1 22S 105 Statistical Methods and Computing 2 Two independent sample problems Goal of inference to compare the characteristics of two different populations to compare responses to two different treatments Two independent sample t tests Lecture 17 Mar 28 2011 Kate Cowles 374 SH 335 0727 kcowles stat uiowa edu 3 Examples of two independent sample problems A medical researcher is interested in the effect on blood pressure of added dietary calcium She conducts a randomized comparative experiment in which one group of subjects receives a calcium supplement and a control group gets a placebo A climatologist wishes to test whether seeding with silver nitrate affects the amount of rainfall produced from clouds He randomly selects 26 clouds to seed and measures the rain output by each of them as well as the rain output by 26 other randomly selected unseeded clouds 4 Which study design The following situations require inference about a population mean or means Identify which type of problem each one is one sample paired sample two independent samples 5 1 To check a new method of chemical analysis a chemist gets a reference specimen of known concentration from the National Institute of Standards and Technology She then makes 20 measurements of the concentration of this specimen using the new method and checks for bias by comparing the mean of her 20 measurements with the known concentration 6 Comparing means from two different populations Assumptions We have two simple random samples from two distinct populations The samples are independent The selection of one sample has no influence on the selection of the other In particular there is no matching The sizes of the two samples need not be the same We measure the same variable for both samples 2 Another chemist is checking the same new method He has no reference specimen but a familiar analytic method is available He wants to know if the new and old methods agree He takes a specimen of unknown concentration and measures the concentration 10 times with the new method and 10 times with the old method 7 The set up for two independent sample t tests Group 1 Group 2 Population Mean 1 2 1 2 Standard deviation Sample Mean x 1 x 2 Standard deviation s1 s2 n1 n2 Sample size The populations are normally distributed 8 Example Cloud seeding We wish to use our sample data to test whether the population mean of rainfall produced per cloud is the same for unseed clouds as for seeded clouds We will use a two sided test assuming that we don t know in advance in what direction a difference is likely to go H0 u s or u s 0 Ha u 6 s or u s 6 0 We will conduct our test at 05 9 Thus the quantity we really want to estimate is the difference between the two population means u s As usual we will use the corresponding sample statistics x u x s as our best guess of the unknown population value of interest Now we need to standardize x u x s in order to find out whether it is different enough from 0 to provide strong evidence against H0 10 Suppose we knew the standard deviations u and s in the populations of rainfall amounts from unseeded and seeded clouds u s some known value Then the standard error of x s x u would be v u u u u u u u t And the z statistic is z That is we need to compute x u x s u0 s0 standard error of x u x s 11 3000 2500 2000 1500 1000 500 0 seeded The UNIVARIATE Procedure Variable rainfall 0 0 S U 2 2 nu ns x u x s u0 s0 v u u u t 2 2 nu ns 12 The UNIVARIATE Procedure Variable lograin 8 7 6 5 4 3 2 0 0 1 0 0 seeded S U 13 Example 14 The results of measuring and logtransforming the rainfall are as follows We will log transform the rainfall amounts to get more symmetrical distributions of sample values Suppose we knew that the population standard deviation of log transformed rainfall amount was 1 5 log acre feet for both seeded and unseeded clouds Variable LOGRAIN SEEDED N Mean Std Dev Std Error S 26 5 13418678 1 59951361 0 31369043 U 26 3 99040563 1 64184748 0 32199278 The z statistic is u s 1 5 z 3 99 5 13 0 v u u u t 1 52 1 52 26 26 1 14 0 416 2 74 The cutoff values of z for a two sided hypothesis test are 1 96 and 1 96 Because 2 74 is farther from 0 than either of these cutoffs we can reject the null hypothesis and conclude that the pop 15 ulation mean of log rainfall is different in seeded clouds from unseeded clouds 16 If we do not know in the two populations we need to use the data to estimate the standard error of x u x s This can be done under two different assumptions 1 The standard deviations in the two populations are known to be equal 2 The standard deviations in the two populations are not known or assumed to be equal 17 Two sample t test when variances are assumed to be equal 18 For the cloud seeding example s2p 1 622 We must estimate the common variance 2 using the pooled variance s2p from the two samples nu 1 s2u ns 1 s2s nu ns 2 s2p Then we compute the t statistic by substituting s2p for 2 in the formula for the z statistic t t x u x s u0 s0 v u u u u t When we cannot assume that the variances in the two populations are equal When we do not assume that the standard deviations in the two populations are equal i e when we think u2 with s2u s2 with s2s v u u u t 1 622 1 622 26 26 1 144 0 458 2 54 We would compare this to the 025 cutoff for a t distribution with nu ns 2 50 degrees of freedom Because the value we obtained is farther from 0 than ths we can reject the null hypothesis 19 then we estimate 3 99 5 13 0 According to Table C this cutoff would be 2 009 s2p s2p nu ns u 6 s 20 Then we compute the t statistic as t x u x s u0 s0 v u u u t s2u s2s nu ns For the cloud seeding example this gives 3 99 5 13 0 t vuu 2 u 1 64 1 602 t 26 26 1 144 0 458 2 54 This form of t statistic does not come from an exact t distibution Statistical software uses an approximation to compute the p value in this case Generally the p value is very close to that obtained under the assumption of equality of variance 21 22 SAS for two independent sample t tests data 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