Unit 4 · Topic 4.2 · about 30 minutes
Constructing a Confidence Interval for a Population Mean or Population Mean Difference
Estimate a population mean, or the mean difference in paired data, with a t-interval you can justify and interpret in context.
Predict first
In Topic 3.3, a 95% interval for a proportion used the critical value . For a mean you almost never know the population standard deviation , so you estimate it with the sample standard deviation . With standing in for , what should happen to the critical value for 95% confidence?
t-distributions
When you replace with , the standardized value no longer follows the standard normal curve. It follows a t-distribution, also called Student's t-distribution.
The t-distributions are a family of symmetric, bell-shaped, standardized curves centered at 0. Each one is identified by its degrees of freedom (df), which depend on the sample size. For one sample, . Compared with the standard normal curve, a t-distribution has a lower, narrower peak and fatter tails, so more of its area sits far from 0. With few degrees of freedom the difference is large. As the degrees of freedom increase, the t-distribution looks more and more like the standard normal curve, and the critical values close in on the familiar .
| Sample size | Degrees of freedom | for 95% confidence |
|---|---|---|
| 3 | 2 | 4.303 |
| 6 | 5 | 2.571 |
| 11 | 10 | 2.228 |
| 21 | 20 | 2.086 |
| 31 | 30 | 2.042 |
| 121 | 120 | 1.980 |
| Standard normal | none | 1.960 |
The one-sample t-interval
To estimate a population mean when is unknown, use the one-sample t-interval for a population mean:
The point estimate is the sample mean . The standard error, , estimates how far sample means typically land from . It is the formula from Topic 4.1, with in place of . The margin of error is , where is the critical value for the central C% of the t-distribution with . On a TI-84, invT(0.975, df) gives for 95% confidence; the printed t table gives the same values.
State the parameter in context: the mean of the response variable for the population you sampled. "The mean caffeine content of all 16-ounce cold brews sold by the chain" names a parameter. "The mean" does not.
Three conditions
A one-sample t-interval for a population mean requires three conditions:
- Randomization condition: the data come from a random sample or a randomized experiment.
- 10% condition: when sampling without replacement, .
- Sample data condition: the population distribution is approximately normal, or , or, if , the sample data are free from strong skewness and outliers.
The third one is new. With proportions you counted successes and failures. With means you look at the data. For a small sample, graph it with a dotplot, boxplot or histogram and look for strong skew or outliers. Mild unevenness is fine, because small samples from normal populations rarely look perfectly symmetric. An outlier or a long tail means the condition is not met, and the interval may not deserve the confidence level printed next to it.
Caffeine in 12 randomly chosen 16-ounce cold brews
No strong skew and no outliers, so the sample data condition is met even though .
Worked exampleA 95% interval for mean caffeine
A food-safety lab buys one 16-ounce cold brew at each of 12 randomly selected stores in a large coffee chain and measures its caffeine. The dotplot above shows the results: mg and mg. Construct and interpret a 95% confidence interval for the mean caffeine content of all 16-ounce cold brews sold by the chain.
Name it. One-sample t-interval for , the mean caffeine content, in mg, of all 16-ounce cold brews sold by the chain.
Check conditions. Random: the cups came from 12 randomly selected stores. 10%: 12 cups is far less than 10% of all the 16-ounce cold brews the chain sells. Sample data: , but the dotplot shows no strong skewness and no outliers.
Calculate. , so . mg, so the margin of error is mg. The interval is , which is about mg.
Interpret in context. We are 95% confident that the interval from 202.45 mg to 219.39 mg contains the true mean caffeine content of all 16-ounce cold brews sold by the chain.
About 202.45 mg to 219.39 mg. The interval estimates the mean for all of the chain's 16-ounce cold brews, not the caffeine in any single cup.
Paired data: one sample of differences
Sometimes each value in one set is matched with exactly one value in the other, such as the same student measured before and after a program, or twins split between two treatments. This is a matched pairs design, and the two sets of values are dependent. Do not treat them as two separate samples. Subtract within each pair to get one sample of differences, and use the one-sample t-interval for a population mean difference:
Here and are the mean and standard deviation of the differences, is the number of pairs, and . The parameter is , the population mean difference, and you must say which way you subtracted: after minus before is a different parameter from before minus after. The three conditions are checked on the differences, so for the sample data condition you graph the differences, not the two original lists.
| Student | Before | After | Difference (after minus before) |
|---|---|---|---|
| 1 | 182 | 196 | 14 |
| 2 | 205 | 214 | 9 |
| 3 | 167 | 185 | 18 |
| 4 | 221 | 226 | 5 |
| 5 | 194 | 207 | 13 |
| 6 | 176 | 190 | 14 |
| 7 | 210 | 213 | 3 |
| 8 | 188 | 201 | 13 |
| 9 | 199 | 216 | 17 |
| 10 | 172 | 181 | 9 |
Differences in reading speed, after minus before
The 10 differences show no outliers and no strong skew, so the sample data condition is met for the differences.
Worked exampleA paired interval for the mean gain
The table shows reading speeds for a random sample of 10 of the roughly 600 students in a district's summer reading program, before and after the program. Construct and interpret a 95% confidence interval for the mean change in reading speed.
Name it. One-sample t-interval for , the mean difference in reading speed (after minus before), in words per minute, for all students in the district's summer program.
Check conditions. Random: the 10 students were randomly selected. 10%: 10 is less than 10% of the 600 students. Sample data: there are only 10 differences, and their boxplot shows no strong skewness and no outliers.
Calculate. The differences have and . With , , so the interval is , which is about .
Interpret in context. We are 95% confident that the interval from 7.99 to 15.01 words per minute contains the true mean difference in reading speed (after minus before) for all students in the district's summer program.
About 7.99 to 15.01 words per minute, for the population mean of after minus before. Topic 4.3 takes up what this interval can and cannot support as a claim.
Check your understanding
How does a t-distribution with 4 degrees of freedom compare with the standard normal distribution?
A food scientist measures the sodium content of a random sample of 16 cans from a large production run of soup. The sample shows no strong skew or outliers. Which critical value should she use for a 90% confidence interval for the mean sodium content of all cans in the run?
A researcher wants a confidence interval for the mean commute distance of employees at a large company. She records the distances for a random sample of 14 employees. A dotplot of the 14 distances shows most values between 3 and 15 miles and one value at 62 miles. Which statement about the conditions is correct?
A sports scientist measures the vertical jump of 15 randomly selected players from a large volleyball league, once in their usual shoes and once in a new shoe, in random order. She wants to estimate the mean improvement from the new shoe. Which procedure and parameter fit?
A random sample of 20 households in a large city used a mean of 312 gallons of water per day, with standard deviation 64 gallons. The sample shows no strong skew or outliers. Find the margin of error for a 95% confidence interval for the mean daily water use of all households in the city. Round to one decimal place.
Course alignment, for teachers
AP Statistics topic 4.2, Unit 4: Inference for Quantitative Data: Means.
- Skill 2.C: Identify appropriate statistical inference methods.
- Skill 3.E: Calculate appropriate statistical inference method results.
- Skill 4.C: Describe distributions and compare relative positions of points within a distribution.
- Skill 4.E: Justify the use of a chosen statistical inference method by verifying conditions.