Given the following table showing sample means and confidence intervals for the average test scores of two groups, Group A: sample mean = $75$, 95% confidence interval = $[72,78]$; Group B: sample mean = $80$, 95% confidence interval = $[77,83]$. Which statement is best supported by the data?

• A. The true mean scores for Group A and Group B are equal because their confidence intervals are both $6$ units wide.
• B. The sample means are not reliable because the confidence intervals are too narrow.
• C. There is evidence that the true mean score for Group B is higher than that for Group A because the confidence intervals do not overlap.
• D. We can conclude that all individuals in Group B scored higher than all individuals in Group A.

Answer: (C)

Verified step by step guidance

1. Understand what a 95% confidence interval (CI) represents: it is a range of values within which we are 95% confident the true population mean lies, based on the sample data.
2. Note the given confidence intervals: Group A has a CI of [72, 78] and Group B has a CI of [77, 83]. These intervals represent the uncertainty around each group's sample mean.
3. Check if the confidence intervals overlap: since Group A's upper bound is 78 and Group B's lower bound is 77, the intervals overlap slightly between 77 and 78.
4. Interpret the overlap: overlapping confidence intervals suggest that the difference between the true means might not be statistically significant at the 95% confidence level, but slight overlap does not always mean no difference; further statistical testing would be needed for a definitive conclusion.
5. Evaluate the given statements: the statement that the true mean for Group B is higher than Group A because the confidence intervals do not overlap is incorrect here since the intervals do overlap; also, equal width of intervals or narrow intervals alone do not prove equality or unreliability of means, and we cannot conclude individual scores without more data.