Question

The administrator of a nursing home would like to do a time-and-motion study of staff time spend per day performing nonemergency tasks. Prior to the introduction of some efficiency measures, the average number of person-hours per day spent on these task was 휇=16. The administrator wants to test whether the efficiency measures have reduced the value of 휇. How many days must be sampled to test the proposed hypothesis if she wants a test having 훼=0.05and the probability of type II error of at most 0.1 when the actual values of 휇is 12 hours or less? Assume 휎=7.64

Answer 1

Answer:

32 days must be sampled to test the proposed hypothesis.

Step-by-step explanation:

Consider the provided information.

The average number of person-hours per day spent on these task was $\mu_0=16$.

The actual value of μ is 12 hours or less i.e μ=12

She wants a test having α=0.05

The probability of a type II error of at most β=0.10 and σ=7.64.

δ = μ-μ0

   = 12-16

  = -4

Now use the formula for calculating the sample size:

$N=\frac{(Z_{\alpha}+Z_{\beta})^2\sigma^2}{\delta^2}$

Substitute the respective values.

$N=\frac{(Z_{0.05}+Z_{0.1})^2(7.64)^2}{(-4)^2}$

$N=\frac{(1.6449+1.2816)^2(7.64)^2}{16}$

$N=31.2437958482\approx 32$

Hence, 32 days must be sampled to test the proposed hypothesis.