Estimate the population mean by finding a $99 \%$ confidence interval given a sample of size 8 , with a mean of 57.4 and a standard deviation of 15.7 . Preliminary: Is $n \geq 30$ ? Yes No Confidence Interval: What is the $99 \%$ confidence interval to estimate the population mean? Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place. \[ 99 \% \text { C.I. }= \]
Solution
Step 1 ：First, we need to calculate the t-score for a 99% confidence level and 7 degrees of freedom (since the sample size is 8, the degrees of freedom is 8 - 1 = 7).
Step 2 ：Given values are: sample size \(n = 8\), sample mean \(= 57.4\), and standard deviation \(= 15.7\). So, degrees of freedom \(df = n - 1 = 7\).
Step 3 ：Calculate the t-score for 99% confidence level. The t-score is approximately \(t = 3.499\).
Step 4 ：Substitute these values into the formula to calculate the confidence interval. The lower bound of the confidence interval is \(mean - t \times \frac{std\_dev}{\sqrt{n}} = 37.98\) and the upper bound is \(mean + t \times \frac{std\_dev}{\sqrt{n}} = 76.82\).
Step 5 ：Final Answer: The 99% confidence interval to estimate the population mean is \(\boxed{(37.98, 76.82)}\).
