The true average diameter of ball bearings of a cer- tain type is supposed to be 0.5 in. What conclusion is ap- propriate when testing Ho: µ = 0.5 versus H„: µ ± 0.5 in each of the following situations: a. n = 13, t = 1.6, a = .05 b. n = 13, t = -1.6, a = .05 c. n = 25, t = -2.6, a = .01 d. n = 25, t = -3.6 %3D %3D
The true average diameter of ball bearings of a cer- tain type is supposed to be 0.5 in. What conclusion is ap- propriate when testing Ho: µ = 0.5 versus H„: µ ± 0.5 in each of the following situations: a. n = 13, t = 1.6, a = .05 b. n = 13, t = -1.6, a = .05 c. n = 25, t = -2.6, a = .01 d. n = 25, t = -3.6 %3D %3D
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:The true average diameter of ball bearings of a cer-
tain type is supposed to be 0.5 in. What conclusion is ap-
propriate when testing H,: µ = 0.5 versus H: µ + 0.5 in
each of the following situations:
a. n = 13, t = 1.6, a = .05
b. n = 13, t = -1.6, a = .05
c. n = 25, t = -2.6, a = .01
d. n = 25, t = -3.6
%3D
%3D
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