Nurses and Hospital Stays. In the article "A Multifactorial Intervention Program Reduces the Duration of Delirium. Length of Hospitalization, and Mortality in Delirious Patients (Journal of the American Geriatrics Society, Vol. 53. No. 4. pp. 622-628), M. Lundstrom et al. investigated whether education programs for nurses improve the outcomes for their older patients. The standard deviation of the lengths of hospital stay on the intervention ward is $8.3$days.

a. For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of $80$patients on the intervention ward.

b. The distribution of the length of hospital stay is right-skewed. Does this invalidate your result in part (a)? Explain your answer.

c. Obtain the probability that the sampling error made in estimating the population means length of stay on the intervention ward by the mean length of stay of a sample of $80$patients will be at most $2$days.

Let the population mean for hospital stay length be $\mu$days.

Given that the population s. d.,$\sigma =8.3$ days, the population distribution of hospital stay lengths is unknown.

Formula used:

Here the sample size is $80$i.e., $n=80$

Because the sample size of $80$is larger than $30$, we can consider the sample to be substantial.

Mean of the sample mean $$\begin{gathered} {\mu }_{\overline{X}} \\ =\mu Days \end{gathered}$$

S.D of sample mean ${\sigma }_{\overline{X}}$

$$\begin{gathered} =\frac{\sigma }{\sqrt{n}} \\ =\frac{8.3}{\sqrt{80}} \\ =0.92796\text{Days} \end{gathered}$$

As a result of using CLT sampling, the distribution of all potential sample means of size $80$ samples is normal, with a mean of $\mu$ and a standard deviation of $0.92796$ days.

The distribution of hospital stay lengths is right-skewed. Partially, this knowledge does not invalidate the result (a). We can use CLT to approximate the distance of the sample mean as a normal distribution because the sample size is relatively large.

We have to find $P(\mu -2\le \overline{X}\le \mu +2)$

Where $\overline{X}~N\left(\mu ,{\sigma }_{\overline{X}}^2\right)$

Where ${\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}},n=80$

$$\begin{gathered} =0.92796days \\ \therefore P(\mu -2\le \overline{X}\le \mu +2) \\ =P\left(\frac{\mu -2-\mu }{{\sigma }_{\overline{X}}}\le \frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}\le \frac{\mu +2-\mu }{{\sigma }_{\overline{X}}}\right) \\ =P\left(\frac{-2}{0.92796}\le z\le \frac{2}{0.92796}\right),z=\frac{\overline{X}-\mu }{{\sigma }_{\overline{X}}}~N(0,1) \\ =P(-2.1553\le z\le 2.1583) \\ =\Phi (2.1553)-\Phi (-2.1553) \\ =2\Phi (2.1553)-1 \\ =2\times .9844308-1 \\ =0.9660896 \end{gathered}$$

As a result, the probability of guessing the population mean by the value of the sample mean of a sample of size $80$ with a sampling error of at most $2$ days is $0.9660896$