MATH 1040 - Introduction to Statistics

Practice

1. A nutritionist wants to estimate the average number of cups of coffee consumed per week by graduate students. A random sample of 16 students is taken from a normally-distributed population. The sample shows a mean of 9.3 cups with a sample standard deviation of 2.1 cups. Construct a 90% confidence interval for the true mean weekly coffee consumption of graduate students.

Solution

From the problem we get the following information:

• A confidence level of 90%
• Sample Size is \(n=16\)
• The sample mean is \(\bar{x} = 9.3\)
• The sample standard deviation is \(s=2.1\)

First, we verify the central limit theorem.

• Is the sample random? Yes (stated in the problem)
• Is the sample large enough? No (sample size is smaller than 30)
  • If it’s not large enough, is the population normally distributed? Yes (stated in the problem)

Now, we need to find the critical value. Since we only have a sample standard deviation, we need to find a critical t-score. With a 90% confidence interval and \(n-1 = 15\) degrees of freedom, we find a critical value of \(t_c = 1.753\).

\[E = t_c\frac{s}{\sqrt{n}} = 1.753\frac{2.1}{\sqrt{16}} = 0.92\] \[\bar{x} + E = 9.3 + 0.92 = 10.22\] \[\bar{x} - E = 9.3 - 0.92 = 8.38\]

Solution:

We are 90% confident that the average number of cups of coffee consumed by graduate students per week is between 8.38 and 10.22.

Return back to Lesson 18.5