Pooled estimate of variance?

s^2=\frac{(nX-1)sx^2+(nY-1)sy^2}{nX+nY-2}

Test statistic under H0:\muX=\muY?

t=\frac{\bar{x}-\bar{y}}{s\sqrt{\frac1{nX}+\frac1{nY}}}

Confidence interval for \muX-\muY?

(\bar{x}-\bar{y})\pm t\,s\sqrt{\frac1{nX}+\frac1{nY}} with t the two-tailed critical value.

Which assumptions justify the pooled test?

Both populations Normal, population variances equal, and the two samples independent.

Why is a t distribution used and not a Normal one?

Because the common variance is unknown and is estimated from the samples.

Pooled variance from sums of squares?

s^2=\frac{S{xx}+S{yy}}{nX+nY-2} with S{xx}=\sum x^2-\frac{(\sum x)^2}{nX}

A confidence interval for \muX-\muY contains 0. What does that suggest?

No evidence of a difference in the means at the matching significance level.

Where should the pooled variance lie?

Between the two sample variances, closer to that of the larger sample.

Which critical value for a 95\% confidence interval?

The two-tailed 5\% point, leaving 2.5\% in each tail, of t{nX+nY-2}.

Difference between two means: pooled t-testEdexcel A-Level Further Maths: Flashcards

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When is the pooled $t$-test used?

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When is the pooled $t$ -test used?: For independent samples from Normal populations with equal but unknown variances.
Pooled estimate of variance?: $s^2=\frac{(n_X-1)s_x^2+(n_Y-1)s_y^2}{n_X+n_Y-2}$
Degrees of freedom for the pooled test?: $n_X+n_Y-2$
Test statistic under $H_0:\mu_X=\mu_Y$ ?: $t=\frac{\bar{x}-\bar{y}}{s\sqrt{\frac1{n_X}+\frac1{n_Y}}}$
Estimated standard error of $\bar{X}-\bar{Y}$ ?: $s\sqrt{\frac1{n_X}+\frac1{n_Y}}$
Confidence interval for $\mu_X-\mu_Y$ ?: $(\bar{x}-\bar{y})\pm t\,s\sqrt{\frac1{n_X}+\frac1{n_Y}}$ with $t$ the two-tailed critical value.
Which assumptions justify the pooled test?: Both populations Normal, population variances equal, and the two samples independent.
Why is a $t$ distribution used and not a Normal one?: Because the common variance is unknown and is estimated from the samples.
Pooled variance from sums of squares?: $s^2=\frac{S_{xx}+S_{yy}}{n_X+n_Y-2}$ with $S_{xx}=\sum x^2-\frac{(\sum x)^2}{n_X}$
A confidence interval for $\mu_X-\mu_Y$ contains $0$ . What does that suggest?: No evidence of a difference in the means at the matching significance level.
Where should the pooled variance lie?: Between the two sample variances, closer to that of the larger sample.
Which critical value for a $95\%$ confidence interval?: The two-tailed $5\%$ point, leaving $2.5\%$ in each tail, of $t_{n_X+n_Y-2}$ .

Exam questions on Difference between two means: pooled t-test

Independent random samples are taken from two Normal populations whose variances are equal but unknown. Sample $X$ has size $10$ and sample variance $s_x^2=3.6$ . Sample $Y$ has size $12$ and sample variance $s_y^2=6.0$ . The pooled estimate of the common variance is $s^2$ .
Find the estimated standard error of $\bar{X}-\bar{Y}$ , that is $s\sqrt{\frac{1}{10}+\frac{1}{12}}$ , to 3 significant figures.2 marks
A pooled $t$ test of $H_0:\mu_X=\mu_Y$ is carried out using independent random samples of sizes $n_X=6$ and $n_Y=9$ from two Normal populations. The sample means are $\bar{x}=52.4$ and $\bar{y}=49.1$ , and the pooled estimate of the common variance is $s^2=11.7$ . The test is two-tailed at the $5\%$ significance level.
Calculate the value of the test statistic.2 marks
Students solve the same puzzle using one of two methods. Eight students use method A, with times $x$ seconds, where $\sum x=96$ and $\sum x^2=1180$ . Ten different students use method B, with times $y$ seconds, where $\sum y=150$ and $\sum y^2=2310$ . Times for each method may be assumed to be Normally distributed with a common variance.
Find the pooled estimate of the common variance of the two populations.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).