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

To test the proposed hypothesis, the administrator needs to sample 36 days to determine if the efficiency measures have reduced the average person-hours spent on nonemergency tasks.

Step-by-step explanation:

To determine the number of days that need to be sampled to test the proposed hypothesis, we need to use the formula for sample size in hypothesis testing:

n = [(Z_1-alpha/2 + Z_beta) * (sigma / (Mu0 - Mu))^2]

Where:

n is the sample size

Z_1-alpha/2 is the Z-score corresponding to the desired level of significance (alpha)

Z_beta is the Z-score corresponding to the desired probability of Type II error (beta)

sigma is the standard deviation of the population

Mu0 is the hypothesized population mean

Mu is the actual population mean

In this case, we have:

Z_1-alpha/2 = Z_0.05/2 = 1.96 (from the Z-table)

Z_beta = Z_0.1 = 1.28 (from the Z-table)

sigma = 7.64

Mu0 = 16

Mu = 12

Substituting these values into the formula, we get:

n = [(1.96 + 1.28) * (7.64 / (16 - 12))^2] ≈ 35.739

Since we cannot have a fractional number of days, we round up to the nearest whole number to get a sample size of 36 days.

Answer 2

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=((Z_(\alpha)+Z_(\beta))^2\sigma^2)/(\delta^2)`

Substitute the respective values.

`N=((Z_(0.05)+Z_(0.1))^2(7.64)^2)/((-4)^2)`

`N=((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.