Review of Statistics

Estimation of the Population Mean Hypothesis Testing Critical value P-value Confidence Intervals Comparing Means from Different Populations Scatterplots and Sample Correlation

Estimation of the Population Mean One natural way to estimate the population mean, , is simply to compute the sample average from a sample of n i.i.d. observations. This can also be motivated by law of large numbers. But, is not the only way to estimate . For example, Y1 can be another estimator of .

Key Concept 3.1 Copyright © 2003 by Pearson Education, Inc.

In general, we want an estimator that gets as close as possible to the unknown true value, at least in some average sense. In other words, we want the sampling distribution of an estimator to be as tightly centered around the unknown value as possible. This leads to three specific desirable characteristics of an estimator.

Three desirable characteristics of an estimator. Let denote some estimator of , Unbiasedness: Consistency: Efficiency. Let be another estimator of , and suppose both and are unbiased. Then is said to be more efficient than if

Properties of It can be shown that E( )= and (from law of large numbers), is both unbiased and consistent. But, is efficient?

Examples of alternative estimators. Example 1: The first observation Y1? Since E(Y1)= , Y1 is an unbiased estimator of . But, if n≥2, is more efficient than Y1.

Example 2: where n is assumed to be an even number Example 2: where n is assumed to be an even number. The mean of is and its variance is Thus is unbiased and, because Var( ) → 0 as n→∞, is consistent. However, is more efficient than .

In fact, is the most efficient estimator of among all unbiased estimators that are weighted averages of Y1, … , Yn. (Weighted average implies that the estimators are all unbiased.)

Hypothesis Testing The hypothesis testing problem (for the mean): make a provisional decision, based on the evidence at hand, whether a null hypothesis is true, or instead that some alternative hypothesis is true. That is, test

Key Concept 3.5 Copyright © 2003 by Pearson Education, Inc.

Sampling Distribution Basic Idea Sample Means m = 50 H0 Sampling Distribution It is unlikely that we would get a sample mean of this value ... 20 ... therefore, we reject the hypothesis that  = 50. ... if in fact this were the population mean

Rejection Region (One-Tail Test) Ho Value Critical a Sample Statistic Rejection Region Nonrejection Sampling Distribution 1 –  Level of Confidence Observed sample statistic Rejection region does NOT include critical value.

Rejection Region (One-Tail Test) Sampling Distribution Ho Value Critical a Sample Statistic Rejection Region Nonrejection Sampling Distribution 1 –  Level of Confidence Level of Confidence Observed sample statistic Rejection region does NOT include critical value.

Rejection Regions (Two-Tailed Test) Ho Value Critical 1/2 a Sample Statistic Rejection Region Nonrejection Sampling Distribution 1 –  Level of Confidence Observed sample statistic Rejection region does NOT include critical value.

Rejection Regions (Two-Tailed Test) Ho Value Critical 1/2 a Sample Statistic Rejection Region Nonrejection Sampling Distribution 1 –  Level of Confidence Rejection region does NOT include critical value. Observed sample statistic

Rejection Regions (Two-Tailed Test) Ho Value Critical 1/2 a Sample Statistic Rejection Region Nonrejection Sampling Distribution 1 –  Level of Confidence Sampling Distribution Level of Confidence Rejection region does NOT include critical value. Observed sample statistic

Rejection Regions (Two-Tailed Test) Ho Value Critical 1/2 a Sample Statistic Rejection Region Nonrejection Sampling Distribution 1 –  Level of Confidence Sampling Distribution Level of Confidence Rejection region does NOT include critical value. Observed sample statistic

Hypothesis Testing The hypothesis testing problem (for the mean): make a provisional decision, based on the evidence at hand, whether a null hypothesis is true, or instead that some alternative hypothesis is true. That is, test

Key Concept 3.5 Copyright © 2003 by Pearson Education, Inc.

Terminology Significance level; p-value; critical value Confidence interval; acceptances region, rejection region Size; power

Two-Tailed Z Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified  to be 15 grams. Test at the .05 level of significance. 368 gm.

Two-Tailed Z Test Solution H0: Ha:   n  Critical Value(s):  = 368   368 Test Statistic: Decision: Conclusion: .05 25 Z 1.96 -1.96 .025 Reject H

Two-Tailed Z Test Solution Test Statistic: Decision: Conclusion: Do not reject at  = .05 No evidence average is not 368

One-Tailed Z Test Example Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified  to be 15 grams. Test at the .05 level of significance. 368 gm.

One-Tailed Z Test Solution H0: Ha:  = n = Critical Value(s):  = 368  > 368 Test Statistic: Decision: Conclusion: .05 25 Z 1.645 .05 Reject

One-Tailed Z Test Solution Test Statistic: Decision: Conclusion: Do not reject at  = .05 No evidence average is more than 368

Type I Error (α) and Type II error (β)

Type I and Type II Error Type I Error (Red) ，Type II Error (Blue) Alternative α/2 Size=2*α/2 β Power=1-β

Power Function

P value p-value = probability of drawing a statistic (e.g. ) at least as adverse to the null as the value actually computed with your data, assuming that the null hypothesis is true. The significance level of a test is a pre-specified probability of incorrectly rejecting the null, when the null is true.

Calculating the p-value based on : where is the value of actually observed. To compute the p-value, you need to know the distribution of . If n is large, we can use the large-n normal approximation.

where denotes the standard deviation of the distribution of .

Calculating the p-value with Y known

Two-Tailed Z Test p-Value Example Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified  to be 15 grams. Find the p-Value. 368 gm.

Two-Tailed Z Test p-Value Solution 1.50  Z value of sample statistic (observed)

Two-Tailed Z Test p-Value Solution p-value is P(Z  -1.50 or Z  1.50) .5000 - .4332 .0668  Z 1.50 -1.50 1/2 p-Value From Z table: lookup 1.50 .4332   Z value of sample statistic (observed)

Two-Tailed Z Test p-Value Solution p-value is P(Z  -1.50 or Z  1.50) = .1336  1/2 p-Value 1/2 p-Value .0668 .0668 .5000 - .4332 .0668 Z -1.50 1.50  From Z table: lookup 1.50  Z value of sample statistic

Two-Tailed Z Test p-Value Solution (p-Value = .1336)  ( = .05). Do not reject H0. Test statistic is in ‘Do not reject’ region 1/2 p-Value = .0668 1/2 p-Value = .0668 Reject H0 Reject H0 1/2  = .025 1/2  = .025 Z -1.50 1.50

How to estimate when is unknown In practice, is unknown—it must be estimated. Estimator of the variance of Y: Fact: If (Y1, … , Yn ) are i.i.d. and E(Y4)< ∞ , then Why does the law of large numbers apply? Because is a sample average. Technical note: we assume E(Y4)< ∞ because here the average is not of Yi , but of its square.

For the first term, Define , then , and W1, … , Wn are i.i.d. . Thus W1, … , Wn are i.i.d. and Var(Wi) < ∞ , so Therefore, For the second term, because

Computing the p-value with estimated:

The p-value and the significance level With a prespecified significance level (e.g. 5%): reject if |t| > 1.96. equivalently: reject if p ≤ 0.05. The p-value is sometimes called the marginal significance level.

What happened to the t-table and the degrees of freedom? Digression: the Student t distribution If Yi, i = 1,…, n is i.i.d. N(Y, ), then the t-statistic has the Student t-distribution with n – 1 degrees of freedom. The critical values of the Student t-distribution is tabulated in the back of all statistics books. Remember the recipe? Compute the t-statistic Compute the degrees of freedom, which is n – 1 Look up the 5% critical value If the t-statistic exceeds (in absolute value) this critical value, reject the null hypothesis.

t Test for Mean ( Unknown) Assumptions Population is normally distributed If not normal, only slightly skewed & large sample (n  30) taken Parametric test procedure t test statistic

Two-Tailed t Test Example Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372.5 and a standard deviation of 12 grams. Test at the .05 level of significance. 368 gm.

Two-Tailed t Test Solution  = 368   368 H0: Ha:  = df = Critical Value(s): Test Statistic: Decision: Conclusion: .05 36 - 1 = 35 t 2.030 -2.030 .025 Reject H

Two-Tailed t Test Solution Test Statistic: Decision: Conclusion: Reject at  = .05 There is evidence population average is not 368

Comments on this recipe and the Student t-distribution The theory of the t-distribution was one of the early triumphs of mathematical statistics. It is astounding, really: if Y is i.i.d. normal, then you can know the exact, finite-sample distribution of the t-statistic – it is the Student t. So, you can construct confidence intervals (using the Student t critical value) that have exactly the right coverage rate, no matter what the sample size. But….

Comments on Student t distribution, ctd. If the sample size is moderate (several dozen) or large (hundreds or more), the difference between the t-distribution and N(0,1) critical values are negligible. Here are some 5% critical values for 2-sided tests:

Comments on Student t distribution, ctd. So, the Student-t distribution is only relevant when the sample size is very small; but in that case, for it to be correct, you must be sure that the population distribution of Y is normal. In economic data, the normality assumption is rarely credible. Here are the distributions of some economic data. Do you think earnings are normally distributed? Suppose you have a sample of n = 10 observations from one of these distributions – would you feel comfortable using the Student t distribution?

The Student-t distribution – summary The assumption that Y is distributed N(Y, ) is rarely plausible in practice (income? number of children?) For n > 30, the t-distribution and N(0,1) are very close (as n grows large, the tn–1 distribution converges to N(0,1)) The t-distribution is an artifact from days when sample sizes were small and “computers” were people For historical reasons, statistical software typically uses the t-distribution to compute p-values – but this is irrelevant when the sample size is moderate or large. For these reasons, in this class we will focus on the large-n approximation given by the CLT

Confidence Intervals A 95% confidence interval for is an interval that contains the true value of Y in 95% of repeated samples. Digression: What is random here? the confidence interval— it will differ from one sample to the next; the population parameter, , is not random.

A 95% confidence interval can always be constructed as the set of values of not rejected by a hypothesis test with a 5% significance level.

This confidence interval relies on the large-n results that is approximately normally distributed and

Tests for Difference between Two Means Let be the mean hourly earning in the population of women recently graduated from college and let be population mean for recently graduated men. Consider the null hypothesis that earnings for these two populations differ by certain amount d, then

Sampling Distribution Population 1   Select simple random sample, n1. Compute X1 Population 2    Select simple random sample, n2. Compute X2 Compute X1 – X2 for every pair of samples In this diagram, do the populations have equal or unequal variances? Unequal. Astronomical number of X1 – X2 values m 1 - 2 Sampling Distribution 32

One Population Case

Two Population Case (independent) So far we do not know the sampling distribution of , If these two populations are independent, then we have

Comments on Student t distribution, ctd. Consider the t-statistic testing the hypothesis that two means (groups s, l) are equal Even if the population distribution of Y in the two groups is normal, this statistic doesn’t have a Student t distribution! There is a statistic testing this hypothesis that has a normal distribution, the “pooled variance” t-statistic – see SW (Section 3.6) – however the pooled variance t-statistic is only valid if the variances of the normal distributions are the same in the two groups. Would you expect this to be true, say, for men’s v. women’s wages?

Replace population variances by sample variances, we have the standard error and the t-statistic is If both nm and nw are large, the t-statistic has a standard normal distribution.

Walmart v.s. Dukes In 2000 Betty Dukes, a 54-year-old Wal-Mart worker in California, claimed sex discrimination. Despite six years of work and positive performance reviews, she was denied the training she needed to advance to a higher salaried position. Wal-Mart argued Dukes clashed with a female Wal-Mart supervisor and was disciplined for admittedly returning late from lunch breaks.[3] In June 2001, the lawsuit began in U.S. District Court in San Francisco. The plaintiffs sought to represent 1.6 million women, including women who were currently working or who had previously worked in a Wal-Mart store since December 26, 1998.[4]

Summarize the relationship between variables Scatterplots:

The population covariance and correlation can be estimated by the sample covariance and sample correlation. The sample covariance is The sample correlation is

It can be shown that under the assumptions that (Xi , Yi) are i. i. d It can be shown that under the assumptions that (Xi , Yi) are i.i.d. and that Xi and Yi have finite fourth moments,

It is easy to see that the second term converges in probability to zero because so by Slutsky’s theorem.

By the definition of covariance, we have To apply the law of large numbers on the first term, we need to have which is satisfied since

The second inequality follows by applying the Cauchy-Schwartz inequality, and the last inequality follows because of the finite fourth moments for (Xi , Yi). The Cauchy-Schwartz inequality is

Applying the law of large numbers, we have Also, , therefore

Scatterplots and Sample Correlation

Summary: From the assumptions of: (1) simple random sampling of a population, that is, (2) we developed, for large samples (large n): Theory of estimation (sampling distribution of ) Theory of hypothesis testing (large-n distribution of t-statistic and computation of the p-value). Theory of confidence intervals (constructed by inverting test statistic). Are assumptions (1) & (2) plausible in practice? Yes