Lecture 17: Small-Sample Inferences for Normal Populations Confidence intervals for µ when σ is unknown If the population distribution is normal, then X µ σ/ n has a standard normal distribution. If σ is known, this justifies the use of X ± z α/2 σ n as a confidence interval for µ, where z α/2 comes from the standard normal table. For example, z α/2 = 1.96 for a 95% confidence interval. What if σ isn t known? Would like to replace σ by an estimate S computed from the data.
Question: Can we use the interval X ± z α/2 S n? Answer: Not when n is small. The intervals won t have the correct confidence levels. Idea: Replacing σ (a number) by S (a random variable) introduces more uncertainty, so we need to use a bigger multiplier c. How do we find c? Fact: If the population distribution is normal, then X µ S/ n has a t distribution with n 1 degrees of freedom. Some algebra will justify using the interval X ± c S n where c is computed not from the standard normal table, but from the t table.
Still we want c to satisfy the area between c and c is equal to the confidence level, but now it s area under a t density curve. Using the t table The t table is a bit different from the standard normal table. Each row of the table corresponds to a different degree of freedom. If the sample size is n, use the n 1 degrees of freedom row. Each column in the table corresponds to a different area α to the right. Each number in the table has area α to its right. A few examples will make this much more clear. Examples: We ll solve the kind of problem needed for a confidence interval. Problem: Find the number c such that the area between c and c under the t density with 7 degrees of freedom is equal to 0.90.
Degrees of Area α in right tail Freedom 0.25 0.1 0.05 0.025 1 1.000 3.078 6.314 12.706 2 0.816 1.886 2.920 4.303 3 0.765 1.638 2.353 3.182 4 0.741 1.533 2.132 2.776 5 0.727 1.476 2.015 2.571 6 0.718 1.440 1.943 2.447 7 0.711 1.415 1.895 2.365 8 0.706 1.397 1.860 2.306 9 0.703 1.383 1.833 2.262 10 0.700 1.372 1.812 2.228 60 0.679 1.296 1.671 2.000 Table 1: A portion of the t table
Solution: The basic picture is the same as before. Since we want the area between c and c to be 0.90, the area to the right of c is 0.05. So we look in the 0.05 column of the t table, and get the answer 1.895. We denote t 0.05 = 1.895. (Note that the corresponding answer from the standard normal table is 1.64.) Problem: Find the number c such that the area between c and c under the t density with 5 degrees of freedom is equal to 0.95. Solution: Since we want the area between c and c to be 0.95, the area to the right of c is 0.025. That is, we are looking for t 0.025. So we look in the 0.025 column of the t table, and get the answer 2.571. (Note that the corresponding answer from the standard normal table is 1.96.)
Putting it all together To find a t confidence interval at level 100(1 α)%: Find the degrees of freedom, which is n 1. Find t α/2 from the t table. Plug t α/2, S, X, and n into the formula (X t α/2 S n, X + t α/2 S n ). Remember that this procedure is valid if the population distribution is normal. Confidence interval example In 1882, Michelson performed an experiment to measure the speed of light. He obtained 23 measurements of the speed of light in km/sec. The mean of his measurements is 299715.1 km/sec. The standard deviation of his measurements is 158.2 km/sec. Problem: Compute a 98% confidence interval for the true speed of light.
Solution: First, since there are n = 23 measurements, we will use the 22 degrees of freedom row of the t table. Since we want a 98% confidence interval, so α = 0.02. From the t table we find t α/2 = 2.508. Plug everything into the formula: 299715.1 ± 2.508 158.2 23. This is 299715.1 ± 82.7 Example 3. See page 386. Testing hypotheses about means: t-test State the hypotheses (in terms of µ), say H 0 : µ = µ 0 vs H 1 : µ > µ 0. We use the test statistic T = X µ 0 S/ n. and the rejection region { T t α }.
Collect data and compute the value of the test statistic T. Draw conclusion: state whether or not H 0 is rejected. Example 4. See page 387. Example: Question: How accurate are radon detectors? Study: Twelve radon detectors were placed in a chamber which exposed them to 105 picocuries/liter of radon. The mean of the 12 readings was 101.13 and the standard deviation of the 12 readings was 9.40. We want to test H 0 : µ = 105 versus H 1 : µ 105. We ll choose α = 0.10. From Table 4, we get t α/2 = 1.796. Hence the rejection region is { } T 1.796 or T 1.796. Test statistic is T = 101.13 105 9.40/ 1.43. 12
Since 1.43 does not fall into the rejection region, we will not reject H 0. Applicability The t-testing procedure for means can be used For any sample size if the population is normally distributed. For large enough sample sizes if the population is not normally distributed.