Cardiac output was measured by thermo dilution in normal sample of 10 post cardiac surgical patients in the left lateral position. the results are 4.91, 4.10, 6.74, 7.27, 7.42, 7.50, 6.56, 4.64, 5.98, 3.14can we conclude on the basis of this data that population mean is different from 5.05.

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Mathematics Tejvir singh 3 months 1 Answer 79 views 0

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  1. We will use the Wilcoxon signed ranks test by assuming that all the requirement of that test are met.

     

    We have the given hypothesis as mean = 5.05.

    Let α = 0.05

    Cardiac output Di = Xi – 5.05 Rank of |Di| Signed rank of |Di|
    4.91 -0.14 1 -1
    4.10 -0.95 4 -4
    6.74 +1.69 6 +6
    7.27 +2.22 8 +8
    7.42 +2.37 9 +9
    7.50 +2.45 10 +10
    6.56 +1.52 5 +5
    4.64 -0.41 2 -2
    5.98 +0.93 3 +3
    3.14 -1.91 7 -7

    Here T+ = 41 and T- = -14 T = 14

    For n = 1o and α/2 = 0.0245

    The decision rule is to reject the assumed mean if the computed value of T is less than or equal to 8 which is the critical value for n = 1o and α/2 = 0.0245 in the K table.

    Since T = 14 which is greater than 8 so we are unable to reject the mean.

    Hence 5.05 can be the mean.

     

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