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06_StatisticalInference/03_01_TwoGroupIntervals/index.Rmd
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,186 +1,186 @@ | ||
| --- | ||
| title : Two group intervals | ||
| subtitle : Statistical Inference | ||
| author : Brian Caffo, Jeff Leek, Roger Peng | ||
| job : Johns Hopkins Bloomberg School of Public Health | ||
| logo : bloomberg_shield.png | ||
| framework : io2012 # {io2012, html5slides, shower, dzslides, ...} | ||
| highlighter : highlight.js # {highlight.js, prettify, highlight} | ||
| hitheme : tomorrow # | ||
| url: | ||
| lib: ../../libraries | ||
| assets: ../../assets | ||
| widgets : [mathjax] # {mathjax, quiz, bootstrap} | ||
| mode : selfcontained # {standalone, draft} | ||
| --- | ||
| ```{r setup, cache = F, echo = F, message = F, warning = F, tidy = F, results='hide'} | ||
| # make this an external chunk that can be included in any file | ||
| options(width = 100) | ||
| opts_chunk$set(message = F, error = F, warning = F, comment = NA, fig.align = 'center', dpi = 100, tidy = F, cache.path = '.cache/', fig.path = 'fig/') | ||
| options(xtable.type = 'html') | ||
| knit_hooks$set(inline = function(x) { | ||
| if(is.numeric(x)) { | ||
| round(x, getOption('digits')) | ||
| } else { | ||
| paste(as.character(x), collapse = ', ') | ||
| } | ||
| }) | ||
| knit_hooks$set(plot = knitr:::hook_plot_html) | ||
| runif(1) | ||
| ``` | ||
|
|
||
| ## 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 | ||
| ```{r} | ||
| 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 | ||
| ``` | ||
|
|
||
| --- | ||
| ```{r} | ||
| 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 | ||
| t.test(x1, x2, paired = FALSE, var.equal = TRUE)$conf | ||
| t.test(x1, x2, paired = TRUE)$conf | ||
| ``` | ||
|
|
||
| --- | ||
| ## Ignoring pairing | ||
| ```{r, echo = FALSE, fig.width=5, fig.height=5} | ||
| plot(c(0.5, 2.5), range(x1, x2), type = "n", frame = FALSE, xlab = "group", ylab = "Extra", axes = FALSE) | ||
| axis(2) | ||
| axis(1, at = 1 : 2, labels = c("Group 1", "Group 2")) | ||
| for (i in 1 : n1) lines(c(1, 2), c(x1[i], x2[i]), lwd = 2, col = "red") | ||
| for (i in 1 : n1) points(c(1, 2), c(x1[i], x2[i]), lwd = 2, col = "black", bg = "salmon", pch = 21, cex = 3) | ||
| ``` | ||
|
|
||
| --- | ||
|
|
||
| ## Unequal variances | ||
|
|
||
| - Under unequal variances | ||
| $$ | ||
| \bar Y - \bar X \sim N\left(\mu_y - \mu_x, \frac{\sigma_x^2}{n_x} + \frac{\sigma_y^2}{n_y}\right) | ||
| $$ | ||
| - The statistic | ||
| $$ | ||
| \frac{\bar Y - \bar X - (\mu_y - \mu_x)}{\left(\frac{\sigma_x^2}{n_x} + \frac{\sigma_y^2}{n_y}\right)^{1/2}} | ||
| $$ | ||
| 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. | ||
| --- | ||
| title : Two group intervals | ||
| subtitle : Statistical Inference | ||
| author : Brian Caffo, Jeff Leek, Roger Peng | ||
| job : Johns Hopkins Bloomberg School of Public Health | ||
| logo : bloomberg_shield.png | ||
| framework : io2012 # {io2012, html5slides, shower, dzslides, ...} | ||
| highlighter : highlight.js # {highlight.js, prettify, highlight} | ||
| hitheme : tomorrow # | ||
| url: | ||
| lib: ../../librariesNew | ||
| assets: ../../assets | ||
| widgets : [mathjax] # {mathjax, quiz, bootstrap} | ||
| mode : selfcontained # {standalone, draft} | ||
| --- | ||
| ```{r setup, cache = F, echo = F, message = F, warning = F, tidy = F, results='hide'} | ||
| # make this an external chunk that can be included in any file | ||
| options(width = 100) | ||
| opts_chunk$set(message = F, error = F, warning = F, comment = NA, fig.align = 'center', dpi = 100, tidy = F, cache.path = '.cache/', fig.path = 'fig/') | ||
| options(xtable.type = 'html') | ||
| knit_hooks$set(inline = function(x) { | ||
| if(is.numeric(x)) { | ||
| round(x, getOption('digits')) | ||
| } else { | ||
| paste(as.character(x), collapse = ', ') | ||
| } | ||
| }) | ||
| knit_hooks$set(plot = knitr:::hook_plot_html) | ||
| runif(1) | ||
| ``` | ||
|
|
||
| ## 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 | ||
| ```{r} | ||
| 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 | ||
| ``` | ||
|
|
||
| --- | ||
| ```{r} | ||
| 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 | ||
| t.test(x1, x2, paired = FALSE, var.equal = TRUE)$conf | ||
| t.test(x1, x2, paired = TRUE)$conf | ||
| ``` | ||
|
|
||
| --- | ||
| ## Ignoring pairing | ||
| ```{r, echo = FALSE, fig.width=5, fig.height=5} | ||
| plot(c(0.5, 2.5), range(x1, x2), type = "n", frame = FALSE, xlab = "group", ylab = "Extra", axes = FALSE) | ||
| axis(2) | ||
| axis(1, at = 1 : 2, labels = c("Group 1", "Group 2")) | ||
| for (i in 1 : n1) lines(c(1, 2), c(x1[i], x2[i]), lwd = 2, col = "red") | ||
| for (i in 1 : n1) points(c(1, 2), c(x1[i], x2[i]), lwd = 2, col = "black", bg = "salmon", pch = 21, cex = 3) | ||
| ``` | ||
|
|
||
| --- | ||
|
|
||
| ## 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. |
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