A clinical psychologist wished to compare three methods for reducing hostility levels in university students using a certain psychological test (HLT). High scores on this test were taken to indicate great hostility, and 11 students who got high and nearly equal scores were used in the experiment. Five were selected at random from among the 11 students and treated by method A, three were taken at random from the remaining six students and treated by method B, and the other three students were treated by method C. All treatments continued throughout a semester, when the HLT test was given again. The results are shown in the table.
Method | Scores on the HLT Test | ||||
A | 72 | 84 | 77 | 67 | 79 |
B | 55 | 75 | 72 | ||
C | 79 | 96 | 88 |
Let μA and μB, respectively, denote the mean scores at the end of the semester for the populations of extremely hostile students who were treated throughout that semester by method A and method B.
(a) Find a 95% confidence interval for μA.
(Round your answers to two decimal places.)
_ to _
(b) Find a 95% confidence interval for μB.
(Round your answers to two decimal places.)
_ to _
(c) Find a 95% confidence interval for (μA −
μB). (Round your answers to two decimal
places.)
_ to _
a)
95% confidence interval for μA =estimate -/+ t*sqrt(MSE/n) =67.26 , 84.34
b)
95% confidence interval for μB =56.31 , 78.35
c)
95% confidence interval for (μA − μB) =(-5.47 , 22.41)
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