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Unformatted text preview: MN1025 Business Statistics 24 Lecture 6Friday 15/2/2008 COMPARING THE MEANS OF TWO POPULATIONS Reference: Lind et al. , Chapter 11. Note that they deal with the material in this lecture in a different order. 6.1 Revision: testing the mean In all examples covered last week, the null hypoth- esis H was of the form equals something. We looked at three different cases for the alternative hy- pothesis, namely less than, greater than and not equal to something. In the following, we will use the T-statistic and the 5% level of significance. We will assume sample data of size n , mean x and standard deviation s . Less than: Testing H : = 88 against H 1 : < 88. The critical value c is given by P ( T < c ) = 0 . 05. c=-1.76 probability density T n=15 0.05 0.1 0.2 0.3 0.4-4-2 2 4 Compute T observed = x 88 s/ n . Reject H if T observed < c . The P-value is given by P ( T < T observed ). Greater than: Testing H : = 12 against H 1 : > 12. The critical value c is given by P ( T > c ) = 0 . 05, which means P ( T < c ) = 0 . 95. area=0.95 0.05 T probability density n=20 c=1.729 0.1 0.2 0.3 0.4-4-2 2 4 Compute T observed = x 12 s/ n . Reject H if T observed > c . The P-value is given by P ( T > T observed ). Not equal: Testing H : = 520 against H 1 : negationslash = 520. The critical value c is given by P ( T < c ) + P ( T > c ) = 0 . 05, which means P ( T < c ) = 0 . 975. area=0.95 probability density-2.306 2.306 area=0.025 area=0.025 n=9 T 0.1 0.2 0.3 0.4-4-2 2 4 Compute T observed = x 520 s/ n . Reject H if T observed < c or T observed > c . The P-value is given by P ( T < | T observed | )+ P ( T > | T observed | ). 6.2 Testing for Change We consider first the case where two populations consist of the same objects with two different treat- ments, or possibly before and after some treat- ment has been applied. The samples will consist of the same objects: they are measured before and after. The question is: has the treatment had any effect? or has it led to an improvement? This is not a new problem; all that we need to do is ap- ply last weeks methods to the differences between the populations. Are these differences, on average, zero (that is, no change), or positive (an improve- ment), or negative (a worsening)? Examples might be: sales in a particular area before and after a sales campaign, or performance of staff before and after training. 6.3 Example: solutions for seeds Solutions A and B for seeds are tested. The data are for 9 types of seed and give the number of days for growth to 6 inches. Question: is there significant evidence that there is a difference (at the 5% level of significance)? Here we are testing the difference between the times for A and B....
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