EDUC/PSY 7610

Chapter 6 talks about experimental and statistical control. The following examples help illustrate a few items that were discussed. Follow all instructions to complete Chapter 6.

- Let’s start by loading the
`tidyverse`

package (you can ignore the notes that you see below that it gives you once you load it) and the`furniture`

package.

```
library(tidyverse)
library(furniture)
```

- We are going to use a ficticious, experimental data set that is inputted below.
`posttest`

is the posttest scores regarding words recognized accurately from a person with a motor speech disorder;`pretest`

is the initial accurately recognized words;`therapy`

is the experimental group where`1`

is the intervention group and`0`

is the control group.

```
## Don't change this code :)
set.seed(843)
df <- data_frame(
pretest = c(2,4,6,6,9,10,12, 6,7,9,9,12,12,15),
posttest = c(1,3,7,10,13,17,19, 1,5,7,9,13,16,19),
therapy = c(1,1,1,1,1,1,1, 0,0,0,0,0,0,0)
) %>%
mutate(gain = posttest - pretest)
```

- Let’s take a look at this visually.

```
df %>%
mutate(therapy = factor(therapy, labels = c("No Therapy", "Therapy"))) %>%
ggplot(aes(pretest, posttest, group = therapy, color = therapy)) +
geom_point() +
geom_smooth(method = "lm", se = FALSE) +
scale_color_manual(values = c("darkorchid", "firebrick1"))
```

- Let’s use a t-test to assess if there are differences between the therapy group in the gain scores. Is this difference significant?

```
df %>%
t.test(gain ~ therapy,
data = .,
var.equal = TRUE)
```

```
##
## Two Sample t-test
##
## data: gain by therapy
## t = -1.6672, df = 12, p-value = 0.1213
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -6.9207101 0.9207101
## sample estimates:
## mean in group 0 mean in group 1
## 0 3
```

- We could do the same analysis using regression with
`gain`

and`therapy`

.

```
df %>%
lm(gain ~ therapy,
data = .) %>%
summary()
```

```
##
## Call:
## lm(formula = gain ~ therapy, data = .)
##
## Residuals:
## Min 1Q Median 3Q Max
## -5.00 -2.00 0.50 3.25 4.00
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.121e-16 1.272e+00 0.000 1.000
## therapy 3.000e+00 1.799e+00 1.667 0.121
##
## Residual standard error: 3.367 on 12 degrees of freedom
## Multiple R-squared: 0.1881, Adjusted R-squared: 0.1204
## F-statistic: 2.779 on 1 and 12 DF, p-value: 0.1213
```

- Since we have information on the pretest, we could use that to increase the precision of our estimates. We can do that by using multiple regression with
`pretest`

as a covariate. What did it do to the estimate? Why?

```
df %>%
lm(gain ~ therapy + pretest,
data = .) %>%
summary()
```

```
##
## Call:
## lm(formula = gain ~ therapy + pretest, data = .)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.2687 -1.0168 -0.6642 0.8992 2.1343
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -9.3284 1.2521 -7.450 1.28e-05 ***
## therapy 5.7985 0.7888 7.351 1.45e-05 ***
## pretest 0.9328 0.1147 8.132 5.59e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.328 on 11 degrees of freedom
## Multiple R-squared: 0.8842, Adjusted R-squared: 0.8632
## F-statistic: 42 on 2 and 11 DF, p-value: 7.084e-06
```

- Did it increase the precision of the estimate on
`therapy`

? - Can random assignment make groups that aren’t equal? If so, what can be done to help?

Regression is well adapted for both experimental and observational research designs. Using both experimental and statistical controls within the same design can increase validity and statistical power of the analyses.