t-test6/10/10Lecture 22 1 Last class: t-distribution: helps us do statistical inference when population SD is unknown Computed confidence intervals Today: Testing hypothesis using t-distribution6/10/10Lecture 22 2 For CI and hypothesis testing about a normal mean , when is unknown, the sample standard deviation sis used to estimate . Note: replaced by When is the mean of a random sample of size n from a normal distribution with mean , thenhas atdistribution with n- 1 degrees of freedom (df).XnsXt/nXz/t6/10/10Lecture 22 3 To determine number of workers required to meet demand, the productivity of new trainees is studied. It is believed trainees can process more than 450 packages per hour within one week of hiring. A sample of productivity of 50 trainees is observed and summarized as: Can we conclude that this belief is correct based on the sample?.83.38,38.460 sx6/10/10Lecture 22 4 Null hypothesis: H0: = 0 Test statistic:Hap-value 02P (T |t| )>0P (T t )<0P (T t )./0nsxt6/10/10Lecture 22 5 The problem objective is to describe the population of the number of packages processed in one hour. The hypotheses areH0: = 450 vs. Ha: > 450 (1-sided) The t statistic d.f. = n - 1 = 49, .025 < P-value < .05. There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at .05 significance level.89.15083.3845038.460nsxt6/10/10Lecture 22 6 Compare two population means based on twosamples Matched pairs study: turn two samples to one sample by taking differences Subjects are matched in pairs and outcomes compared within each pair Also common when observations are taken on same subject under different conditions6/10/10Lecture 22 7 To compare mean reaction times to two types of traffic signs, ◦ prohibitive (I) (e.g. No Left Turn)◦ permissive (II) (e.g. Left Turn Only) 15 subjects were chosen and each subject was presented with 40 traffic signs, 20 prohibitive and 20 permissive, in random order. The mean reaction times of the 15 subjects are as follows:1 2 3 4 5 6 7 8 9 10 11 12 13 14 15I7.6 10.2 9.5 1.3 3.0 6.3 5.3 6.2 2.2 4.8 11.3 12.1 6.9 7.6 8.4 II 7.3 9.1 8.4 1.5 2.7 5.8 4.9 5.3 2.0 4.2 11.0 11.0 6.1 6.7 7.5d0.3 1.1 1.1 -0.2 0.3 0.5 0.4 0.9 0.2 0.6 0.3 1.1 0.8 0.9 0.96/10/10Lecture 22 8 There is noticeable variability between subjects. Some people react more quickly to any type of sign than other people; The correlation within subjects is large. If a person reacts quickly to one type of signs, he/she is more likely to react quickly to another type of signs as well. 6/10/10Lecture 22 9 Null hypothesis: H0: d= 1-2 = 0 Test statistic: d.f. n-1Hap-valued 0 2P (T |t| )d > 0P (T t )d < 0 P (T t )./0nsdtd6/10/10Lecture 22 10 To compare mean reaction times to two types of traffic signs, prohibitive (I) (e.g. No Left Turn) and permissive (II) (e.g. Left Turn Only), 15 subjects were chosen and each subject was presented with 40 traffic signs, 20 prohibitive and 20 permissive, in random order. The mean reaction times of the 15 subjects are as follows:1 2 3 4 5 6 7 8 9 10 11 12 13 14 15I7.6 10.2 9.5 1.3 3.0 6.3 5.3 6.2 2.2 4.8 11.3 12.1 6.9 7.6 8.4 II 7.3 9.1 8.4 1.5 2.7 5.8 4.9 5.3 2.0 4.2 11.0 11.0 6.1 6.7 7.5d0.3 1.1 1.1 -0.2 0.3 0.5 0.4 0.9 0.2 0.6 0.3 1.1 0.8 0.9 0.9 Use the paired ttest to test the appropriate hypotheses at = .05.6/10/10Lecture 22 11Parameter of interest: d= 1- 2 , the difference between the average reaction times.Hypotheses: H0: d= 0 vs. Ha: d 0. So here Test statistic:P-value: 2*P(T>=5.98) < 2*0.0005. Conclusion: p-value < = .05, so H0is rejected. Therefore, the mean reaction times to the two types of traffic signs are significantly different..394.0 ,61.0 ,15 dsdn.98.515/394.061.0/nsdtd006/10/10Lecture 22 12 With paired samples, we can derive a confidence interval for d= 1- 2in the same way as we constructtintervals in one-sample problem. The statistics has a tdistribution with df = n- 1. Manipulation of this tvariable yields the following 100(1-)% CI:nsdTdd/./1,2/nstddn6/10/10Lecture 22 13 Construct a 95% confidence interval for the mean difference in reaction times: d= 1- 2. The 95% CI does NOT contain 0, as expected. Why?].828.0 ,392.0[218.061.015/394.0145.261.015/14,025.dstd6/10/10Lecture 22 14 1-sample t test and CI Assumptions: normal population, unknown SD, small samples Compare two population means via matched pairs turn 2 samples to just 1 sample by focusing on differences Exercises: (don’t turn in) 7.32, 7.39 (a – c)6/10/10Lecture 22
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