Question

A doctor conducts a survey with a random sample of his patients, measuring their cholesterol levels in millimolar per liter.
The data measurements for 14 patients are given below. Assuming the population standard deviation is o = 1.3 and that
the population is normally distributed, compute a 90% confidence interval for the mean cholesterol levels of his patients.
Round your answers to two decimal places and use ascending order.

Answer 1

Using the z-distribution, as we have the standard deviation for the population, it is found that the 90% confidence interval for the mean cholesterol levels of his patients is (5.59, 6.73).

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the sample.

In this problem, we have a 90% confidence level, hence$\alpha = 0.90$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so the critical value is z = 1.645.

Researching on the internet, the other parameters are given by:

$\overline{x} = 6.16, \sigma = 1.3, n = 14$

Hence:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 6.16 - 1.645\frac{1.3}{\sqrt{14}} = 5.59$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 6.16 + 1.645\frac{1.3}{\sqrt{14}} = 6.73$

The 90% confidence interval for the mean cholesterol levels of his patients is (5.59, 6.73).

More can be learned about the z-distribution at https://brainly.com/question/25890103