1 Small Sample CI for a Population Mean µ

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1 Lecture 7: Small Sample Confidence Intervals Based on a Normal Population Distribution Readings: Sections Small Sample CI for a Population Mean µ The large sample CI x ± z α/2 s n was constructed based on Central Limit Theorem (CLT). When sample size is small, CLT does not apply. We will assume instead that the population distribution is normal with mean µ and standard deviation σ. Sampling Distribution of X If population distribution is normal, or the sample size is large, X N(µ X = µ, σ X = σ/ n), i.e., X µ σ/ N(0, 1). n Since σ is often times unknown, we use the sample standard deviation s as the estimate of σ. When sample size is large, s serves as a good estimate of σ and the sampling distribution of is approximately N(0, 1). X µ S/ n X µ However, when sample size is small, S/ doesn t have a standard normal distribution n any more. The multiplier z α/2 is no longer appropriate. Properties of t Distributions T = X µ S/ n follows a t distribution with n 1 degrees of freedom. The t distribution is symmetric about zero and bell-shaped. The t distribution has more variability than the standard normal distribution. As the degrees of freedom increase, the t distribution approaches the standard normal distribution (because as n increases, s σ). 1

2 The One-Sample t Confidence Interval for µ Let x and s be the sample mean and sample standard deviation of a random sample of size n from a normal population with mean µ. Then a 100(1 α)% confidence interval for µ is s x ± t α/2,n 1, n where t α/2,n 1 is the value such that P (T > t α/2,n 1 ) = P (T < t α/2,n 1 ) = α/2, where T t(n 1). If the sample size is large, the critical value can be taken from the standard normal table. The t CIs are robust to small or even moderate deviations from normality unless n is quite small. Example 1: How accurate are radon detectors of a type sold to homeowners? To answer this question, university researchers placed 12 detectors in a chamber that exposed them to 105 picocuries per liter of radon. The detector readings were as follows: The sample mean is x = and the sample standard deviation is s = Find the 90% confidence interval for the population mean. 2

3 Assessing Normality Using Normal Quantile Plots Basic idea of normal quantile plots: if you data come from a normal distribution, then the i th smallest observation should roughly correspond to the (i/n) 100 th percentile of a normal distribution. Here is how it works: 1. Order the data from the smallest to the largest. Let x (i) denote the i th smallest value. 2. Take x (i) to be the ((i 0.5)/n) 100 th percentile. 3. Determine the corresponding percentiles for standard normal distributions, i.e., calculate the ((i 0.5)/n) 100 th percentile z i of Z. 4. Plot the data values x (i) against z i. A plot for which the points fall close to some straight line suggests that the assumption of a normal population is plausible. Example 1 (cont d): Check the normality of data. Radon Detector Readings Normal Quantiles data radon; input datalines; ; run; proc univariate data=radon; var reading; QQplot / Normal(mu=est sigma=est); run; 3

4 Prediction Interval for a Single Future Value In many applications, we are interested in predicting a single value of a variable to be observed at some future time, rather than estimating the mean value of that variable. Example 2: Consider the following sample of fat content (in percentage) of n = 10 randomly selected hot dogs: Assuming that these were selected from a normal population distribution, a 95% for the population mean fat content is Suppose, however, you are going to eat a single hot dog of this type and want a prediction of the resulting fat content. 4

5 Prediction Interval for a Single Value Let x and s be the sample mean and sample standard deviation of a random sample of size n from a normal population. Then the prediction interval (PI) for a single observation to be selected from the normal population distribution is The prediction level is 100(1 α)%. x ± t α/2,n 1 s n If the sample size is large, the critical value can be taken from the standard normal table. The validity of a prediction interval is closely tied to the normality assumption. The interval shouldn t be used in the absence of compelling evidence for normality. The interpretation of the prediction intervals is similar to a confidence interval. If the prediction interval is calculated for a large number of samples, in the long run, 100(1 α)% of these intervals will include the corresponding future value. Example 2 (cont d): Find the 95% prediction interval for the fat content of a single hot dog. 5

6 2 Small Sample CI for Difference of Two Population Means µ 1 µ 2 The (unpooled) Two-Sample t Confidence Interval for µ 1 µ 2 If samples of size n 1 and n 2 are taken from two normal populations with means µ 1 and µ 2, then the two-sided 100(1 α)% confidence interval for µ 1 µ 2 is where the value of the degrees of freedom is ( x 1 x 2 ) ± t α/2,k s 2 1 n 1 + s2 2 n 2, k = ( s 2 1 n 1 + s2 2 n 2 ) 2 (s 2 1 /n 1) 2 n (s2 2 /n 2) 2 n 2 1 If the sample sizes are large, the critical value can be taken from the standard normal table. The degree of freedom is usually not an integer. SAS can calculate the critical value for non-integer df s. For example, the critical values for the 90%, 95%, and 99% CIs with degrees of freedom are calculated as follows: data critical_value; df = 10.45; /*critical value for 90% CI*/ t_90 = tinv(0.95, df); /*critical value for 95% CI*/ t_95 = tinv(0.975, df); /*critical value for 99% CI*/ t_99 = tinv(0.995, df); run; proc print data=critical_value; run; SAS OUTPUT: Obs df t_90 t_95 t_ If SAS is not handy, round this df down to the nearest integer. (t 0.05,10 = 1.812, t 0.025,10 = 2.228, t 0.005,10 = 3.169). 6

7 The (pooled) Two-Sample t Confidence Interval for µ 1 µ 2 If samples of size n 1 and n 2 are taken from two normal populations with means µ 1 and µ 2 and a common standard deviation, then the two-sided 100(1 α)% confidence interval for µ 1 µ 2 is x 1 x 2 ± t α/2,k s pooled 1 n n 2, where the value of the degrees of freedom is k = n 1 + n 2 2 and the pooled standard deviation is (n 1 1)s 2 1 s pooled = + (n 2 1)s 2 2 n 1 + n 2 2 The pooled t confidence interval is not robust to violations of the equal standard deviation assumption. We therefore recommend the unpooled t confidence intervals unless there is really compelling evidence for doing otherwise. Example 3: Seedlings were germinated under two different lighting conditions. Their lengths (in cm) were measured after a specified time period. The data are as follows Dark Light n x s Calculate a 95% C.I. for the difference in the mean length under different lighting conditions, assuming that the lengths under different lighting conditions follow normal distributions with different standard deviations. 7

8 Calculate a 95% C.I. for the difference in the mean length under different lighting conditions, assuming that the lengths under different lighting conditions follow normal distributions with a common standard deviation. 3 CI for µ 1 µ 2 from Paired Data Sometimes we have observations in pairs such as: as identical twins two observations on the same individual (two days, pre- and post-tests, before and after measurements) Confidence intervals for paired data are based on the difference obtained between the 2 measurements Find the difference for each of the n pairs, that is d i = x i1 x i2. Find the sample mean d and sample standard deviation s d of these differences. Perform one-sample procedures for these differences. That is, The 100(1 α)% CI for µ d = µ 1 µ 2 is given by d ± t α/2,n 1 s d n If the sample size n is large, the critical value can be taken from the standard normal table: z α/2. Example 4: Researchers are interested in whether Vitamin C is lost when wheat soy blend (CSB) is cooked as gruel. Samples of gruel were collected, and the vitamin C content was measured (in mg per 100 grams of gruel) before and after cooking. Here are the results: 8

9 Sample x s Before After Before - After Find a 90% confidence interval for the mean vitamin C content loss 9