statistics
Find the critical values to test the claim that μ1 = μ2 . Two samples are random, independent, and come from populations that are normal. The sample statistics are given below. Use α = 0.05.
$n_1 = 14,\; n_2 = 12$
$\bar X = 15,\; \bar Y = 16$
$S_1 = 2.5,\; S_2 = 2.8$
Would appreciate if anyone can give me some guidance on how to start solving this problem. Thank you very much.
To start, let's clear up two points of confusion:
(1) "[W]e use the t-distribution if the sample size is small."
Not exactly, if variances $\sigma_1^2,\, \sigma_2^2$ are unknown and estimated by $S_1^2,\, S_2^2,$ respectively, then you always use the t-distribution. (If sample sizes are large enough for degrees of freedom to exceed 30, then in some circumstances it is OK to use a normal approximation. But with modern software or printed t tables, the normal approximation is not necessary. The approximation works best for tests at the 5% level, not so well at 1%.)
(2) "[A]ssuming that the true standard deviations are not equal, ... then the degrees of freedom is given [by the Welch–Satterthwaite equation]."
No. This equation works whether or not $\sigma_1 = \sigma_2.$ However, if variances are not equal, you must use the Welch–Satterthwaite equation (not the pooled-variance equation with degrees of freedom $\nu = n_1 + n_2 - 2.)$
Pooled 2-sample t test: If data are normal and population variances are equal, then the test statistic for testing $H_0: \mu_1 = \mu_2$ against $H_a: \mu_1 \ne \mu_2$ is:
$$T = \frac{\bar X_1 - \bar X_2}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}{}},$$ where $S_p^2 =\frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1 + n_2 - 2}.$ If $H_0$ is true, then $T$ has Student's t distribution with degrees of freedom $\nu = n_1 + n_2 - 2.$
Welch 'separate variances' 2-sample t test. However, more generally if $H_0$ is true, the test statistic
$$T^\prime = \frac{\bar X_1 + \bar X_2}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}}}.$$
is approximately distributed according to Student's t distribution with degrees of freedom $\nu$ given by the Welch-Satterthwaite equation. This is true whether or not the population variances are equal.
One can show that that degrees of freedom $\nu$ according to the Welch-Satterthwaite equation satisfies $$\min(n_1 - 1, n_2 - 1) \le \nu \le n_1 + n_2 - 2.$$ So if the smaller of the two sample sizes exceeds 30, then $\nu \ge 30$ and (testing at the 5% level) it is OK to use a normal approximation for the distribution of $T^\prime.$
Whatever the sample size, $T^\prime$ has very nearly Student's t distribution with the the Welch-Satterthwaite degrees of freedom. (This is known from probability theory and from many simulation studies.)
Which to use? The bottom line is that most statisticians use the $T^\prime$-statistic and the Welch-Satterthwaite degrees of freedom to do 2-sample t tests unless they have very strong prior evidence that population variances are equal (rarely the case). Most modern statistical software packages use the Welch 2-sample t test by default. Some programs will use $T$ with the pooled SD $S_p$ if the user overrides the default.
Notes: (a) If $n_1 = n_2,$ then one can show that $T = T^\prime$ numerically, but one should still use the Welch-Satterthwaite degrees of freedom unless the population variances are known to be equal.
(b) If sample variances $S_1^2$ and $S_2^2$ are nearly equal, then the Welch-Satterthwaite $\nu$ is near $n_1 + n_2 - 2.$ If the sample variances are far apart then $\nu$ may be considerably smaller---perhaps as small as $n_1 -1$ or $n_2 - 1.$
(c) Especially if $n_1 << n_2$ and $\sigma_2 << \sigma_1,$ then results from the pooled 2-sample test using $T$ and $S_p$ can be very misleading. (The notation $<<$ means 'much smaller than'.)
(d) It is not a good idea to test whether $\sigma_1^2 = \sigma_2^2$ in order to decide whether to use $T$ or $T^\prime.$ The test for equal variances has poor power, and simulation studies have shown that the 'hybrid' test (using $T^\prime$ only if the equal-variances test rejects) can give misleading results.
Demonstration of note (c). Using R statistical software:
Small sample from $\mathsf{Norm}(\mu_1=150,\sigma_1=30);$ larger sample from $\mathsf{Norm}(\mu_2=150,\sigma_2=5.)$ The null hypothesis is true, and so should not be rejected.
x1 = rnorm(10, 150, 30); x2 = rnorm(50, 150, 5)
mean(x1); sd(x1)
[1] 139.3158
[1] 31.34551
mean(x2); sd(x2)
[1] 150.1088
[1] 5.246149
Welch 2-sample test properly fails to reject:
t.test(x1, x2)
Welch Two Sample t-test
data: x1 and x2
t = -1.0858, df = 9.1011, p-value = 0.3055
alternative hypothesis: true difference in means is not equal to 0
sample estimates:
mean of x mean of y
139.3158 150.1088
Pooled two-sample t test improperly rejects at the 5% level, 'finding' a difference in population means that does not actually exist. (The small sample with the large SD gives a misleading sample mean.)
t.test(x1, x2, var.eq=T)
Two Sample t-test
data: x1 and x2
t = -2.3504, df = 58, p-value = 0.02217
alternative hypothesis: true difference in means is not equal to 0
sample estimates:
mean of x mean of y
139.3158 150.1088
The central limit theorem tells you that $\dfrac{\overline X_i - \mu_i}{\sigma_i/\sqrt{n_i}}$ is approximately normally distributed for $i=1,2.$
Without the central limit theorem you know that this random variable has expected value $0$ and variance $1.$
If $\operatorname{var}(\overline X_i-\mu_i) = \sigma_i^2/n_i$ for $i=1,2,$ then $\operatorname{var}\big((\overline X_1 - \overline X_2) - (\mu_1 - \mu_2)\big) = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}$ (and you don't need C.L.T. for that).
And the sum of two independent approximately normally distributed random variables is approximately normally distributed.
Best Answer
There are two versions of the two-sample t test. (1) The pooled test makes the assumption that the two population variances are equal, computes the 'pooled' variance estimate as $$S_p^2 = \frac{(n_1 - 1)S_1^2 + (n_2-1)S_2^2}{n_1 + n_2 -2},$$ and uses the statistic $$T = \frac{\bar X_1 - \bar X_2}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}},$$ which has Student's t distribution with $n_1 + n_2 -2$ degrees of freedom under the null hypothesis $H_0: \mu_1 = \mu_2.$
(2) Because the pooled test can perform quite badly if the two population variances are not equal, most practitioners prefer to routinely use the Welch (separate variances) test which does not assume equal variances and uses the test statistic $$T^\prime = \frac{\bar X_1 - \bar X_2}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}},$$ which has approximately Student's t distribution with degrees of freedom $\nu$ found according to a formula. The result of the formula is that $\min(n_1 - 1, n_2 - 1) \le \nu \le n_1 + n_2 - 2,$ with $\nu$ nearer the larger value if $S_1^2 \approx S_2^2$ and nearer the smaller value if the sample variances differ greatly. If the Welch test is not discussed in your textbook, you can read about it and find the formula for the degrees of freedom on Wikipedia.
The bad behavior of the pooled t test can be especially serious if sample sizes are unequal and the smaller sample has the larger variance.
Results from Minitab for the pooled test are as follows:
Because the P-value exceeds 5%, you would not reject the null hypothesis at the 5% level of significance. (This P-value is for a two-sided test; that is the alternative hypothesis is $H_a: \mu_1 \ne \mu_2.)$ I will leave it to you to consult a printed t table to find the critical value for a test at the 5% level, verify the value of $S_p.$ and to do the other intermediate computations required to find the $T$ statistic.
For the Welch test the Minitab output is as follows:
Notice that $\nu = 22;$ you should use the formula to verify this result. Then find the critical value for a 5% level test, and perform the intermediate computations necessary to get $T^\prime.$