6.5 - Testing a population mean

6.5 - Testing a population mean

A sample average can differ from a claimed population mean even when that claim is correct. You will judge the size of that difference using the distribution of a sample mean, carry out one-tailed and two-tailed normal tests, and connect probability comparisons with critical boundaries. The spread of sample means is the key scale for the comparison.

A known population variance

Nine readings have average 53.653.6, compared with a claimed population mean of 5050. Is a difference of 3.63.6 large? If individual readings have standard deviation 66, the average of nine independent readings has standard deviation 6/9=26/\sqrt9=2. The difference is 1.81.8 standard errors, not 0.60.6 individual standard deviations.

You already know

For independent observations from X∼N(μ,σ2)X\sim\mathrm N(\mu,\sigma^2), their mean has exact distribution X‾∼N(μ,σ2/n)\overline X\sim\mathrm N(\mu,\sigma^2/n). The standard deviation σ/n\sigma/\sqrt n of the sample mean is its standard error. MF19 gives the sample-mean normal form under its Central Limit Theorem heading; exact normality for a normal population is also a fact to know.

The hypotheses concern μ\mu, the population mean. Under H0:μ=μ0H_0:\mu=\mu_0, centre the sample-mean distribution at μ0\mu_0 and use the known population variance. A normal population makes this exact even for a small sample.

Normal test statistic for a mean

For a normal population with known variance σ2\sigma^2,

Z=X‾−μ0σ/n∼N(0,1)under H0:μ=μ0.Z=\frac{\overline X-\mu_0}{\sigma/\sqrt n}\sim\mathrm N(0,1) \quad\text{under }H_0:\mu=\mu_0.

Use the observed sample mean to calculate zz. Not in the formula booklet as a testing rule: derive it by standardising the sample-mean distribution. Normal probability and critical-value tables are supplied.

Use this standardisation to measure the observed difference on the correct scale.

Worked example 1: testing for a larger mean

Question. Independent readings come from a normal population with known variance 3636. A random sample of 99 readings has mean 53.653.6. Test H0:μ=50H_0:\mu=50 against H1:μ>50H_1:\mu>50 at the 5%5\% level.

Plan. Standardise the sample mean under the null and find the upper-tail probability.

Under H0H_0, X‾∼N(50,36/9)=N(50,4)\overline X\sim\mathrm N(50,36/9)=\mathrm N(50,4). Hence

z=53.6−506/9=1.8.z=\frac{53.6-50}{6/\sqrt9}=1.8.

The upper-tail probability is 1−Φ(1.8)=0.0359<0.051-\Phi(1.8)=0.0359<0.05.

Answer. Reject H0H_0. There is sufficient evidence at the 5%5\% level that the population mean reading exceeds 5050.

Check. The upper-tail critical value is 1.6451.645, and 1.8>1.6451.8>1.645 gives the same decision.

Now reflect the departure below the null mean and investigate a decrease. The relevant probability changes sides.

Worked example 2: testing for a smaller mean

Question. With the same normal population variance and sample size, another sample has mean 46.446.4. Test H0:μ=50H_0:\mu=50 against H1:μ<50H_1:\mu<50 at the 5%5\% level.

Plan. Use the same standard error, but calculate a lower-tail probability.

z=46.4−506/9=−1.8,P(Z≤−1.8)=1−Φ(1.8)=0.0359<0.05.z=\frac{46.4-50}{6/\sqrt9}=-1.8,\qquad \mathrm P(Z\le-1.8)=1-\Phi(1.8)=0.0359<0.05.

Answer. Reject H0H_0. There is sufficient evidence at the 5%5\% level that the population mean reading is below 5050.

Check. −1.8<−1.645-1.8<-1.645 places the statistic in the lower critical tail. Symmetry explains why the evidence matches the reflected previous example.

The population variance remains known even though its mean is being tested. The two parameters describe different features.

Watch out

Use σ/n\sigma/\sqrt n in the denominator. Dividing by σ\sigma tests the scale of one reading, while dividing by σ2/n\sigma^2/n uses a variance as though it were a standard deviation. No continuity correction is needed for these continuous normal readings.

Apply the method with a fresh measurement context.

Large samples

What if the population is not normal? For a sufficiently large independent random sample from a population with finite variance, the Central Limit Theorem makes the sample mean approximately normal. It does not make the individual observations normal. The resulting test is approximate.

If the population variance is known, use it. If it is unknown, a large sample lets us estimate it with the unbiased sample variance s2s^2. The estimated standard error is s/ns/\sqrt n. A small sample from an unspecified non-normal population does not justify this method; neither does a small sample with unknown variance simply because its histogram looks roughly normal.

Large-sample normal test

For a sufficiently large independent random sample,

X‾≈N(μ0,σ2n)under H0.\overline X\approx\mathrm N\left(\mu_0,\frac{\sigma^2}{n}\right) \quad\text{under }H_0.

When σ2\sigma^2 is unknown, estimate it using

s2=1n−1(∑x2−(∑x)2n),z≈x‾−μ0s/n.s^2=\frac1{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right), \qquad z\approx\frac{\overline x-\mu_0}{s/\sqrt n}.

The CLT sample-mean form and unbiased variance estimator are in the formula booklet. Their use in a normal test with estimated standard error must be understood. The required sample size depends on the population's shape; there is no universal cutoff guaranteeing accuracy.

Begin with a large sample whose population spread is known, then change how the spread is obtained.

Worked example 3: a large sample with known spread

Question. A population of task completion times has known standard deviation 1010 seconds and an unspecified distribution. A random independent sample of 100100 times has mean 8282 seconds. Test at the 5%5\% level for an increase from the historical mean of 8080 seconds.

Plan. Use the CLT for the sample mean, then standardise with the known standard error.

Let μ\mu be the population mean time. Use H0:μ=80H_0:\mu=80, H1:μ>80H_1:\mu>80. With the large sample, under H0H_0,

X‾≈N(80,100/100)=N(80,1),z=82−8010/100=2.\overline X\approx\mathrm N(80,100/100)=\mathrm N(80,1), \qquad z=\frac{82-80}{10/\sqrt{100}}=2.

The upper-tail probability is 1−Φ(2)=0.0228<0.051-\Phi(2)=0.0228<0.05.

Answer. Reject H0H_0. The approximate test provides sufficient evidence at the 5%5\% level that the population mean completion time has increased.

Check. A mean 22 seconds above the claim is 22 standard errors above it; 2>1.6452>1.645 confirms the upper-tail decision.

Keep the sample size, mean and investigation, but suppose the population variance is unknown. Sample information must now provide the spread.

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