hw6-solution.pdf - Statistics 106 Solutions for Homework 6...

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Statistics 106 Solutions for Homework 6 Due : Nov. 30, 2018, In Class 19.31 Refer to Eye contact e ff ect Problems 19.12 and 19.13. a. Estimate μ 21 with a 99 percent confidence interval. Interpret your interval estimate. b. Estimate μ 1 . with a 99 percent confidence interval. Interpret your interval estimate. c. Prepare a bar graph of the estimated factor B level means. What does this plot suggest about the factor B main e ff ects? d. Obtain confidence intervals for μ . 1 and μ . 2 , each with a 99 percent confidence coe ffi cient. Interpret your interval estimates. e. Prepare a bar graph of the estimated factor A level means. What does this plot suggest about the factor A main e ff ects? f. Obtain confidence intervals for D 1 = μ 2 . - μ 1 . and D 2 = μ . 2 - μ . 1 ; use the Bonferroni procedure and a 95 percent family confidence coe ffi cient. Summarize your findings. Are your findings consistent with those in parts (c) and (e)? g. Is the Bonferroni procedure used in part (f) the most e ffi cient one that could be used here? Explain. (Hint: Sche ff e’s procedure is also applicable for those comparisons in part (f), so calculate S and compare it with B ) h. How many pairwise comparisons are there between treatment means? Construct Tukey’s confidence intervals for these pairwise comparisons at family-wise confidence coe ffi cient 95%. Based on these intervals, which pairs of treatment means are significantly di ff erent from each other at level 0.05? Solution:
c. The bar graph of the estimated factor B level means is shown below: 1
The bar graph shows that ¯ Y . 1 . and ¯ Y . 2 . are di ff erent, thus μ . 1 and μ . 2 are very likely to be di ff erent, which means the factor B main e ff ects probably exist. d. ¯ Y . 1 . = 11 . 1, ¯ Y . 2 . = 15, s ( ¯ Y . 1 . ) = s ( ¯ Y . 2 . ) = MS E / ( an ) = . 7794, t (1 - α/ 2; ( n - 1) ab ) = t ( . 995; 16) = 2 . 921. Thus, the 99% CI for μ . 1 and μ . 2 are 11 . 1 ± 2 . 921( . 7794) = [8 . 823 , 13 . 377] and 15 ± 2 . 921( . 7794) = [12 . 723 , 17 . 277]. e. The bar graph of the estimated factor A level means is shown below: 2
The bar graph shows that ¯ Y 1 .. and ¯ Y 2 .. are di ff erent, thus μ 1 . and μ 2 . are very likely to be di ff erent, which means the factor A main e ff ects probably exist. f. ˆ D 1 = ¯ Y 2 .. - ¯ Y 1 .. = 3 . 3, ˆ D 2 = ¯ Y . 2 . - ¯ Y . 1 . = 3 . 9, s ( ˆ D 1 ) = 2 MS E / ( bn ) = 1 . 1023, s ( ˆ D 2 ) = 2 MS E / ( an ) = 1 . 1023, B = t (1 - α/ (2 g ); ( n - 1) ab ) = t ( . 9875; 16) = 2 . 473.

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