Comparing Two Population Parameters Equal Variance t-test for Means, Section 11.1 - 11.2 Unequal Variance t-test Means, Section 11.3

Chapter Objectives Select and use the appropriate hypothesis test in comparing Means of two independent samples Continuous variables Means of two dependent samples Proportions of two independent samples Discrete variables - count Variances of two independent samples PP 5

Differences between Means, Independent Samples The two samples are independent Selection of one sample is in no way affected by the selection of the other sample We are interested in whether men and women with a college education and working full-time have the same annual earnings Random variable is annual earnings; two populations Draw a sample from each population; calculate the respective sample means Observe Why do we observe a difference between the sample means? The samples are drawn from populations with different means, i.e., 1  2 The samples are drawn from populations with the same population means, but because of sampling error, we observe different sample means, 1 = 2 The test that we will develop and the sampling distribution assume that the two samples are independent. That is, the selection of one sample is in no way affected by the selection of the other sample. We are interested in whether men and women with a college education and working full-time have the same annual earnings. The random variable is annual earnings. The two populations are not related to each other. Each is a random sample. We draw a sample from each population and we calculate the respective sample means and we observe: The problem here is, why do we observe a difference between the sample means? The samples are drawn from populations with different means, i.e., 12. The samples are drawn from populations with the same population means, but because of sampling error, we observe different sample means. PP 5

Sampling Distribution of To answer this question, need to know how is distributed Difference between sample means is a statistic Need to know the characteristics of the sampling distribution of the difference between the sample means PP 5

Sampling Distribution of Let the random variable, X, follow a normal distribution in the populations or the sample sizes, n1 and n2, are both greater than 30 Imagine drawing all possible samples from each of the two populations Form all possible pairs of differences in sample means from population 1 and population 2 The distribution of the differences between these pairs of sample means is the sampling distribution PP 5

Characteristics of Sampling Distribution Distribution is normal X follows normal distribution in the populations or CLT Mean of the distribution Unbiased estimator Standard error is Sampling Distribution of the Difference in Sample Means, normally distributed 1 - 2 Z PP 5

Assumptions about Population Variances Two different scenarios arise in the comparison of the population means of independent samples Variances of the underlying populations are either known to be or assumed to be equal Use Equal (or pooled) variance t-test Variances are not assumed to be equal Use Unequal (or separate) variance t-test Two different scenarios arise in the comparison of independent samples. In the first, the variances of the underlying populations are either known to be or assumed to be equal. This leads to the pooled (or equal) variance t-test. In the second scenario, the variances are not assumed to be equal. In this case, there is considerable discussion among statisticians about the preferred approach. We will use a separate (or unequal) variance t-test. PP 5

Equal Variance t-Test for Differences in Means Two-sided test H0:1 - 2 = 0 There is no difference in the population means H1:1 - 2  0 There is a significant difference in the population means One-sided tests H0:1 - 2 ≤ 0 H1:1 - 2 > 0 H0:1 - 2 ≥ 0 H1:1 - 2 < 0 PP 5

Under the Null Hypothesis What do we expect to observe under the null hypothesis? The center of the sampling distribution is zero If we observe a large absolute difference between the sample means, then it is unlikely that the samples came from populations with equal means Sampling Distribution of the Difference in Sample Means, under the null normal What does the sampling distribution of the difference in the sample means look like under the null hypothesis? We are given the two sample means. Where does the observed lie? PP 5

Equal Variance t-Test for Differences in Means We need to convert into a standardized variable If we knew the population variances, the test statistic would be Realistically, we do not know the population variances PP 5

Equal Variance t-Test for Differences in Means Pool both samples together Use all information to produce a more reliable estimate of the unknown population variance Weighted average of the sample variances Weights are the respective degrees of freedom Substitute for unknown population variances Pooled variance estimate of unknown population variances PP 5

Equal Variance t-Test for Differences in Means Note: the degrees of freedom are n1 + n2 - 2 Compare the value of this t-test statistic with a critical t value DR: if (Crit. Value t ≤ t-test statistic ≤ Crit. Value t) do not reject Or use p-value approach PP 5

Problem - Negative Income Tax Would this welfare program cause the recipients to stop working? An experiment in the 1960s was done to find out Target population consisted of 10,000 low-income families in three New Jersey cities From these families, 400 were chosen at random for the control group Another 225 were chosen at random for the treatment group - and put on the negative income tax. All 625 families were followed for three years The control families averaged 7,100 hours of paid work and their standard deviation was 3,900 hours The treatment families averaged 6,200 hours and their standard deviation was 3,400 hours Is the difference between the sample means statistically significant at ⍺ = 0.05? In the 1960s, a “negative income tax” was proposed, supplementing low incomes instead of taxing them PP 5

Problem - Negative Income Tax H0: 1 - 2 = 0 The NIT has no effect on hours of work H1: 1 - 2  0 The NIT has an effect on hours of work Assume the population variances are equal Calculate the pooled sample variance PP 5

Problem - Negative Income Tax Ask how far the observed difference in the sample means lies from the center of the sampling distribution (0) if the null hypothesis is true Find the critical t value at ⍺ = 0.05 PP 5

Problem - Negative Income Tax Degrees of freedom n1 +n2 - 2 = 400 + 225 -2 = 623 = Compare the t-test statistic with the critical value 2.90 (test statistic) > 1.96 (critical value) Reject null hypothesis There is a significant difference in mean hours worked between the control and the treatment group Consumer theory would predict that as income increases the quantity demanded increases for normal goods. If we view leisure as a good, then we would expect people to demand more leisure as their income rises. Hence, their work hours should decrease as their leisure hours increase. Given consumer theory, a one-tail test would be appropriate here. PP 5

Online Homework covers section 11.1-11.2 CengageNOW fifth assignment Chapter 11:Intro to Two Samples CengageNow sixth assignment Chapter 11:Tests about Two Means PP 5

Unequal Variance t-Test for Differences in Means Assume variances of the two populations are not equal How do we estimate the standard error of the difference in sample means? Substitute sample variances The Separate (Unequal) Variances Test Next consider the situation in which the variances of the two populations are not assumed to be equal. In this case a modification of the previous test must be used. Instead of using as an estimate of the common population variance 2, we substitute the sample variances for the respective unknown population variances. The appropriate test statistic is PP 5

Unequal Variance t-Test for Differences in Means Test Statistic Use an approximation for the degrees of freedom Behrens Fischer solution Unlike the case in which we have equal variances, the exact distribution of t is difficult to derive. We use an approximation for the degrees of freedom (the Behrens Fischer problem) PP 5

Calculating the Degrees of Freedom Round df down to the nearest integer Compare the test statistic with the critical t value using estimated df degrees of freedom DR: if (Crit. Value t ≤ t-test statistic ≤ Crit. Value t) do not reject Round v down to the nearest integer. We are approximating the distribution of t by a t distribution with v degrees of freedom. We can then proceed to compare the test statistic with the critical t value using v degrees of freedom. PP 5

Two Possible Tests for Differences in Means, Independent Samples Variances of the underlying populations are either known to be or assumed to be equal Use Equal (or pooled) variance t-test Variances are not assumed to be equal Use Unequal (or separate) variance t-test Can test whether population variances are equal PP 5

P-value Approach with t - Distribution Tables Calculate the p-value for the test statistic, 2.90 Look at the t table, infinite degrees of freedom Find the table entries closest to the test statistic, 2.90 2.576 cuts off  = .005 It is the value closest to our test statistic Test statistic is more extreme So the p-value must be < .005 in one tail of the distribution or <.01 for both tails Compare the p-value with 0.05 .01 (p-value) <.05 (level of significance) Therefore we reject the null hypothesis PP 5

T-table

t - Distribution for Infinite Degrees of Freedom 0.10 0.05 0.025 0.01 1.28 1.64 1.96 2.33 0.005 2.58 Find appropriate distribution in terms of degrees of freedom Find boundaries for your particular test statistic, 2.90 2.58 cuts off 0.005 of t values 2.90 > 2.58 2.90 cut off less than 0.005 of t values, p-value < 0.005 PP 5

T Distribution for Infinite Degrees of Freedom 0.10 0.05 0.025 0.01 1.28 1.64 1.96 2.33 0.005 2.58 2.20 Suppose test statistics is 2.20 1.96 < 2.20 < 2.33 0.01 < area in tail < 0.025 = p-value for one-tail test 0.02 < p-value < 0.05 for two-tail test - multiply area in tail by two PP 5