Perform an independent t-test

In this exercise, you'll manually perform an independent t-test the same way you did for the dependent t-test in the previous chapter. Continuing with the working memory example, our null hypothesis is that the difference in intelligence score gain between the group that trained for 8 days and the group that trained for 19 days is equal to zero. If our observed t-value is sufficiently large, we can reject the null in favor of the alternative hypothesis, which would imply a significant difference in intelligence gain between the two training groups.

Calculation of the observed t-value for an independent t-test is similar to the dependent t-test, but involves slightly different formulas. The t-value is now

$$t = \frac{(\bar{x_1} - \bar{x_2})}{se_p}$$

where $\bar{x_1}$ and $\bar{x_1}$ are the mean intelligence gains for group 1 and group 2, respectively. $se_p$ is the pooled standard error, which is equivalent to

$$se_p = \sqrt{ \frac{var_1}{n_1} + \frac{var_2}{n_2} }$$

where $var_1$ and $var_2$ are the variances and $n_1$ and $n_2$ are the sample sizes of groups 1 and 2, respectively.

This exercise is part of the course

Intro to Statistics with R: Student's T-test

View Course

Exercise instructions

The subsets of data for both 8 and 19 days of training, wm_t08 and wm_t19, are available in your workspace. Recall that the gain column contains the gain in intelligence score before and after training for each subject.

Compute the mean intelligence score gain for each of the two training groups and store the results in mean_t08 and mean_t19. Use the mean() function to do this.
Use the objects mean_t08 and mean_t19 to find the difference in means. Subtract the lowest mean from the highest mean and store the result in mean_diff.
Determine the number of participants in each sample using the code provided for you. Nothing to change here.
Calculate the degrees of freedom for the relevant t-distribution. The formula for degrees of freedom in an independent t-test is $n1 + n2 - 2$. Use the sample sizes you created above and save the result to df.
Create var_t08 and var_t19 by applying the var() function to the intelligence gain for each of the two training groups. These objects represent the variance for each group.
Compute the pooled standard error se_pooled using the formula outlined above. You will need to use the sqrt() function in addition to the other variables you've created in this exercise. Mind your brackets!

Hands-on interactive exercise

Have a go at this exercise by completing this sample code.

# The subsets wm_t08 and wm_t19 are still loaded in the console

# Find the mean intelligence gain for both the 8 and 19 training day group
mean_t08 <- ___
mean_t19 <- ___

# Calculate mean difference by subtracting t08 by t19
mean_diff <- ___

# Determine the number of subjects in each sample
n_t08 <- nrow(wm_t08)
n_t19 <- nrow(wm_t19)

# Calculate degrees of freedom
df <- ___

# Calculate variance for each group
var_t08 <- ___
var_t19 <- ___

# Compute pooled standard error
se_pooled <- ___

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