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MAXImum [283]
3 years ago
6

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
Mathematics
1 answer:
WINSTONCH [101]3 years ago
5 0

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.

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