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Inferential statistics — chapter review — 9231 FPS

Use sample variance and the appropriate t degrees of freedom. Estimate a common variance and compare two independent means. Reduce a paired design to a one-sample analysis of differences.

Notes and key methods

For independent observations from a normal population of unknown variance, \(\displaystyle t= \frac{\bar{x}- \mu_{0}}{\frac{s}{\sqrt{n}}}\) has n−1 degrees of freedom under \(\displaystyle H_{0}\). The sample variance \(\displaystyle s^{2}\) uses denominator n−1.

A confidence interval uses \(\displaystyle \bar{x}\pm \frac{t_{n- 1, 1- \frac{\alpha }{2}}s}{\sqrt{n}}\). The t distribution allows uncertainty from estimating σ; replacing its quantile by z can give an interval that is too narrow for a small sample.

For independent normal samples with equal unknown population variances, pool the unbiased sample variances: \(\displaystyle s_{p}^{2}= \frac{[(n_{1}- 1)s_{1}^{2}+ (n_{2}- 1)s_{2}^{2}]}{n_{1}+ n_{2}- 2}\).

The standard error of the mean difference is \(\displaystyle s_{p}\sqrt{\frac{1}{n_{1}}+ \frac{1}{n_{2}}}\), and the pooled t statistic uses \(\displaystyle n_{1}+n_{2}-2\) degrees of freedom. A confidence interval for the difference uses the same standard error.

For matched observations form differences in a stated order, such as after minus before. Apply a one-sample t procedure to the differences: \(\displaystyle t= \frac{\bar{d}- \delta_{0}}{\frac{s_{d}}{\sqrt{n}}}\), with n−1 degrees of freedom.

The pairs must be independent of one another; the two observations within a pair need not be independent. The normality assumption concerns the population of differences, rather than the separate before/after measurements.

Check the conditions

Three worked examples

Example 1 · One-sample t inference

A random sample of 9 independent normal observations has mean 25.5 and unbiased sample standard deviation 5. Test \(H_0:\mu=23\) against an increase; the population variance is unknown.

Use a 5% significance level for a written conclusion. A distribution calculator may be used for numerical p-values.

  1. Find the observed t statistic.
  2. Find the degrees of freedom.
  3. Find the one-sided p-value.
  4. Find the half-width of a two-sided 95% t confidence interval.
Show the reasoning

(a)

\(\displaystyle t= \frac{\bar{x}- \mu_{0}}{\frac{s}{\sqrt{n}}}= 1.5\)

Answer: \(1.5\).

(b)

\(n-1=8\)

Answer: \(8\).

(c)

\(P(T_{8}\ge 1.5)=0.0860016459759556\)

Answer: \(0.0860016459759556\).

(d)

\(\displaystyle \frac{t_{n- 1, 0.975}s}{\sqrt{n}}= 3.84334022534028\)

Answer: \(3.84334022534028\).

The p-value is \(0.0860016459759556\). Since it exceeds 0.05, do not reject \(\displaystyle H_{0}\): the sample gives insufficient evidence that the population mean exceeds its null value. This does not give the probability that \(\displaystyle H_{0}\) is true.

Example 2 · Two-sample pooled t inference

Two independent normal populations have a common unknown variance. Sample sizes are 7 and 12, their unbiased variances are 4 and 6, and \(\displaystyle \bar x_1-\bar x_2=\frac{7}{2}\).

Use \(\displaystyle H_{1}: \mu _{1}>\mu _{2}\) and a 5% significance level for the written test conclusion.

  1. Find the pooled variance estimate.
  2. Find the standard error of the mean difference.
  3. Find the t statistic for equal population means.
  4. Find the half-width of a two-sided 95% confidence interval for the mean difference.
Show the reasoning

(a)

\(\displaystyle s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}=\frac{90}{17}\)

Answer: \(\displaystyle \frac{90}{17}\).

(b)

\(\displaystyle \sqrt{s_{p}^{2}(\frac{1}{n_{1}}+ \frac{1}{n_{2}})}= 1.09429383238536\)

Answer: \(1.09429383238536\).

(c)

\(\displaystyle t= \frac{\bar{x}_{1}- \bar{x}_{2}}{S}E= 3.1984096925509\)

Answer: \(3.1984096925509\).

(d)

\(t_{17,0.975}SE=2.30875817429355\)

Answer: \(2.30875817429355\).

With 17 degrees of freedom the upper-tail p-value is \(0.00263277157273379\). Reject \(\displaystyle H_{0}\): the first mean is greater.

Example 3 · Paired t-tests

For six independent matched pairs, after-minus-before differences are \(2,8,11,17,20,26\). Assume that population differences are normal. Test a positive mean difference.

Use a 5% significance level for a written conclusion. A distribution calculator may be used for numerical p-values.

  1. Find the mean difference.
  2. Find the unbiased variance of the differences.
  3. Find the t statistic for null mean difference zero.
  4. Find the one-sided p-value.
Show the reasoning

(a)

\(\displaystyle \bar{d}= \sum \frac{d}{6}= 14\)

Answer: \(14\).

(b)

\(\displaystyle s_{d}^{2}= \sum \frac{(d- \bar{d})^{2}}{5}= \frac{378}{5}\)

Answer: \(\displaystyle \frac{378}{5}\).

(c)

\(\displaystyle t= \frac{\bar{d}}{\frac{s_{d}}{\sqrt{6}}}= 3.94405318873308\)

Answer: \(3.94405318873308\).

(d)

\(P(T_5\ge 3.94405319)=0.0054577501915573\)

Answer: \(0.0054577501915573\).

The p-value is \(0.0054577501915573\). Since it is at most 0.05, reject \(\displaystyle H_{0}\): there is evidence that the mean paired difference is positive. This does not give the probability that \(\displaystyle H_{0}\) is true.

Common mistakes

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