This is the formula for calculating the two-sample t-test statistic in hypothesis testing. Let's break it down step by step:

The general formula is:

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

Where:

1. $\bar{x}_1$ and $\bar{x}_2$ are the sample means of the two groups.
2. $n_1$ and $n_2$ are the sample sizes of the two groups.
3. $S_p$ is the pooled standard deviation, which accounts for variability across both groups.

Step-by-step explanation of the provided calculation:

• Numerator: The difference in sample means: $0.55 - 0.57 = -0.02$.

• Denominator:

  • Pooled standard deviation: $S_p = 0.1823$ (given).
  • Scaling factor for sample sizes: $\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} = \sqrt{\frac{1}{5215} + \frac{1}{6071}}$.
    • $\frac{1}{5215} \approx 0.0001917$, and $\frac{1}{6071} \approx 0.0001647$, so: $\frac{1}{5215} + \frac{1}{6071} \approx 0.0003564$
    • Taking the square root: $\sqrt{0.0003564} \approx 0.01888$.
  • The denominator becomes: $0.1823 \cdot 0.01888 \approx 0.00344$.

• The t-statistic: $t = \frac{-0.02}{0.00344} \approx -5.81$

The calculation in the image gives $-5.88$, which may involve additional rounding adjustments in intermediate steps.

Interpretation of $t = -5.88$:

• A very large magnitude of $t$ (whether positive or negative) indicates a significant difference between the two sample means.
• To determine the statistical significance, compare $t = -5.88$ to the critical $t$-value from the $t$-distribution table (based on degrees of freedom).

Would you like me to:

1. Explain the pooled standard deviation formula in detail?
2. Show how to find the critical $t$-value?
3. Discuss hypothesis testing steps in more depth?
4. Provide an example with new sample data?
5. Illustrate a real-life application of the two-sample $t$-test?

Tip:

Always check assumptions for a $t$-test: normality, equal variances, and independent samples.