Step 1 :Given a random sample from a normal distribution: \(3,8,10,4,6,5\)
Step 2 :Calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\)
Step 3 :The sample mean \(\bar{x}\) is 6.0 and the sample standard deviation \(s\) is approximately 2.61
Step 4 :Construct a 99% confidence interval for the population mean \(\mu\) using the formula: \(\bar{x} \pm Z \frac{s}{\sqrt{n}}\), where \(Z\) is the Z-score (which is 2.33 for a 99% confidence interval), \(s\) is the sample standard deviation, and \(n\) is the sample size
Step 5 :The 99% confidence interval for the population mean \(\mu\) is approximately \((3.26, 8.74)\)
Step 6 :Assume that the sample mean \(\bar{x}\) and sample standard deviation \(s\) remain exactly the same as those you just calculated but that are based on a sample of \(n=25\) observations
Step 7 :Repeat the calculation of the confidence interval with a sample size of 25
Step 8 :The 99% confidence interval for the population mean \(\mu\) with a sample size of 25 is approximately \((4.66, 7.34)\)
Step 9 :Compare the results of the confidence intervals with sample sizes of 6 and 25
Step 10 :Increasing the sample size decreases the width of the confidence interval, making our estimate of the population mean more precise
Step 11 :Final Answer: The 99% confidence interval for the population mean \(\mu\) with a sample size of 6 is \(\boxed{(3.26, 8.74)}\). When the sample size is increased to 25, the confidence interval becomes \(\boxed{(4.66, 7.34)}\)