Form and interpret hypothesis tests for a population mean

Specify the population parameter and the question

Let μ be the mean of a named target population. H₀ specifies the reference μ=μ₀; H₁ states μ<μ₀, μ>μ₀ or μ≠μ₀ according to the investigation question. The hypotheses concern μ, not the realised x̄. Choose the alternative, significance level and sampling procedure before looking at the test data

One-sided questions use a boundary reference

A claim of an increase uses H₁: μ>μ₀. We calculate with the boundary null μ=μ₀. For a known-sigma normal model with fixed variance, the same upper-tail rejection region has its largest rejection probability over μ≤μ₀ at that boundary. This justifies the usual boundary calculation for the composite no-increase null; it does not permit choosing a direction after data

A study asks in advance whether mean processing time differs from 12 minutes. Which alternative matches?

Build the statistic from the null sampling model

For iid normal observations with known positive σ, Z=(X̄−μ₀)/(σ/√n) is standard normal under H₀. With unknown σ and a suitably large iid sample, use (x̄−μ₀)/(s/√n) as an approximate normal-reference statistic when the approximation is adequate. A sample size of 30 is not a universal guarantee, and large n does not remove bias or dependence

Work a right-sided known-sigma test

Prespecify H₀: μ=50 and H₁: μ>50 at α=0.01. For an iid normal sample, σ=8, n=64 and x̄=52.4. The p-value is below 0.01, so reject H₀: there is evidence that the population mean exceeds 50 under the stated model

For an iid normal sample with known σ=8 and n=64, what is the standard error of the sample mean?

For H₀: μ=50, observed x̄=52.4 and SE=1, what is the observed z statistic?

Choose the extreme event from H₁

For a left alternative use p=Φ(zobs). For a right alternative use p=1−Φ(zobs). For a two-sided normal test use p=2Φ(−|zobs|), including both equally extreme tails. A right-sided test with a negative zobs can have p>0.5: the data points away from the prespecified alternative. Do not replace it with the smaller tail after seeing the result

With observed z=2, which null event defines a two-sided p-value?

For a right-sided test with z=−1, give p to four decimal places

The same data can answer a different prespecified question

For the data z=2.4, a two-sided alternative gives p=2[1−Φ(2.4)]≈0.0164. At α=0.01 do not reject H₀. This differs from the right-sided conclusion because the question changed; it is not permission to try both and report whichever is significant. Retain unrounded probabilities for the decision

For a two-sided normal test with observed z=2.4, give the p-value to four decimal places

Separate the decision from proof

Use the declared rule p≤α to reject H₀; otherwise do not reject. Rejection is evidence against the specified null under the model. Non-rejection means insufficient evidence at that level, not proof that μ=μ₀, absence of a meaningful effect, or acceptance that the sampling assumptions are correct. A p-value is not Pr(H₀ is true | data)

An exact supplied p-value equals α=0.05. Under the stated p≤α convention, what is the decision?

Use critical regions as the same prespecified procedure

For a right-sided exact normal test, reject when zobs≥Φ⁻¹(1−α). For a left-sided test use zobs≤−Φ⁻¹(1−α). A two-sided test splits α equally and rejects when |zobs|≥Φ⁻¹(1−α/2). Convert z boundaries to sample-mean boundaries using μ₀±cSE, retaining full precision

Work critical boundaries on the mean scale

With μ₀=50, σ=8, n=64 and α=0.05, SE=1. A right-sided test rejects when x̄≥51.6448536… . A two-sided test rejects when x̄≤48.0400360… or x̄≥51.9599640… . The central region between the two-sided thresholds is a non-rejection region under the null, not the range of individual observations

For a right-sided normal test with μ₀=50, known σ=8, n=64 and α=0.05, report the upper sample-mean rejection boundary to three decimal places

What probability is assigned to each rejection tail in a two-sided 5% test?

Use an estimated standard error honestly

Suppose H₀: μ=40, H₁: μ>40, n=100, x̄=41 and s=4. Under an adequate iid large-sample approximation, estimated SE=0.4, z=2.5 and p≈0.0062. The approximate test rejects at 5%. State the sampling and approximation conditions and that σ was estimated. This calculation does not claim an exact small-sample z law

For a sample of n=100 with sample SD s=4, give the estimated standard error of the sample mean

Statistical significance is not practical effect

The preceding observed difference is only 1 measurement unit. If the practical threshold was 2 units, statistical rejection of μ=40 does not establish that the improvement exceeds that threshold. Report the observed effect, uncertainty, units and substantive criterion. A tiny p-value can accompany a small effect in a precise large sample

What extra information is needed to judge practical importance?

Define errors using the true generating state

A Type I error rejects a true H₀. For an exact level-α test, Pr(reject | H₀ true)=α under its specified model (or at most α for a level bound). A Type II error fails to reject at a specified false mean μ₁; β(μ₁)=Pr(do not reject | μ=μ₁). Power at that mean is 1−β(μ₁). These are repeated-sampling conditional probabilities, not probabilities that the observed decision is wrong

Which denominator gives empirical Type II error in this supplied table?

Of 200 null-population trials, 10 rejected H₀. Give the empirical Type I error fraction

Of 100 trials at the specified alternative, 80 rejected H₀. Give empirical power at that alternative

In this mixture, 10 null-population trials and 80 alternative-population trials rejected H₀. What fraction of rejections came from the null population?

Power depends on the specified alternative and design

For the right-sided known-sigma test, let c=Φ⁻¹(1−α). The displayed formula gives power at a specified true mean μ₁. For μ₀=50, σ=8, n=64, α=0.05 and μ₁=52, power≈0.6388 and β≈0.3612. At μ₁=50 the rejection probability returns to 0.05. At the same positive effect, larger n reduces SE and raises power under this fixed normal model

Report a full test in context, ungraded

A random iid normal sample of 64 package masses has x̄=52.4 g and known population SD 8 g. Before collection, the investigation specified whether mean mass exceeds 50 g, at 1% significance. Write the parameter, hypotheses, assumptions, statistic, p-value, decision and contextual conclusion. Explain what has not been proved

Model: connect the calculation to the parameter

Let μ be the mean mass in the target production population. Set H₀: μ=50 g and H₁: μ>50 g as prespecified. Under iid normal observations and known σ=8 g, SE=1 g and z=(52.4−50)/1=2.4. The right-tail null probability is about 0.0081975, below 0.01. Reject H₀: the data support a mean above 50 g under these assumptions. This does not give a posterior probability for H₀, prove every package is heavier than 50 g, establish practical importance, or verify the sampling design

Derive the rejection rule and power, ungraded

For iid normal observations with known σ, derive the level-α right-sided rejection boundary for X̄. Then derive its power at a specified μ₁. Check the expression at μ₁=μ₀ and explain the role of n without claiming that every realised larger sample must reject

Model: use the correct conditioning distribution

Under H₀, (X̄−μ₀)/SE is standard normal. Choose c=Φ⁻¹(1−α), so the event X̄≥μ₀+cSE has null probability α. Under a specified μ₁, standardise that same fixed boundary by subtracting μ₁: power=Pr(Z≥c−(μ₁−μ₀)/SE)=1−Φ(c−(μ₁−μ₀)/SE). Setting μ₁=μ₀ gives α. For μ₁>μ₀ and fixed σ, increasing n decreases SE, lowers the standardised boundary and raises the probability of rejection. This is a repeated-sampling statement, not a guarantee about each new sample

Interpret conditional errors without reversing them, ungraded

In 200 null-population trials, 10 rejected H₀ and 190 did not. In 100 trials at the specified alternative, 80 rejected H₀ and 20 did not. Calculate empirical Type I error, Type II error, power and the null-origin fraction among all rejections. Explain why the last quantity is not the test significance and how the mixture of null and alternative trials affects it

Model: retain each conditioning set

Of 200 null-population trials, 10 rejected H₀ and 190 did not. Of 100 trials at the specified alternative, 80 rejected H₀ and 20 did not. The row-conditional fractions are 10/200=0.05, 20/100=0.20 and 80/100=0.80. The reverse conditional is 10/90=1/9. Its denominator combines rejections from both generating populations and depends on their mixture. The supplied counts are an empirical fixture, not exact population rates or a real-world prior over hypotheses. A Type II rate must name the alternative mean at which it is evaluated

Assess an approximate test and practical claim, ungraded

A convenience sample of 100 repeated readings from one device gives x̄=41, s=4 and an approximate right-sided z-test p-value 0.0062 against μ=40. The report claims a practically important improvement of at least 2 units and proof that H₀ is false. Critique the sampling assumptions, effect claim and conclusion

Model: separate calculation, model and substantive effect

The arithmetic estimated SE=0.4 and z=2.5 is correct for the stated summaries. Convenience sampling may not represent the target, and repeated readings may be dependent, so the nominal normal-reference p-value needs justification. Even under suitable assumptions, rejection supplies evidence rather than proof. The observed difference is 1 unit; testing μ=40 does not establish an improvement of at least 2 units. Report uncertainty and test the appropriate scientific claim using a design and analysis selected in advance

Assessment boundary

Fixed checks and practice grade defined hypotheses, calculations, decisions and conditional probabilities. Full reports, derivations, design critiques and practical judgments remain ungraded. Do not infer proof mastery from recognition answers