE · QUESTION

A large box contains discs, each labelled 1, 2 or 5, in the proportions 60%, 30% and 10% respectively. A random sample of three discs is taken (the box is large enough to treat the draws as independent). (a) List all the possible samples. (b) Find the sampling distribution of the median, M, of the sample.

SOLUTION

(a) The different samples, ignoring order (10 of them): {1,1,1}, {1,1,2}, {1,1,5}, {1,2,2}, {1,2,5}, {1,5,5}, {2,2,2}, {2,2,5}, {2,5,5}, {5,5,5}.

Probability of each (count the orders: 1, 3 or 6): {1,1,1}: 0.6³=0.216; {1,1,2}: 3(0.6²)(0.3)=0.324; {1,1,5}: 3(0.6²)(0.1)=0.108; {1,2,2}: 3(0.6)(0.3²)=0.162; {1,2,5}: 6(0.6)(0.3)(0.1)=0.108; {1,5,5}: 3(0.6)(0.1²)=0.018; {2,2,2}: 0.027; {2,2,5}: 0.027; {2,5,5}: 0.009; {5,5,5}: 0.001.

(b) The median is the middle value; add the probabilities for each median:

Check: 0.648+0.324+0.028=1. E(M)=1.436: a sample median is near the population median, 1, but not always equal to it.

A hypothesis test asks "could this result just be chance?". Assume nothing has changed, and abandon that assumption only if the data would be surprising enough.

A hypothesis is a statement about a population parameter: the probability p in B(n,p), or the mean λ in Po(λ).

The null hypothesis H₀ is the claim being tested, assumed true unless the data give enough evidence against it. It always contains "=": H₀: p=0.15.

The alternative hypothesis H₁ says how the parameter is believed to differ. Read the question: "increased" → p>0.15; "decreased / lower" → p<0.15 (both one-tailed); "changed / different" → p≠0.15 (two-tailed).

The test statistic is the statistic used to decide, e.g. X = the number of successes in the sample.

The significance level (e.g. 5%) sets how unlikely a result must be, assuming H₀, before we reject H₀; a two-tailed test splits it between the tails.

Hypotheses are about the population, never the sample. Every test follows the same five-step routine.

WORKED EXAMPLE · QUESTION

A machine fills bags of flour. Over a long period 15% of the bags have been underweight. After the machine is serviced, an engineer takes a random sample of 20 bags to test whether the proportion of underweight bags has changed. (a) Define a suitable test statistic and state its distribution under the null hypothesis. (b) Write down suitable hypotheses and state whether the test is one-tailed or two-tailed.

SOLUTION

(a) A count of "successes" in a fixed number of independent trials: X = the number of underweight bags in the sample of 20; under H₀, X~B(20,0.15)

(b) Hypotheses about the population proportion p of underweight bags: H₀: p=0.15, H₁: p≠0.15; two-tailed, because the engineer is looking for a change in either direction.

Writing "H₀: X=3" or "H₀:

15% are underweight in the sample" is wrong: hypotheses are about the parameter p.

The critical region (CR) is the set of values of the test statistic that would lead you to reject H₀. The critical value is the boundary value: the first value inside the critical region.

For a discrete distribution the tail probabilities jump, so the question tells you which rule to use. Read it and follow it:

"as close as possible to 5% without exceeding 5%" (or "should not exceed 5%"): take the largest region whose probability is at most the level;

"as close as possible to 5%": take the region whose probability is nearest to the level, even if it is slightly above it — compare the distances of the two probabilities either side of the level.

The two rules often agree, but not always (Worked example 4). In a two-tailed test apply the rule to each tail with half the level. Lower tail: read P(X≤c) directly; upper tail: P(X≥c)=1-P(X≤c-1).

WORKED EXAMPLE · QUESTION

For the test in Worked example 3, using a 10% level of significance: (a) find the critical region, given that the probability in each tail should be as close as possible to 5% without exceeding 5%; (b) state the actual significance level; (c) the engineer finds 6 underweight bags. Comment on this result.

SOLUTION

(a) Lower tail, X~B(20,0.15), compare with 0.05: P(X=0)=0.0388≤0.05, P(X≤1)=0.1756>0.05, so the lower part is X=0. Upper tail: P(X≥6)=1-0.9327=0.0673>0.05, P(X≥7)=1-0.9781=0.0219≤0.05, so the upper part is X≥7. Critical region: X=0 or X≥7

(b) Add the two tails: 0.0388+0.0219= 0.0607 (6.07%)

(c) 6 is not in the critical region, so do not reject H₀: there is insufficient evidence to suggest that the proportion of underweight bags has changed.

6 out of 20 is double the old rate, yet with only 20 bags it would occur fairly often by chance.

The rule matters: under "as close as possible to 5%" the upper tail would be X≥6 (0.0673 is 0.0173 from 0.05; 0.0219 is 0.0281 away), and 6 would then be significant.

There are two equivalent ways to carry out step 3 of the routine, and they always give the same decision:

Probability method: find the probability of the observed value or anything more extreme in the direction of H₁, and compare it with the significance level (with half the level in a two-tailed test).

Critical region method: find the CR as in the previous section and see whether the observed value lies in it.

If the question does not say which, the probability method is usually quicker.

"More extreme" means in the direction of H₁: for H₁: p<0.45 with 8 observed, find P(X≤8).

If the probability is less than or equal to the significance level, the result is significant: reject H₀. Otherwise do not reject H₀.

p greater than 0.5: the tables only go up to p=0.5. Count failures instead: if X~B(n,p) then Y=n-X~B(n,1-p), and P(X≤k)=P(Y≥n-k).

WORKED EXAMPLE · QUESTION

The owner of a café knows that, in the past, 45% of her customers ordered a hot drink. After introducing a range of iced drinks she believes this proportion has fallen. In a random sample of 30 customers, 8 ordered a hot drink. Test, at the 5% level of significance, whether there is evidence to support her belief. State your hypotheses clearly.

SOLUTION

Hypotheses (one-tailed: "fallen"): H₀: p=0.45, H₁: p<0.45.

Distribution under H₀: X = number of hot-drink orders in the sample, X~B(30,0.45).

Probability of 8 or fewer (the direction of H₁): P(X≤8)=0.0312.

0.0312<0.05, so the result is significant: reject H₀. There is sufficient evidence to suggest that the proportion of customers ordering a hot drink has fallen.

Critical region method: P(X≤8)=0.0312≤0.05 but P(X≤9)=0.0694>0.05, so the CR is X≤8, and 8 lies in it.

Same decision.

The parameter is the mean rate λ. Hypotheses look like H₀: λ=0.8 (per hour), H₁: λ>0.8, or <, or ≠. Scale the mean to the period observed: events at a mean rate of 0.8 per hour give, over 5 hours, X~Po(5×0.8)=Po(4), and you read the tables for the scaled mean. The hypotheses may be written for the original rate (λ=0.8) or the scaled one (λ=4), as long as you are consistent and say which period it refers to. Otherwise the routine is identical.

WORKED EXAMPLE · QUESTION

Calls to a garage's breakdown line arrive at random at a mean rate of 0.8 per hour. After a local advertising campaign, the manager records 8 calls in a 5-hour period. (a) Test, at the 5% level of significance, whether the mean rate of calls has increased. (b) Find the critical region for this test.

SOLUTION

(a) H₀: λ=0.8, H₁: λ>0.8 (per hour); over 5 hours, under H₀: X~Po(4).

P(X≥8)=1-P(X≤7)=1-0.9489=0.0511.

0.0511>0.05: not significant, do not reject H₀. There is insufficient evidence to suggest that the mean rate of calls has increased.

(b) P(X≥8)=0.0511>0.05 and P(X≥9)=1-0.9786=0.0214≤0.05, so CR: X≥9

When n or λ is beyond the tables, approximate the test statistic by a Normal distribution with the same mean and variance (S2.3):

B(n,p)≈N(np,np(1-p)) when n is large and p is close to 0.5 (in practice, np and n(1-p) both comfortably above 5);

Po(λ)≈N(λ,λ) when λ is large (say above 10).

Apply a continuity correction, because the original distribution is discrete: P(X≥72)≈P(Y>71.5) and P(X≤25)≈P(Y<25.5). Then standardise, round z to 2 d.p. and use Φ(z); for a critical region use the percentage points table (1.6449 for 5% in one tail, 1.9600 for 2.5%). Hypotheses are still about p or λ.

WORKED EXAMPLE · QUESTION

In the past, 40% of visitors to a museum bought something from the museum shop. After the shop is redesigned, a random sample of 150 visitors is taken and 72 of them buy something. Using a suitable approximation, test at the 5% level of significance whether the proportion of visitors buying something has increased.

SOLUTION

Hypotheses and exact distribution: H₀: p=0.4, H₁: p>0.4; under H₀, X~B(150,0.4).

Normal approximation (n large, p not far from 0.5): np=60, np(1-p)=36⇒Y~N(60,6²).

Continuity correction and standardise:

0.0274<0.05: significant, reject H₀. There is sufficient evidence to suggest that the proportion of visitors buying something from the shop has increased.

Critical region version: (x-0.5-60)/6≥1.6449⇒x≥70.37, so CR X≥71. The exact tail is about 0.028.

A long question often combines a critical region from the tables with a later test where the mean is too large for the tables. Only the source of the probabilities changes.

WORKED EXAMPLE · QUESTION

Requests for a technician at an office building arrive at random at a mean rate of 3 per day. After a new computer system is installed, the facilities manager wishes to test whether the rate of requests has changed. (a) State suitable null and alternative hypotheses. (b) Using a 2-day period and a 5% level of significance, find the critical region for the test. The probability in each tail should be as close as possible to 2.5%. (c) Find the actual significance level of the test. (d) In a 2-day period there are 11 requests. State, with a reason, the conclusion of the test. (e) The manager now suspects that the rate has increased. Over the next 20 days there are 75 requests. Using a suitable approximation, test at the 5% level whether the rate has increased.

SOLUTION

(a) Two-tailed, because the question says "changed": H₀: λ=3, H₁: λ≠3 (requests per day)

(b) Under H₀, over 2 days X~Po(6). Lower tail, nearest to 0.025: P(X≤1)=0.0174 (0.0076 from 0.025) and P(X≤2)=0.0620 (0.0370 from 0.025), so X≤1. Upper tail:

P(X≥11)=1-0.9574=0.0426 (0.0176 from 0.025) and P(X≥12)=1-0.9799=0.0201 (0.0049 from 0.025), so X≥12. Critical region: X≤1 or X≥12

(c) Add the tails: 0.0174+0.0201= 0.0375

(d) 11 is not in the critical region, so do not reject H₀: there is insufficient evidence that the rate of requests has changed.

(e) One-tailed now; over 20 days the mean is 60, too large for the tables. H₀: λ=3, H₁: λ>3; X~Po(60) under H₀, approximated by Y~N(60,60).

0.0307<0.05: significant, reject H₀. There is sufficient evidence to suggest that the rate of requests for a technician has increased.

Equivalently, the CR is X≥74, since 60+0.5+1.6449√60=73.24. 11 in two days was not enough evidence; 75 in twenty days is: larger samples can detect smaller changes.

S2.1 Binomial and Poisson distributions

Every test here assumes a binomial or Poisson model; the conditions, the cumulative tables and the switch to n-X for p>0.5 come from S2.1.

S1.6 and S2.3 — the Normal distribution and approximations

Standardising, Φ(z) and the percentage points table from S1 are exactly what the approximate tests use, together with the continuity correction from S2.3.

P2 binomial expansion

Counting arrangements when listing samples uses the same coefficients C(n, r) as (a+b)ⁿ.

1. A factory tests the burning time of its candles by burning them until they go out. Why must it use a sample rather than a census?

2. Faults on a cable occur at random at a mean rate of 2.5 per kilometre. What is the distribution of the number of faults on a 4 km length?

3. H₀: p=0.8, H₁: p<0.8, n=10, and 5 successes are observed. Find the probability needed for the test.

4. A random sample X₁,X₂,X₃ is taken from a population with unknown mean μ. Is X₁+X₂+X₃ a statistic? Is X₁-μ?

5. A test at the 5% level has critical region X≥9 with P(X≥9)=0.0214. What is the probability of rejecting H₀ when it is true?

1. The test destroys each candle. A census would leave no candles to sell, so only a sample can be tested.

2. Po(10): the mean scales with the length, 2.5×4=10.

3. Count failures: Y=10-X~B(10,0.2) under H₀. P(X≤5)=P(Y≥5)=1-P(Y≤4)=1-0.9672=0.0328, which is significant at 5%.

4. The sum is a statistic: it uses only the sample values. X₁-μ is not, because μ is unknown and so it cannot be calculated from the data.

5. 0.0214: the actual significance level, the probability of falling in the critical region when H₀ is true. It is less than 5% because the distribution is discrete.