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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
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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 • Continuous variables • 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 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 Sampling Distribution of the Difference in Sample Means, normally distributed • Distribution is normal • X follows normal distribution in the populations or CLT • Mean of the distribution • Unbiased estimator • Standard error is 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 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
0 Where does the observed lie? 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 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 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? 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 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 PP 5
Unequal Variance t-Test for Differences in Means • Test Statistic • Use an approximation for the degrees of freedom • Behrens Fischer solution 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 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
0.10 0.05 0.025 0.005 0.01 1.28 1.64 1.96 2.33 2.58 0 t - Distribution for Infinite Degrees of Freedom Find boundaries for your particular test statistic, 2.902.58 cuts off 0.005 of t values2.90 > 2.58 2.90 cut off less than 0.005 of t values, p-value < 0.005 PP 5
0.10 0.05 0.025 0.005 0.01 1.28 1.64 1.96 2.33 2.58 0 T Distribution for Infinite Degrees of Freedom 2.20 Suppose test statistics is 2.201.96 < 2.20 < 2.330.01 < area in tail < 0.025 = p-value for one-tail test0.02 < p-value < 0.05 for two-tail test - multiply area in tail by two PP 5