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# 669 Class Note for STAT 51100 with Professor Levine at Purdue

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Date Created: 02/06/15

Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Lecture 15 Tests about a Population Mean Devore Section 83 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 A Normal Population with known 0 o This case is not common in practice We will use it to illustrate basic principles of test procedure design a Let X1 Xn be a sample size n from the normal population The null value of the mean is usually denoted MO and we consider testing either of the three possible alternatives ugtuoultuoanduuo o The test statistic that we will use is X Mo Z a It measures the distance of X from 0 in standard deviation units AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 0 Consider Ha u gt MO as an alternative The outcome that would allow us to reject the nulkl hypothesis H0 u MO is Z 2 CforsomeC gt 0 o How do you select C We need to control the probability of Type Error For a test of level 04 we have 05 P Type Error PZ 2 CZ N NO 1 0 Therefore we need to choose C 2a Such a test procedure is called uppertailed o It is easy to understand that for Ha u lt LLO we will have the rejection region of the form z g C For the test to have the level 04 we need to choose C za Such a test is called a lowertailed test AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 0 Now considerthe case of HG u 7 MO The rejection region here consists of z 2 c and 2 g C o For simplicity consider the case 05 005 Then 005 PZ 2 corZ g cz N M0 1 ltIgt C 1 c 20 c 0 Therefore we select C Such that 1 ltIgtc P Z 2 c 0025 it is 20025 196 This test is called a twotailed test AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Summary X uo ax 39 1 Ha u gt uo has the rejection region Z 2 2a and is called a Let H0 u Lto define the test statistic Z an uppertailed test 39239 Ha u lt 0 has the rejection region z 3 za and is called an lowertailed test 393 Ha u 75 do has the rejection region 2 Z 2042 or 239 g Za2 and is called a twotailed test AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Recommended Steps for Testing Hypotheses about a Parameter 1 Identify the parameter of interest and describe it in the context of the problem situation Determine the null value and state the null hypothesis State the alternative hypothesis Give the formula for the computed value of the test statistic State the rejection region for the selected significance level 93914er Compute any necessary sample quantities substitute into the formula for the test statistic value and compute that value a The formulation of hypotheses steps 2 and 3 should be done before examining the data AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Example 0 Consider Ex 86 in Devore39 0 Parameter of interest is u true average activation temperature a H0M130Hau7 130 a Test statistic is E u0 53 130 Z 0W 15 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 o Rejection region is 2 g Za2 orz 2 zaZ Ifoz 001 we have Z S Z0005 and Z 2 20005 a With n 9 and a 13108 Z 13108 130 15 0 Since 216 is outside the rejection region we fail to reject H0 at 216 significance level 001 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Type II Error 0 As an example consider an uppertailed test with the rejection region Z 2 2a 0 H0 is not rejected when a lt uo 2a a o For a particular u gt u the probability of Type II error is then lu PO lt 0 Ha ax lu M X M Z Mo M Pa lt a 39 Z M ay AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 0 Similar derivations can help us to derive Type II error probabilities for a lowertailed test and a twotailed test Results can be summarized as follows 1 Ha u gt uo has the probability of Type II Error Mo u 1 2a ONE 2 Ha u lt uo has the probability of Type II Error Ito ll 1 ltIgt 2a IAH 3 Ha u 7 0 has the probability of Type II Error I 2042 13 za2 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Sample Size Determination 0 Sometimes we want to bound the value of Type II error for a u u Sp eCIfIC value u 0 Consider Ex 86 again fix oz and specify for such an alternative value For M 132 we may want to require 132 01 in addition to Oz 01 o The sample size required for that purpose is such that 0 q M Z 3 a AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 0 Solving for n we obtain n 0 my lm M and the same answer is true for a lowertailed test a For a twotailed test it is only possible to give an approximate solution It is 2 0 z ilt 2 wi lm H AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Large Sample Tests 0 When the sample size is large the 2 tests described earlier are modified to yield valid test procedures without requiring either a normal population distribution or a known 0 a Let us assume n gt 40 Then the test statistic X Z J0 S is approximately standard normal 0 The use of the same rejection regions as before results in a test procedure for which the significance level is approximately a AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 A Normal POpulation Distribution 0 When n is small we can no longer invoke CLT as a justification for the large sample test 0 Remember that for a normally distributed random sample X1 7Xn the statistic XM Sx has a 75 distribution with n 1 df T 0 Therefore we have the test with H0 u39 MO and a test statistic value if S AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Summary of the Three Possible tTests o If Ha u gt uo the rejection region of the level oz test is t Z 7john 1 o If Ha u lt 0 the rejection region of the level 04 test is t S tozn 1 o If Ha u 3e M0 the rejection region of the level 05 test is t Z taZ L l ort toz2n 1 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Example 0 The Edison Electric Institute publishes figures on the annual number of kilowatt hours expended by various home appliances a It is claimed that a vacuum cleaner expends an average of 46 kilowatt hours per year 0 Suppose a planned study includes a random sample of 12 homes and it indicates that VC s expend an average of 42 kilowatt hours per year with s 119 kilowatt hours 0 Assuming the population normality design a 005 level test to see whether VC s spend less than 46 kilowatt hours annually AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 0 H0 u 46 kilowatt hours and Ha u lt 46 kilowatt hours 0 Assuming Oz 005 we have a critical region 75 lt 1796 where 513 M0 Sx t with 11 df o The value of the statistic is 42 4 13 6 116 119 0 Since I is not in the rejection region we fail to reject H0 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 A curve for the ttest o It is much more difficult to compute the probability of the Type II Error in this case than in the normal case 0 The reason is that it requires the knowledge of distribution of T under the alternative Ha To do it precisely we must compute mo PT lt ta7n1 when u ul 0 There exist extensive tables of these probabilities for both one and twotailed tests AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Figure 1 A Typical Cum for the tTest 3 suite far it 1 11 ii when 4 i H Li Value nfd cmi39eiipunding t0 specified alternative to ti39 AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 Calculating 0 First we select u and the estimated value for unknown 0 Then we find an estimated value of d u0 u0 Finally the value of is the height of the n 1 df curve above the value of d a If n 1 is not the value for which the corresponding curve appears visual interpolation is necessary AUG 2006 Statistics 511 Statistical Methods Purdue University Dr Levine Fall 2006 a One can also calculate the sample size it needed to keep the Type II Error probability below for specified 04 1 First we compute d 2 Then the point d is located on the relevant set of graphs 3 The curve below and closest to the point gives n 1 and thus n 4 Interpolation of course is often necessary AUG 2006

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