Setting Up a Test for a Mean · Mise en place d'un test pour une moyenne
A fill label gives a claim to test
- A fictional random sample of 16 containers has mean fill 496 mL and sample standard deviation 8 mL. The label claims a population mean of 500 mL.
- The question is whether the population mean is below 500 mL. The difference in sample means alone does not settle that question.
Write hypotheses about the population
- Set · Définir $H_0:\mu=500$ and the preplanned lower-sided alternative $H_a:\mu<500$. The symbol $\mu$ names the population mean.
- Do not put the observed sample mean into a null hypothesis about the population. Do not change the direction because the sample happens to fall below the label.
The null hypothesis for a one-mean test has the form...
H0 states a specific claimed mean μ0.
Match the preplanned question to its alternative.
Choose the direction from the research question before inspecting the sample.
Check when this reference model is useful
- Check random sampling and independence, including the 10% condition when sampling without replacement. With 16 observations, inspect skew and outliers carefully.
- The t model is exact for independent normal observations and can be approximate in suitable other cases. Passing an arithmetic check does not establish the sampling assumptions.
μ0 = 500, x-bar = 496, s = 8, n = 16. Compute t = (x-bar − μ0)/(s/√n).
(496−500)/(8/4) = −4/2 = −2.
Standardize using sample variability
- Use $t=(\bar x-\mu_0)/(s/\sqrt n)$. Here $SE=8/\sqrt{16}=2$ mL.
- The observed statistic is $t=(496-500)/2=-2$. The t reference distribution has $df=16-1=15$.
With the same t = -2 and df = 15, a two-sided alternative has p ≈ 0.0639. Its decision at 0.05 differs from the preplanned lower-sided test.
For n = 16, what are the degrees of freedom for the one-sample t-test?
df = n − 1 = 15.
Select the tail that answers the question
- The lower-sided alternative uses the area below $t=-2$ in the t distribution with 15 degrees of freedom. It is approximately 0.0320.
- The standard normal left-tail area at -2 is about 0.0228, a different value. Substituting a normal plot for this t reference changes the displayed evidence.
The reference curve, degrees of freedom and tail rule must match the selected procedure.
For a mean with unknown σ, you should use a z-test if n is large.
For the normal-theory mean procedure taught here, use t with sample SD, including at large n; the t curve then approaches normal.
For df = 15 and t = -2, the t and standard-normal left-tail probabilities are identical.
The t left tail is about 0.0320; the standard-normal left tail is about 0.0228.
Report evidence rather than certainty
- At a prechosen significance level of 0.05, the approximate p-value 0.0320 leads to rejection of the 500 mL null claim in favour of a lower population mean.
- State the population, design, statistic, degrees of freedom, p-value and decision. The conclusion is not that every container is underfilled or that the true mean is certainly exactly 496 mL.
At a prechosen significance level of 0.05, the approximate p-value 0.0320 leads to rejection of the 500 mL null claim in favour of a lower population mean.
The denominator of the one-sample t statistic is...
Use the sample SD over √n; σ is unknown.
The lower-sided test rejects the claim μ = 500 mL. Which statements are justified?
The test concerns a population mean with uncertainty. It does not determine each individual fill or an exact population mean.