MATH 1040 - Introduction to Statistics

Lesson 18.5 Confidence Interval when you don’t know \(\sigma\)

Reading

Reading sections are from the Introductory Statistics Textbook

• 7.1.2 Introducing the t-distribution (pages 278-281)
• 7.1.3 The t-distribution and the standard error of a mean (page 281)
• 7.1.5 One sample t-intervals (pages 282-285)

Lesson

Consider this problem:

A health researcher wants to estimate the average number of hours of sleep that college students get on weeknights. A random sample of 25 students is taken from a normally-distributed population, and the sample yields a mean of 6.8 hours with a sample standard deviation of 1.2 hours.

Construct a 95% confidence interval for the true mean number of hours of sleep college students get on weeknights.

Notice that in this case, we do not have a population standard deviation (\(\sigma\)). So, our Margin of Error calculation won’t work.

Fortunately, we have a solution. The normal distribution depends on the population. Another distribution that we call the t-distribution depends instead on the sample. This is a good approximation of the normal distribution.

With the t-distribution, we find a critical value from t-scores and use the sample standard deviation to find the margin of error.

\[E = t_c\frac{s}{\sqrt{n}}\]

However, since the t-distribution relies on the sample, we also have to consider one more variable:

• The Degrees of Freedom is calculated as \(DF = n-1\)
• As \(n\) increases, the t-distribution becomes a better approximation of the normal distribution.

To find the critical t-score, use a T-Table or a calculator (Note that the TI-83 will not work for this step, so be familiar with the table). In the problem above,

• The Sample Size is \(n=25\)
• There are \(DF = n-1 = 24\) degrees of freedom
• The confidence level is 95%

Using these on the T-Table, we get a critical t-score of \(t_c = 2.064\)

Other than that, the calculations for the Margin of Error and the Confidence Interval are exactly the same. \(E = t_c\frac{s}{\sqrt{n}} \qquad \mu = \bar{x} \pm SE\)

Example

Here is the problem from earlier:

A health researcher wants to estimate the average number of hours of sleep that college students get on weeknights. A random sample of 25 students is taken from a normally-distributed population, and the sample yields a mean of 6.8 hours with a sample standard deviation of 1.2 hours.

Construct a 95% confidence interval for the true mean number of hours of sleep college students get on weeknights.

Here is the information given from the problem.

• Confidence Level is 95%
• Sample Size is \(n = 25\)
• Degrees of Freedom is \(DF = n-1 = 24\)
• Sample mean is \(\bar{x} = 6.8\)
• Sample standard deviation is \(s = 1.2\)

Before we do anything, we must verify that the Central Limit Theorem holds.

• 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)

So, the Central Limit Theorem holds.

Using the T-Table or the TI-84 with a confidence level of 95% and 24 degrees of freedom, we get a critical value of \(t_c = 2.064\).

\[E = t_c\frac{s}{\sqrt{n}} = 2.064\frac{1.2}{\sqrt{25}} = 0.495\] \[\bar{x} + E = 6.8 + 0.495 = 7.295 \approx \mathbf{7.3}\] \[\bar{x} - E = 6.8 - 0.495 = 6.305 \approx \mathbf{6.3}\]

Solution: We are 95% confident that the true mean for the number of hours of sleep college students get on weeknights is between 6.3 and 7.3 hours.

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 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.
   • After solving on your own, see solution here
2. A tech company tests a new phone battery. A sample of 30 phones, coming from a normally-distributed population, shows an average battery life of 22.4 hours with a sample standard deviation of 3.5 hours. Construct a 99% confidence interval for the true mean battery life of the new phone model.
   • After solving on your own, see solution here
3. A professor surveys 41 students about how many hours they studied for the last exam. The sample mean is 14.2 hours, and the sample standard deviation is 4.6 hours. Construct a 95% confidence interval for the true mean number of hours students studied for the exam.
   • After solving on your own, see solution here

Technology

TI-83/84

Confidence Interval with t-values

• STAT –> [TESTS]
• 8:TInterval
• Set Input to ‘Stats’
• Add mean, standard deviation, and size of your sample
• Set C-Level to the requested confidence level
• Select ‘Calculate’

Alternatively, you can set Input to ‘Data’ and add your data to a list. The calculator will calculate the mean and standard deviation for you.

Finding a t-value

Note that this method will not work on the TI-83. Use the T-Table instead.

To find the t-score:

• Find the confidence level and the degrees of freedom
• From the confidence level, find the area of the left tail
  • If confidence level is 95%, the remaining 5% is divided between the 2 tails, so the left tail has 2.5%.
• Press 2nd –> [DISTR]
• Select 4:invT(
• Input the “area” of the left tail
• Input “df” (the degrees of freedom)
• “Paste” and then press “Enter” again