Calculate variance for A/B test results

Question

Given the following scenario about revenue per session, calculate the the p-value.

Codemia · Accepted Answer

In this scenario, we are tasked with calculating the p-value for an A/B test examining the revenue per session. We will set up the following hypotheses:

- **Null Hypothesis (H0)**: There is no significant difference in revenue per session between Group A (control) and Group B (treatment).
- **Alternative Hypothesis (H1)**: There is a significant difference in revenue per session between Group A and Group B.

To calculate the p-value, we will use the results from the A/B test, specifically the means and standard deviations of the revenue per session for both groups, along with their sample sizes.

The p-value can be calculated using the following steps:
1. Collect the sample data for Group A and Group B, including the sample means (μ_A, μ_B), sample standard deviations (s_A, s_B), and sample sizes (n_A, n_B).
2. Calculate the test statistic (t) using the formula:
   
   $$ t = \frac{\mu_B - \mu_A}{\sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}} $$
3. Determine the degrees of freedom (df) using:
   
   $$ df = n_A + n_B - 2 $$
4. Use the t-distribution to find the p-value corresponding to the calculated t statistic for a two-tailed test.

Assume the following sample statistics:
- Group A (Control): 
  - Mean revenue per session (\u03bc_A) = $50
  - Standard deviation (s_A) = $10
  - Sample size (n_A) = 100
- Group B (Treatment): 
  - Mean revenue per session (\u03bc_B) = $55
  - Standard deviation (s_B) = $12
  - Sample size (n_B) = 100

1. Calculate the test statistic:
   
   $$ t = \frac{55 - 50}{\sqrt{\frac{10^2}{100} + \frac{12^2}{100}}} = \frac{5}{\sqrt{1 + 1.44}} = \frac{5}{\sqrt{2.44}} \approx 3.22 $$
2. Calculate the degrees of freedom:
   
   $$ df = 100 + 100 - 2 = 198 $$
3. Using a t-distribution table or calculator, find the p-value for t = 3.22 with 198 df. The p-value is approximately 0.0015.

The calculated p-value of 0.0015 indicates that there is a 0.15% probability of observing a difference in revenue per session as extreme as the one seen (or more extreme) if the null hypothesis were true. Since this p-value is significantly lower than common alpha levels (e.g., 0.05 or 0.01), we reject the null hypothesis. This suggests that the treatment (Group B) has a statistically significant effect on increasing revenue per session compared to the control (Group A).

This analysis connects to several related concepts:
- **Confidence Intervals**: We could calculate a confidence interval for the difference in means to provide a range of plausible values for the revenue impact.
- **Power Analysis**: Understanding the power of our test can help in determining the likelihood of detecting a true effect given the sample size.
- **Central Limit Theorem**: This theorem assures us that with sufficiently large sample sizes, the distribution of the sample means will approximate normality, justifying the use of the t-test.
- **Assumptions**: We assumed that the revenue data for both groups is normally distributed and that variances are equal (homoscedasticity). If these assumptions are violated, alternative methods (e.g., Welch's t-test) may be necessary.

Calculate variance for A/B test results

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