A data scientist is conducting an A/B test to compare the mean session duration for two different versions of a website's user interface (UI), 'A' and 'B'. They collect data from two independent groups of 50 users each. An initial exploratory data analysis reveals that the session durations for both groups are approximately normally distributed. However, a Levene's test for equality of variances returns a p-value of 0.02. Given these findings, which statistical test is the most appropriate to determine if there is a significant difference between the mean session durations of the two UIs?
The correct option is Welch's t-test. This test is an adaptation of the two-sample t-test specifically designed for comparing the means of two independent groups when their variances are unequal. The scenario states that a Levene's test, which assesses the equality of variances, yielded a p-value of 0.02. Since this p-value is below the common alpha threshold of 0.05, the null hypothesis of equal variances is rejected, indicating the variances are significantly different. In this situation, Welch's t-test is more reliable and robust than Student's t-test.
Student's t-test is incorrect because it operates under the assumption that both groups have equal variances, a condition explicitly violated according to the Levene's test results.
A paired samples t-test is incorrect because it is used for dependent samples, such as measuring the same individuals before and after an intervention, not for two independent groups of users as described.
The Mann-Whitney U test is a non-parametric alternative used when the assumption of normality is violated. The scenario states the data is approximately normal, making a parametric test like the t-test the more powerful and appropriate choice.
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What is Welch's t-test and why is it used in this scenario?
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What does the p-value from Levene's test represent?
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Why is the Mann-Whitney U test not appropriate in this case?