Independent group \( t \) confidence intervals

• Suppose that we want to compare the mean blood pressure between two groups in a randomized trial; those who received the treatment to those who received a placebo
• We cannot use the paired t test because the groups are independent and may have different sample sizes
• We now present methods for comparing independent groups

Notation

• Let \( X_1,\ldots,X_{n_x} \) be iid \( N(\mu_x,\sigma^2) \)
• Let \( Y_1,\ldots,Y_{n_y} \) be iid \( N(\mu_y, \sigma^2) \)
• Let \( \bar X \), \( \bar Y \), \( S_x \), \( S_y \) be the means and standard deviations
• Using the fact that linear combinations of normals are again normal, we know that \( \bar Y - \bar X \) is also normal with mean \( \mu_y - \mu_x \) and variance \( \sigma^2 (\frac{1}{n_x} + \frac{1}{n_y}) \)
• The pooled variance estimator \[ S_p^2 = \{(n_x - 1) S_x^2 + (n_y - 1) S_y^2\}/(n_x + n_y - 2) \] is a good estimator of \( \sigma^2 \)

Note

• The pooled estimator is a mixture of the group variances, placing greater weight on whichever has a larger sample size
• If the sample sizes are the same the pooled variance estimate is the average of the group variances
• The pooled estimator is unbiased \[ \begin{eqnarray*} E[S_p^2] & = & \frac{(n_x - 1) E[S_x^2] + (n_y - 1) E[S_y^2]}{n_x + n_y - 2}\\ & = & \frac{(n_x - 1)\sigma^2 + (n_y - 1)\sigma^2}{n_x + n_y - 2} \end{eqnarray*} \]
• The pooled variance estimate is independent of \( \bar Y - \bar X \) since \( S_x \) is independent of \( \bar X \) and \( S_y \) is independent of \( \bar Y \) and the groups are independent

Result

• The sum of two independent Chi-squared random variables is Chi-squared with degrees of freedom equal to the sum of the degrees of freedom of the summands
• Therefore \[ \begin{eqnarray*} (n_x + n_y - 2) S_p^2 / \sigma^2 & = & (n_x - 1)S_x^2 /\sigma^2 + (n_y - 1)S_y^2/\sigma^2 \\ \\ & = & \chi^2_{n_x - 1} + \chi^2_{n_y-1} \\ \\ & = & \chi^2_{n_x + n_y - 2} \end{eqnarray*} \]

Putting this all together

• The statistic \[ \frac{\frac{\bar Y - \bar X - (\mu_y - \mu_x)}{\sigma \left(\frac{1}{n_x} + \frac{1}{n_y}\right)^{1/2}}}% {\sqrt{\frac{(n_x + n_y - 2) S_p^2}{(n_x + n_y - 2)\sigma^2}}} = \frac{\bar Y - \bar X - (\mu_y - \mu_x)}{S_p \left(\frac{1}{n_x} + \frac{1}{n_y}\right)^{1/2}} \] is a standard normal divided by the square root of an independent Chi-squared divided by its degrees of freedom
• Therefore this statistic follows Gosset's \( t \) distribution with \( n_x + n_y - 2 \) degrees of freedom
• Notice the form is (estimator - true value) / SE

Confidence interval

• Therefore a \( (1 - \alpha)\times 100\% \) confidence interval for \( \mu_y - \mu_x \) is \[ \bar Y - \bar X \pm t_{n_x + n_y - 2, 1 - \alpha/2}S_p\left(\frac{1}{n_x} + \frac{1}{n_y}\right)^{1/2} \]
• Remember this interval is assuming a constant variance across the two groups
• If there is some doubt, assume a different variance per group, which we will discuss later

Example

Based on Rosner, Fundamentals of Biostatistics

• Comparing SBP for 8 oral contraceptive users versus 21 controls
• \( \bar X_{OC} = 132.86 \) mmHg with \( s_{OC} = 15.34 \) mmHg
• \( \bar X_{C} = 127.44 \) mmHg with \( s_{C} = 18.23 \) mmHg
• Pooled variance estimate

sp <- sqrt((7 * 15.34^2 + 20 * 18.23^2) / (8 + 21 - 2))
132.86 - 127.44 + c(-1, 1) * qt(.975, 27) * sp * (1 / 8 + 1 / 21)^.5

[1] -9.521 20.361

data(sleep)
x1 <- sleep$extra[sleep$group == 1]
x2 <- sleep$extra[sleep$group == 2]
n1 <- length(x1)
n2 <- length(x2)
sp <- sqrt( ((n1 - 1) * sd(x1)^2 + (n2-1) * sd(x2)^2) / (n1 + n2-2))
md <- mean(x1) - mean(x2)
semd <- sp * sqrt(1 / n1 + 1/n2)
md + c(-1, 1) * qt(.975, n1 + n2 - 2) * semd

[1] -3.3639  0.2039

t.test(x1, x2, paired = FALSE, var.equal = TRUE)$conf

[1] -3.3639  0.2039
attr(,"conf.level")
[1] 0.95

t.test(x1, x2, paired = TRUE)$conf

[1] -2.4599 -0.7001
attr(,"conf.level")
[1] 0.95

Ignoring pairing

Unequal variances

• Under unequal variances \[ \bar Y - \bar X \sim N\left(\mu_y - \mu_x, \frac{s_x^2}{n_x} + \frac{\sigma_y^2}{n_y}\right) \]
• The statistic \[ \frac{\bar Y - \bar X - (\mu_y - \mu_x)}{\left(\frac{s_x^2}{n_x} + \frac{\sigma_y^2}{n_y}\right)^{1/2}} \] approximately follows Gosset's \( t \) distribution with degrees of freedom equal to \[ \frac{\left(S_x^2 / n_x + S_y^2/n_y\right)^2} {\left(\frac{S_x^2}{n_x}\right)^2 / (n_x - 1) + \left(\frac{S_y^2}{n_y}\right)^2 / (n_y - 1)} \]

Example

• Comparing SBP for 8 oral contraceptive users versus 21 controls
• \( \bar X_{OC} = 132.86 \) mmHg with \( s_{OC} = 15.34 \) mmHg
• \( \bar X_{C} = 127.44 \) mmHg with \( s_{C} = 18.23 \) mmHg
• \( df=15.04 \), \( t_{15.04, .975} = 2.13 \)
• Interval \[ 132.86 - 127.44 \pm 2.13 \left(\frac{15.34^2}{8} + \frac{18.23^2}{21} \right)^{1/2} = [-8.91, 19.75] \]
• In R, t.test(..., var.equal = FALSE)

Comparing other kinds of data

• For binomial data, there's lots of ways to compare two groups
  • Relative risk, risk difference, odds ratio.
  • Chi-squared tests, normal approximations, exact tests.
• For count data, there's also Chi-squared tests and exact tests.
• We'll leave the discussions for comparing groups of data for binary and count data until covering glms in the regression class.
• In addition, Mathematical Biostatistics Boot Camp 2 covers many special cases relevant to biostatistics.