The administrator of a nursing home would like to do a time-and-motion study of

Question

The administrator of a nursing home would like to do a time-and-motion study of staff time spent per day performing non-emergency tasks. Prior to the introduction of some efficiency measures, the average number of person-hours per day spent on these tasks was = 16. The administrator wants to test whether the efficiency measures have reduced the value of , so the relevant hypotheses are Ho: = 16 vs. Ha: < 16. Calculate n , the number of days that must be sampled to test the proposed hypotheses if she wants a test having = 0.05 and = 0.10 when she is trying to detect a difference of 4 person-hours and we assume = 7.64.

Explanation / Answer

Solution:

Test is H0 : µ 16 vs. Ha : µ < 16 at = 0.05

Want = 0.10, whenever µ 12. Given = 7.64

= -0
    = 12-16 = -4

Since z.05 = 1.645 and z.10 = 1.28,

n = (7.64)2 (1.645 + 2.33) /(12 16)2 = 14.5011

giving n = 15

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