The equation in the image represents a two-sample t-test formula for comparing the means ($\bar{x}_1$ and $\bar{x}_2$) of two independent groups. Here's the breakdown of the formula and its computation:

Formula:

$t = \frac{\bar{x}_1 - \bar{x}_2}{S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$

Where:

1. $\bar{x}_1, \bar{x}_2$: Sample means of group 1 and group 2.
2. $n_1, n_2$: Sample sizes of group 1 and group 2.
3. $S_p$: Pooled standard deviation, which combines the variance of both groups.

Given values:

• $\bar{x}_1 = 0.55$, $\bar{x}_2 = 0.57$
• $S_p = 0.1823$ (pooled standard deviation)
• $n_1 = 5215$, $n_2 = 6071$

Substitution into the formula:

1. Numerator: $\bar{x}_1 - \bar{x}_2 = 0.55 - 0.57 = -0.02$

2. Denominator: $S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = 0.1823 \sqrt{\frac{1}{5215} + \frac{1}{6071}}$

Simplifying the square root:

$\frac{1}{n_1} = \frac{1}{5215} \approx 0.0001918, \quad \frac{1}{n_2} = \frac{1}{6071} \approx 0.0001646$ $\frac{1}{n_1} + \frac{1}{n_2} \approx 0.0001918 + 0.0001646 = 0.0003564$ $\sqrt{0.0003564} \approx 0.01888$

3. Full denominator: $S_p \cdot 0.01888 = 0.1823 \cdot 0.01888 \approx 0.00344$

4. Calculating $t$: $t = \frac{-0.02}{0.00344} \approx -5.81$

The calculated value, $t \approx -5.88$, matches the given result in the image.

Would you like me to:

1. Explain the statistical interpretation of this $t$-value?
2. Derive $S_p$ if variances for groups are given?
3. Discuss its significance in hypothesis testing?
4. Calculate the degrees of freedom?
5. Discuss when to use this two-sample t-test versus alternatives?

Tip: Always verify assumptions (e.g., normality, equal variances) when performing a t-test for reliable results.