Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 112 and the sample standard deviation, s, is found to be 10.

a) Construct a 96% confidence interval about μ (population mean) if the sample size, n, is 11.
Provide the confidence interval as (lower bound, upper bound) in ascending order, rounding to one decimal place as needed.

Answer 1

The student wants to construct a 96% confidence interval for a population mean using a normally distributed sample of size 11, with a sample mean of 112 and a sample standard deviation of 10. We use the t-distribution to find the appropriate t-value and then calculate the margin of error and the confidence interval.

Step-by-step explanation:

To construct a 96% confidence interval about the population mean μ when the population standard deviation is unknown and the sample size is small (n < 30), we use the student's t-distribution because the sample standard deviation s is used to estimate the population standard deviation. In this case, the sample mean x is 112, the sample standard deviation s is 10, and the sample size n is 11.

First, we need to find the t-value that corresponds to a 96% confidence level with n - 1 degrees of freedom. We can use a t-distribution table or technology to find this value. Once we have the t-value, the margin of error (EBM) is calculated as follows:

EBM = t * (s / √n)

The confidence interval (CI) can then be calculated as:

(x - EBM, x + EBM)

Let's do the math with the given values:

1. Find the t-value for a 96% confidence level with 10 degrees of freedom.
2. Calculate the EBM using the t-value and the sample standard deviation.
3. Calculate the upper and lower bounds of the confidence interval.
4. Present the confidence interval rounded to one decimal place.

It's important to remember that the confidence interval provides a range within which the true population mean is likely to fall.