SH6 - Hypothesis tests for a mean

SH6 - Hypothesis tests for a mean

SH6 tests a single population mean using a Normal or tt reference distribution. You will choose the distribution from the sampling conditions, distinguish one- and two-sided alternatives, use the right critical-value convention, and state what the evidence suggests in context.

Choose the question and model before calculating

Can the same sample give a significant one-sided result but a non-significant two-sided result? Yes. At the 5%5\% level, a positive standardised statistic 1.81.8 exceeds the one-sided Normal cutoff 1.6451.645, but not the two-sided cutoff 1.9601.960.

This is not permission to choose the alternative after seeing the data. The alternative must reflect the question you set out to investigate.

You already know

The sampling distribution concerns the sample mean. Standard error is σ/n\sigma/\sqrt n, estimated by s/ns/\sqrt n when necessary. Small Normal samples with unknown variance use tn−1t_{n-1}, as in confidence intervals.

Define μ\mu as the population mean of the measurement in context. Use H0:μ=μ0H_0:\mu=\mu_0, and choose H1:μ>μ0H_1:\mu>\mu_0, μ<μ0\mu<\mu_0, or μ≠μ0\mu\ne\mu_0 according to the claim. A hypothesis about xˉ\bar x would concern a number already observed.

Sampling conditionsReference distribution
Normal population, known population varianceNormal, exact under the model, at any sample size
Large independent sample, finite varianceNormal approximation, using s2s^2 if the population variance is unknown
Small independent sample, Normal population, unknown variancett with n−1n-1 degrees of freedom

For a small sample with no justified Normal model, a tt test is not automatically appropriate. A signed rank test addresses a median and needs symmetry; it cannot be substituted without checking what parameter the question asks about.

Watch out

Wrong: “The population is Normal, so use a Normal test even though its variance is unknown and the sample is small.”

Right: use tt in that case. Population Normality and knowledge of the population variance are separate conditions.

Choose a reference distribution for a specified design.

Standardise a mean using the null hypothesis

The null model tells you how surprising the observed mean would be if the proposed population mean were correct. Use μ0\mu_0, not a fitted alternative mean, in the numerator.

Normal test statistic for a mean

For a Normal sampling distribution with known standard error, under H0:μ=μ0H_0:\mu=\mu_0,

Z=X‾−μ0σ/n∼N(0,1)Z=\frac{\overline X-\mu_0}{\sigma/\sqrt n}\sim N(0,1)

Calculate the observed statistic using xˉ\bar x. For an exact small-sample Normal test, the underlying population must be Normal and σ\sigma known.

In the formula booklet.

The booklet's percentage-point table gives total two-tail percentages. For a one-sided 5%5\% upper test, use 1.6451.645, from its 10%10\% total-tail column. For a two-sided 5%5\% test, use ±1.960\pm1.960, from its 5%5\% column. Reverse the sign for a lower-sided test.

Worked example 1: a one-sided Normal test

Question. A Normal population of instrument readings has known standard deviation 88. A random independent sample of 1616 readings has mean 103.6103.6. Test at the 5%5\% level whether the population mean exceeds 100100.

Plan. Standardise using the known spread and compare with the upper one-sided cutoff.

H0:μ=100,H1:μ>100μ is the population mean readingSE⁡=816=2known spreadz=103.6−1002=1.8null-centred statistic1.8>1.645upper 5% rejection region\begin{aligned} H_0&:\mu=100,\qquad H_1:\mu>100 && \mu\text{ is the population mean reading}\\ \operatorname{SE}&=\frac8{\sqrt{16}}=2 && \text{known spread}\\ z&=\frac{103.6-100}{2}=1.8 && \text{null-centred statistic}\\ 1.8&>1.645 && \text{upper }5\%\text{ rejection region} \end{aligned}

Why this step? A larger-mean alternative asks whether the observed mean is unusually far above the null value, so only the upper tail counts.

Answer. Reject H0H_0. There is sufficient evidence at the 5%5\% level to suggest that the population mean reading exceeds 100100.

Check. The one-sided tail probability is approximately 0.0359<0.050.0359<0.05, agreeing with the critical-value decision.

Change only the alternative: suppose the investigation asks whether the mean differs from 100100.

Worked example 2: the same statistic in a two-sided test

Question. For the same sample, test H0:μ=100H_0:\mu=100 against H1:μ≠100H_1:\mu\ne100 at the 5%5\% level.

Plan. Retain the statistic 1.81.8, but split the significance level between both tails.

z=1.8same data and null modelcritical values=±1.9602.5% in each tail∣1.8∣<1.960not in either rejection tail\begin{aligned} z&=1.8 && \text{same data and null model}\\ \text{critical values}&=\pm1.960 && 2.5\%\text{ in each tail}\\ |1.8|&<1.960 && \text{not in either rejection tail} \end{aligned}

Why this step? A two-sided test must allow for a discrepancy in either direction, so its upper cutoff is further from zero.

Answer. Do not reject H0H_0. There is insufficient evidence at the 5%5\% level to suggest that the population mean differs from 100100.

Check. The two-sided pp-value is about 2(0.0359)=0.0719>0.052(0.0359)=0.0719>0.05.

[DIAGRAM: asset_name: One statistic against one-sided and two-sided cutoffs; asset_slug: h645_b107_tails; caption: A statistic of one point eight is in the one-sided five percent rejection region but outside the two-sided five percent regions; description: Two standard Normal curves show upper five percent shading beyond one point six four five in the first panel, and two point five percent shading beyond plus and minus one point nine six in the second. A vertical clay line marks statistic one point eight in both, with its position relative to the cutoffs explicitly labelled.]
A statistic of one point eight is in the one-sided five percent rejection region but outside the two-sided five percent regions

For a lower-sided claim use the lower tail: at the 5%5\% level reject for z<−1.645z<-1.645. A positive statistic cannot support a lower-sided alternative just because its absolute value is large.

Exam tip

Show the hypotheses, population parameter definition, statistic, reference distribution, correct tail cutoff, comparison and non-assertive conclusion. “Accept H0H_0” must not become a claim that the proposed mean has been proved.

Keep a two-sided investigation, but use a new observed mean.

Watch out

Wrong: reject the two-sided 5%5\% test because 1.8>1.6451.8>1.645. That cutoff allocates 5%5\% to just the upper tail. Correct: use ∣z∣>1.960|z|>1.960 for the two-sided test; 1.8<1.9601.8<1.960, so do not reject.

Match the next alternative to its tail cutoff before comparing the statistic.

A large sample allows an estimated Normal standard error

When the population variance is unknown but the independent sample is sufficiently large, use the CLT for the mean and estimate its standard error by s/ns/\sqrt n. This is an approximate Normal procedure, rather than the exact known-variance calculation.

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