Suppose a sample size of 5 has mean 9.60.If the population variance is 5 and the population is normally distributed,the lower limit for a 92% confidence interval is 7.85.

Question

Quizplus · Accepted Answer

Given the sample size (n=5), sample mean (x̄=9.60), population variance (σ^2=5), and assuming a normal distribution, the calculation for a confidence interval involves finding the standard error of the mean (σ/√n) and then using a Z or T distribution to find the critical value for the desired confidence level (92% in this case). Since the sample size is small (n<30), the T distribution is more appropriate. However, without the exact degrees of freedom and T value specified, we cannot calculate the exact confidence interval here. The statement claims the lower limit is 7.85, which could be true given the provided information and assuming the correct critical value from the T distribution was used.

Suppose a Sample Size of 5 Has Mean 9