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Section 7.3 Table 7.3-1 summarizes hypothesis tests to compare two means. These tests assume that each sample is taken from a normal distribution, but.

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Presentation on theme: "Section 7.3 Table 7.3-1 summarizes hypothesis tests to compare two means. These tests assume that each sample is taken from a normal distribution, but."— Presentation transcript:

1 Section 7.3 Table 7.3-1 summarizes hypothesis tests to compare two means. These tests assume that each sample is taken from a normal distribution, but the tests are robust when each sample is taken from a non-normal distribution with finite mean and variance. These tests also assume that the two populations being compared have equal variance, but the tests are robust when the sample sizes are equal; if the variances are different and the sample sizes are different, Welch’s modified t statistic should be used, as described at the end of Section 7.3 of the text. 1. Do Text Exercise 7.3-2 under the assumption that  X 2 =  Y 2. (a) The test statistic is t = The one-sided critical region with  = 0.01 is x – y —————— = s P  1/n + 1/m t  2.473. x – y —————— s P  1/16 + 1/13

2 (b) (c) Since t = 5.570 > t 0.01 (27) = 2.473, we reject H 0. We conclude that the males have higher mean weight. The p-value of the test is P(T  5.570) which is considerably less than 0.005, from Table VI in Appendix B. (15)(1356.75) + (12)(692.21) s 2 P = ———————————— =, t = 27 1061.405.570 a t(27) random variable The results of Welch’s test are almost identical. We find that the test statistic is t = 5.767, the degrees of freedom  26.6  27, and the critical region is still defined by t  2.473.

3 2. (a) (b) Use the Analyze > Compare Means > Independent-Samples T Test options in SPSS to do Text Exercise 7.3-5 under the assumption of equal variances. The test statistic is t = The one-sided critical region with  = 0.05 is x – y —————— = s P  1/n + 1/m t  – 1.706. x – y —————— s P  1/14 + 1/14

4 From the SPSS output, we find t =– 1.714. Since t = – 1.714 < – t 0.05 (26) = – 1.706, we reject H 0. We conclude that the mean birth weight is smaller with mothers who have five or fewer prenatal visits. (c) The p-value of the test is P(T  – 1.714) which is between 0.025 and 0.05, from Table VI in Appendix B. From the SPSS output we find p-value = 0.098 / 2 = 0.049. a t(26) random variable

5 2.-continued (d) The box plots appear to support the conclusion.


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