Confidence Interval for a Population Mean, Standard Deviation Known

Step By Step Example of a Confidence Interval for a Mean—Standard Deviation Known

Suppose scores on exams in statistics are normally distributed with an unknown population mean, and a population standard deviation of 3 points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. To find a 90% confidence interval for the true (population) mean of statistics exam scores, we have the following guidelines:

1. Plan: State what we need to know.
2. Model: Think about the assumptions and check the conditions.
3. State the parameters and the sampling model.
4. Mechanics: $\text{CL} = 0.90$, so $\alpha = 1-\text{CL} = 1-0.90 = 0.10$; $\alpha_{0.05}$ is $1-0.05 = 0.95$; So $z_{0.05} = 1.645$
5. Conclusion: Interpret your result in the proper context, and relate it to the original question.

1. In our example, we are asked to find a 90% confidence interval for the mean exam score, $\mu$, of statistics students.

We have a sample of 68 students.

2. We know the population standard deviation is 3. We have the following conditions:

• Randomization Condition: The sample is a random sample.
• Independence Assumption: It is reasonable to think that the exam scores of 36 randomly selected students are independent.
• 10% Condition: We assume the statistic student population is over 360 students, so 36 students is less than 10% of the population.
• Sample Size Condition: Since the distribution of the stress levels is normal, our sample of 36 students is large enough.

3. The conditions are satisfied and $\sigma$ is known, so we will use a confidence interval for a mean with known standard deviation. We need the sample mean and margin of error (ME):

$\bar { x } =68 \\ \sigma =3 \\ n=36 \\ \text{ME}={ z }_{ \frac { \alpha }{ 2 } }\left( \dfrac { \sigma }{ \sqrt { n } } \right)$

4. below shows the steps for calculating the confidence interval.

$\displaystyle {\text{ME} = 1.645\cdot \frac{3}{\sqrt{36}} = 0.8225 \\ \bar{x} - \text{ME} = 68-0.8225 = 67.1775 \\ \bar{x} + \text{ME} = 68+0.8225 = 68.8225}$

The 90% confidence interval for the mean score is $(67.1775, 68.8225)$.

Graphical Representation

This figure is a graphical representation of the confidence interval we calculated in this example.

5. In conclusion, we are 90% confident that the interval from 67.1775 to 68.8225 contains the true mean score of all the statistics exams. 90% of all confidence intervals constructed in this way contain the true mean statistics exam score.