Question

A simple random sample with n=54 provided a sample mean of 22.5 and a sample standard deviation of 4.4.

a. Develop a 90% confidence interval for the population mean.

b. Develop a 95% confidence interval for the population mean.

c. Develop a 99% confidence interval for the population mean.

d. What happens to the margin of error and the confidence interval as the confidence level is increased?

Answer 1

Answer:

When confidence level increases, margin of error increases thus making confidence interval wider.

Step-by-step explanation:

Given that a simple random sample with n=54 provided a sample mean of 22.5 and a sample standard deviation of 4.4.

Since n >30 but population std deviation is not known we can use only t critical value.

a) t critical = 2.006

Hence 90% confidence interval = Mean ±$2.006*\frac{4.4}{\sqrt{54} }$

=$(22.5-1.201, 22.5+1.201)\\\\= (21.299,23.701)$

b) t critical = 2.30687 = 2.31

Hence 90% confidence interval = Mean ±$2.31*\frac{4.4}{\sqrt{54} }$

=$(22.5-1.3813, 22.5+1.3814)\\\\= (21.117,23.883)$

c) t critical = 2.30687 = 2.31

Hence 90% confidence interval = Mean ±$2.31*\frac{4.4}{\sqrt{54} }$

=$(22.5-1.754, 22.5+1.754)\\\\= (20.746,24.254)$

d) When confidence level increases, margin of error increases thus making confidence interval wider.