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4.2 Inference using normal and t-distributions

Syllabus
9231–2028–2029
Topic
4.2
Level
AS

Learning objectives

4.2.1Hypothesis tests• formulate hypotheses and apply a hypothesis test concerning the population mean using a small sample drawn from a normal population of unknown variance, using a t-test4.2.2A pooled estimate of a population• calculate a pooled estimate of a population variance from two samples Calculations based on either raw or summarised data may be required.4.2.3Hypothesis tests• formulate hypotheses concerning the difference of population means, and apply, as appropriate - a 2-sample t-test - a paired sample t-test - a test using a normal distribution The ability to select the test appropriate to the circumstances of a problem is expected.4.2.4Mean confidence intervals• determine a confidence interval for a population mean, based on a small sample from a normal population with unknown variance, using a t-distribution4.2.5Difference confidence intervals• determine a confidence interval for a difference of population means, using a t-distribution or a normal distribution, as appropriate.

A small-sample unknown-variance mean test uses t with n minus 1 degrees of freedom

For a random sample of size $n$ from a normal population with unknown variance, test $H_0:\mu=\mu_0$ usingT=\frac{\bar X-\mu_0}{S/\sqrt n}\sim t_{n-1}\quad\text{under }H_0.The alternative $H_1$ determines whether one or both tails are critical.

A sample has $n=10$, $\bar x=52.0$ and $s=3.0$. Test $H_0:\mu=50$ against $H_1:\mu>50$ at 5%. The observed statistic ist=\frac{52-50}{3/\sqrt{10}}=2.108.With 9 degrees of freedom, the one-tail critical value is $1.833$, so reject $H_0$.

Conclude in context: there is sufficient evidence at the 5% level to suggest that the population mean exceeds 50. State the assumptions: a random sample and a normally distributed underlying population.

Hypotheses concern the population mean mu, not the sample mean. Failing to reject H0 means insufficient evidence for H1, not proof that H0 is true; statistical significance also does not establish practical importance.

Pool within-sample squared deviations when two variances estimate one common variance

For independent samples of sizes $n_1,n_2$ from populations assumed to share variance $\sigma^2$, the pooled estimate iss_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}.Equivalently, add the two within-sample sums $\sum(x-\bar x)^2$ and divide by the combined degrees of freedom.

Fromrawdata,eachwithin−samplesumcanbecalculatedasFrom raw data, each within-sample sum can be calculated as\sum(x-\bar x)^2=\sum x^2-\frac{(\sum x)^2}{n}.Dothisseparatelyforeachsample;differencesbetweenthetwosamplemeansdonotenterthepooledvariancenumerator.Do this separately for each sample; differences between the two sample means do not enter the pooled variance numerator.

If $n_1=8,s_1^2=9$ and $n_2=12,s_2^2=16$, thens_p^2=\frac{7(9)+11(16)}{18}=\frac{239}{18}=13.28,so $s_p=3.64$ approximately.

Do not average sample standard deviations or sample means. Pooling is justified only under the common population-variance model; the weights are degrees of freedom, not simply equal weights.

Choose the difference-of-means test from the sampling design and variance model

design/information analyse statistic denominator reference
paired observations differences did_i sd/ns_d/\sqrt n tn−1t_{n-1}
independent normal samples, unknown equal variances Xˉ1−Xˉ2\bar X_1-\bar X_2 sp1/n1+1/n2s_p\sqrt{1/n_1+1/n_2} tn1+n2−2t_{n_1+n_2-2}
independent samples with population variances known or a justified normal model Xˉ1−Xˉ2\bar X_1-\bar X_2 σ12/n1+σ22/n2\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2} standard normal

For paired data define one signed difference consistently. Test $H_0:\mu_d=0$ usingT=\frac{\bar d}{s_d/\sqrt n}.If $n=9$, $\bar d=2.0$ and $s_d=2.4$, then $t=2.5$ with 8 df. Against $H_1:\mu_d>0$ at 5%, $2.5>1.860$, so reject $H_0$.

For a paired t-test, the population distribution of differences must be normal; normality of each marginal sample is not the relevant statement. For the pooled two-sample t-test, the two underlying populations are normal with equal variances and the samples are independent.

Pairing is not a cosmetic label: it removes between-pair variation and changes the sample to the differences. Always define the subtraction order so the alternative, statistic sign and contextual conclusion agree.

A small-sample mean interval uses a t margin with n minus 1 degrees of freedom

For a random sample from a normal population with unknown variance, a $100(1-\alpha)\%$ confidence interval for $\mu$ is\bar x\pm t_{n-1,,1-\alpha/2}\frac{s}{\sqrt n},where $s^2$ is the unbiased sample variance.

Suppose $n=13$, $\bar x=55.6$ and $s=13.07$. For a 90% interval, $t_{12,0.95}=1.782$. The margin is1.782\frac{13.07}{\sqrt{13}}=6.46,givingapproximatelygiving approximately(49.1,62.1).Reportendpointswithsuitableaccuracyandunits.Report endpoints with suitable accuracy and units.

The method has 90% long-run coverage under its assumptions. The population mean is fixed; the interval is random before sampling. A necessary small-sample assumption is that the underlying population, not merely the sample mean, is normally distributed.

Do not use a standard normal critical value when the small-sample population variance is unknown. The interval estimates the population mean; it does not contain 90% of individual observations.

A difference interval must match the design and preserve subtraction order

model interval centre standard error and critical reference
independent normal, unknown equal variances xˉ1−xˉ2\bar x_1-\bar x_2 sp1/n1+1/n2s_p\sqrt{1/n_1+1/n_2} with tn1+n2−2t_{n_1+n_2-2}
independent, known variances/normal method xˉ1−xˉ2\bar x_1-\bar x_2 σ12/n1+σ22/n2\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2} with zz
paired dˉ\bar d sd/ns_d/\sqrt n with tn−1t_{n-1}

For independent equal-variance samples, let $n_1=10,n_2=8$, $\bar x_1-\bar x_2=5.7$ and $s_p^2=341.725$. A 95% interval uses $t_{16,0.975}=2.120$:5.7\pm2.120\sqrt{341.725\left(\frac1{10}+\frac18\right)},givingapproximatelygiving approximately(-12.9,24.3).

The interval is for mu1 minus mu2 in that order. Because zero lies inside this 95% interval, the corresponding two-sided 5% test would not reject equality of the population means. This is insufficient evidence of a difference, not proof of equality.

Do not construct separate intervals for the two means and subtract their endpoints: the standard error of a difference must be derived from the joint design. Reversing the subtraction order negates both endpoints but does not change whether zero is included.