Review of Basic Statistical Concepts

Review of Basic Statistical Concepts The purpose of this review is to summarize the basic statistical concepts. Introductory statistics dealt with three main areas: descriptive statistics, probability, and inference. Descriptive Sample data may be summarized graphically or with summary statistics. Sample statistics include Statistics the mean, variance, standard deviation, and median. For the following definitions let x1, x2, … , xn represent the values obtaining from a random sample of size n drawn from a population of interest. Sample Mean x= x + x2 +  + xn 1 n xi = 1  n i =1 n Sample Variance s2 = 1 n ( xi − x)2 n −1  i =1 The sample variance equals the mean squared deviation from x . A small s 2 means that the observed values cluster around the average, while a large variance means that they are more spread out. Thus, the variance is a measure of the “spread” in the sampled values. Sample Standard Deviation s= 1 n ( xi − x ) 2 n −1  i =1 The sample standard deviation, s, is often a more useful measure of spread than the sample variance, s2, because s has the same units (inches, pounds, etc.) as the sampled values and x . The mean is just the average of the n values observed. StatGraphics Common descriptive statistics can be obtained by following: Describe > Numeric Data > OneVariable Analysis > Tabular Options > Summary Statistics Example The file LMF contains the three-year return for a random sample of 26 mutual funds. All of these funds involve a load (a type of sales charge). StatGraphics output is to the right. Summary Statistics for Return Count = 26 Average = 16.2346 Variance = 40.4208 Standard deviation = 6.35773 Minimum = 8.0 Maximum = 32.7 Range = 24.7 Stnd. skewness = 2.26003 Stnd. kurtosis = 1.1129 Random Variables and their Probability Distributions Random Variable A variable whose numerical value is determined by chance. The key elements here are that the variable assumes a number (sales volume, rate of return, test score, etc.) and that the sample selection process generates the numbers randomly, i.e., by a “random” selection. (In these notes, a random variable will be designated by a capital letter, such as X, to differentiate it from observed values x. For instance, X might represent the height of a man to be selected randomly. Once the man has been selected, his height is given by the value x, say x = 68 inches.) Probability Although the values of a random variable are subject to chance, some values are more likely to Distribution occur than others. For instance, the height of a randomly selected man is more likely to measure 6’ than 7’. It is the random variable’s probability distribution that determines the relative likelihood of possible values. Standardized Values For the value x drawn from a population with mean  and standard deviation , the standardized x − μ the number of standard deviations = value z = . For example, if incomes have a mean and above or below the mean that x is σ standard deviation of $48,000 and $16,000, respectively, then someone making $56,000 has a $56,000 − $48,000 $8,000 1 = = because their income is one-half standard standardized income of $16,000 $16,000 2 deviation above the mean income. The advantage of standardizing is that it facilitates the comparison of values drawn from different populations. Standardized Random Variables X − is the Standardized  random variable. (Note: The Standardized Variable always has mean 0 and standard deviation 1.) For the random variable X with mean  and standard deviation , Z = The Normal Distribution In this course we will make use of (at least) four distributions designed to model continuous data: the Normal, t, F, and Chi-Square. Of these, the normal distribution is by far the most important because of its role in statistical inference. Much of the logic behind what we do and why we do it is based upon an understanding of the properties of the normal distribution, and of the theorems involving it, particularly the Central Limit Theorem. Properties 1. Normal distributions are bell-shaped. (In fact, it is sometimes called the “Bell Curve”.) 2. Normal distributions are symmetric about their mean. 3. Normal distributions follow the 68-95-99.7 rule: • (Approximately) 68% of the area under the curve is within one standard deviation  of the mean  • (Approximately) 95% of the area under the curve is within two standard deviations  of the mean  • (Approximately) 99.7% of the area under the curve is within three standard deviations  of the mean  4. If the random variable X is normal with mean  and standard deviation , then the random X − variable Z = is standard normal, i.e., is normal with mean equal 0 and standard deviation  equal 1. the Standard Normal Distribtion, Z 0.4 Mean,Std. dev. 0,1 density 0.3 0.2 0.1 0 -5 -4 -3 -2 -1 0 z 1 2 3 4 5 Finding probabilities in Excel Example Cumulative Probabilities for any normal random variable X, i.e., P(X  x), are easy to find in Excel. Follow: fx > Statistical > NORMDIST and enter TRUE in the Cumulative field. Probabilities of the form P(X > x) or P(a < X < b) can be obtained by subtraction. To find P(-1.2 < Z < 2), note that P(-1.2 < Z < 2) = P(Z < 2) – P(Z  -1.2) and use the Excel output to the right. Answer = 0.9772 – 0.1151 = 0.8621 Critical Values z  is defined by P(Z > z) = . Critical values are used in the construction of confidence intervals and (optionally) in hypotheses testing. To find the critical value associated with the significance level , follow: fx > Statistical > NORMINV and enter 1 -  in the Probability field. Example From the Excel output to the right we see that z0.05 = 1.645 The Distribution of the Sample Mean Because, when we take a random sample, the values of a random variable are determined by chance, statistics such as the sample mean that are calculated from the values are themselves random X + X2 ++ Xn variables. Thus the random variable X = 1 has a probability distribution of its n x + x + + xn own. If we intend to use the sample mean x = 1 2 to estimate the mean  of the n population from which the sample was drawn, then we need to know what values the random variable X can assume and with what probability, i.e., we need to know the probability distribution of X . It can be shown (using advanced calculus) that X has the following properties: • • • The mean of X equals the mean of X, i.e.,  X =  . This just says that the sample mean x is an unbiased estimator of the population mean . 2 2 The variance of X is less than that of X. In fact  X = . This states that there is less variability n in averaged values (and the variability decreases as the size of the sample increases) than there is in individual values. Hence, you might not be surprised if a randomly selected man measured 7’, but you would be suspicious if someone claimed that 100 randomly chosen men averaged 7’! If the variable X is normally distributed, then X will also be normal. The properties above, however, don’t describe the shape of the distribution of X (needed for making inferences about ) except in the special case where X is normal! They only contribute information about the mean and spread of the distribution. In general, the shape of the distribution of X may be difficult to determine for non-normal populations and small samples. However: • Example For large samples the Central Limit Theorem states that X will be at least approximately normal. (Most introductory statistics texts consider a sample large whenever n > 30.) The dean of a business school claims that the average weekly income of graduates of his school 1 year after graduation is $600, with a standard deviation of $100. Find the probability that a random sample of 36 graduates averages less than $570. Solution: Let X = weekly income of a sampled graduate 1 year after graduation. We are asked to find P ( X  $570) for 36 graduates.  X − $600 $570 − $600  P ( X  $570) = P    P ( Z  −1.8) = 0.0359 $100 36   $100 36 Note: Without the Central Limit Theorem we could not have approximated the probability that a sample of graduates average less than $570 because the distribution of incomes is not usually normal. Statistical Inference: Estimation Point Estimate A single number used to estimate a parameter. For example, the sample mean x is typically used to estimate the population mean . Interval Estimate A range of values used as an estimate of a population parameter. The width of the interval provides a sense of the accuracy of the point estimate. Confidence Interval Estimates for  Confidence intervals for  have a characteristic format: x  CV *standard error, where CV stands for Critical Value and the standard error is the (usually estimated) standard deviation of X . Case I: X normal or n >30, and  is known Case II: n  30 and  is unknown Case III: X is normal and  is unknown A (1 - )*100% confidence interval estimate for  is given by  x  z    n 2 A (1 - )*100% confidence interval estimate for  is given by  , with n-1 degrees of freedom x  t  s  n 2  A (1 - )*100% confidence interval estimate for  is given by  , with n-1 degrees of freedom x  t  s  n 2 Case III requires some explanation. When X is normal, and we must use the sample standard deviation s to estimate the unknown population standard deviation , the studentized statistic X − T= has a t distribution with n-1 degrees of freedom. Hence, we must use the critical value S n t /2 from the t distribution with n-1 degrees of freedom. The properties of the t distributions are similar to those for the standard normal distribution Z, except that the t has a larger spread to reflect the added uncertainty involved in estimating  by s. Note: For large samples, where n  30, there is very little difference between the t distribution with n-1 degrees of freedom and the standard normal distribution Z. Therefore, for large samples (Case II in the table above) some texts replace t /2 with z /2 even when X is normal and  is unknown! Student's t Distribution 0.4 Deg. of freedom 2 100 density 0.3 0.2 0.1 0 -8 -4 0 x 4 8 Example A manufacturer wants to estimate the average life of an expensive component. Because the components are destroyed in the process, only 5 components are tested. The lifetimes (in hours) of the 5 randomly selected components are 92, 110, 115, 103, and 98. Assuming that component lifetimes are normal, construct a 95% confidence interval estimate of the component’s life expectancy. Solution: Using Excel, x = 103.6 hours, and s = 9.18 hours. From the discussion above, the critical value is t 0.025 = 2.776. (Note: In Excel, shown below, to find the critical value associated with the t distribution and significance level , follow: fx > Statistical > TINV and enter  in the Probability field.) Thus a 95% CIE for the mean lifetime of the components is given by  9.18  103.6  2.776  or (92.2, 115.0) hours  5  Statistical Inference: Decision Making In hypothesis testing we are asked to evaluate a claim about something, such as a claim about a population mean. For instance, in a previous example a Business dean claimed that the average weekly income of graduates of his school one year after graduation is $600. Suppose that you suspect the dean’s claim may be exaggerated. Hypothesis testing provides a systematic framework, grounded in probability, for evaluating the dean’s claim against your suspicions. Although hypothesis testing uses probability distributions to arrive at a reasonable (and defensible) decision either to reject or "fail to reject" the claim associated with the null hypothesis of the test, H0, it does not guarantee that the decision is correct! The table below outlines the possible outcomes of a hypothesis test. (Note: We avoid "accepting" the null hypothesis for the same reason juries return verdicts of "not guilty" rather than of "innocent") TRUTH H0 True H0 False Decision: Accept H0 Reject H0 correct decision Type I error Type II error correct decision Type I error The error of incorrectly rejecting H0 when, in fact, it's true. In a hypothesis test conducted at the significance level , the probability of making a type I error, if H0 is true, is at most . Type II error The error of incorrectly failing to reject H0 when, in fact, it's false. For a fixed sample size n, you cannot simultaneously reduce the probability of making a Type I error and the probability of making a Type II error. (This is the statistician’s version of “there is no such thing as a free lunch.”) However, if you can afford to take a larger sample, it is possible to reduce both probabilities. Decision Making: Hypothesis Testing density Example Suppose that a sample of 36 graduates of the business school averaged $570 per week one year after graduation. Test the dean’s claim, against your suspicion, at the 5% level of significance. Solution: Normal Distribution (X 0.001) 1. H0:  = $600 (the dean’s claim) 24 Mean,Std. dev. 600,16.67 HA:  < $600 (your suspicion) 20 2.  =  (the probability of rejecting the 16 dean’s claim if she’s right) 12 8 3. Draw some pictures (see box to the right) 4 4. Critical Value: -z 0.05 = -1.645 0 5. From the sample 510 540 570 600 630 660 690 Standardized Test Statistic: x 570 − 600 z= = −1.8 Standard Normal 100 36 0.4 Mean,Std. dev. Distribution 6. Conclusion: There is sufficient evidence to 0,1 0.3 reject the dean’s claim at the 5% level of significance. Accept H0 0.2 -1.8 0.1 0 -5 -3 -1.645 -1 1 3 5 z the P-value Approach to Hypothesis Testing P-value The smallest significance level at which you would reject H0. The p-value is calculated from the test statistic, and is doubled for two-sided tests. Note:  and the p-value are the “before” and “after” significance levels for the test. We can reach a decision to accept or reject H0 by comparing the two significance levels. Rule: If the p-value > , then we "fail to reject" H0 If the p-value  , then we reject H0, i.e., we reject H0 for small p-values Example Suppose that a sample of 36 graduates of the business school averaged $570 per week one year after graduation. Use the p-value to test the dean’s claim, against your suspicion, at the 5% level of significance. Solution: Steps 1-3 are the same as before. 4. Critical Values are not used in this approach. 5. From the sample Standardized Test Statistic: 570 − 600 z= = −1.8 100 36 p-value = P ( Z  −1.8) = 0.0359 < 0.05 = , where we have used the fact that the test is left-tailed! 6. Conclusion: There is sufficient evidence to reject the dean’s claim at the 5% level of significance. Notice that we rejected the Dean’s claim under both the critical value and p-value approaches. This was not a coincidence: the two approaches always lead to the same decision. Since p-values are routinely computed by StatGraphics and Excel, we will usually use p-values to conduct significance tests. Note: Many of the (hypothesis) tests conducted in this course are two-sided, and assume that we are sampling from a normal population with unknown variance. When this is the case, Statgraphics will automatically return the correct p-value for the two-sided t test.