Final answer:
The value of x² left for a 95% confidence interval when n=15 being 26.119 is False, since the appropriate distribution for this sample size and confidence level is the t-distribution, not the chi-squared distribution. The statement is false.
Step-by-step explanation:
The statement that the value for x² left for a 95% confidence interval when n=15 is 26.119 is False. For a 95% confidence interval with a sample size of n=15, we need to use the t-distribution since the sample size is less than 30 and the population standard deviation is unknown.
The degrees of freedom (df) would be n-1, which is 15-1=14 for this scenario. We would look up the critical value of t for 14 degrees of freedom for a 95% confidence interval, which is typically found in a t-distribution table or using statistical software. The provided x² value incorrectly suggests usage of a chi-squared distribution, rather than the applicable t-distribution.
The correct value for a 95% confidence interval with 14 degrees of freedom is closer to 2.145 (or similar, depending on the source), not 26.119.
If anything, 26.119 might be mistaken for a critical value in a chi-squared distribution for a different context, but it is not applicable here for constructing the confidence interval for the mean.