If the assumptions are satisfied, the confidence interval for µ when σ is unknown is found using the following steps:

Compute the sample mean $LaTeX: \bar{x}$ and sample standard deviation, s, if they are not given.
Find the number of degrees of freedom n – 1 and the critical value $LaTeX: t_{\frac{\alpha}{2}}$ .
Compute the standard error $LaTeX: \frac{s}{\sqrt[]{n} }$ and multiply it by the critical value to obtain the margin of error $LaTeX: t_{\frac{\alpha}{2}}\cdot\frac{s}{\sqrt[]{n}}$ .
Construct the confidence interval $LaTeX: \bar{x}\pm t_{\frac{\alpha}{2}}\cdot\frac{s}{\sqrt[]{n}}$
Interpret the result.

Example

A food chemist analyzed the calorie content for a popular type of chocolate cookie. Following are the numbers of calories in a sample of eight cookies.

113, 114, 111, 116, 115, 120, 118, 116

Find a 98% confidence interval for the mean number of calories in this type of cookie.

Solution

We check the assumptions. We have a simple random sample. Because the sample size is small, the population must be approximately normal. A dotplot indicates that there is no evidence of strong skewness and no outliers, therefore we may proceed.

Dotplot of data showing it is not skewed and has no outliers.PNG

Now, we find the sample mean and sample standard deviation. We have $LaTeX: \bar{x}=115.375$ and LaTeX: s=2.8253 .

The number of degrees of freedom is n − 1 = 8 − 1 = 7. Using a calculator (we find invT in distributions, then enter .99 for the area and 7 for the degrees of freedom) we find that the critical value corresponding to a level of 98% is $LaTeX: t_{\frac{\alpha}{2}}=2.998$ .

The margin of error is: $LaTeX: t_{\frac{\alpha}{2}}\cdot\frac{s}{\sqrt[]{n}}=2.998\cdot\frac{2.8253}{\sqrt[]{8}}=2.9947$ .

The 98% confidence interval is: $LaTeX: \bar{x}\pm t_{\frac{\alpha}{2}}\cdot\frac{s}{\sqrt[]{n}}=115.375\pm2.9947$ or $LaTeX: 112.4<\mu <118.4$

We are 98% confident that the mean number of calories per cookie is between 112.4 and 118.4.

Example

A sample of 123 people aged 18–22 reported the number of hours they spent on the Internet in an average week. The sample mean was 8.20 hours, with a sample standard deviation of 9.84 hours. Assume this is a simple random sample from the population of people aged 18–22 in the U.S. Construct a 95% confidence interval for µ, the population mean number of hours per week spent on the Internet by people aged 18–22 in the U.S.

Solution

We have a simple random sample and the sample size is large. We may proceed. Note that $LaTeX: \bar{x}=8.20$ and LaTeX: s=9.84 . The number of degrees of freedom is 122, because this is 1 less than the sample size of 123. Using the calculator to compute InvT(.975, 122), we get the critical value corresponding to a level of 95% is $LaTeX: t_{\frac{\alpha}{2}}=1.980$ .

Margin of error: $LaTeX: t_{\frac{\alpha}{2}}\cdot\frac{s}{\sqrt[]{n}}=1.980\cdot\frac{9.84}{\sqrt[]{123}}=1.7564$

The 95% confidence interval: $LaTeX: \bar{x}\pm t_{\frac{\alpha}{2}}\cdot\frac{s}{\sqrt[]{n}}=8.20\pm1.7564$ or $LaTeX: 6.44<\mu <9.96$

We are 95% confident that the mean number of hours per week spent on the Internet by people 18 to 22 years old is between 6.44 and 9.96.