Questions: The standard deviation of the lengths of hospital stay on the intervention ward is 6.1 days. Complete parts (a) through (c) below. a. For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 62 patients. The sampling distribution will be approximately normal with mean μ and standard deviation σ̄ₓ= days. (Round to four decimal places as needed.)

The standard deviation of the lengths of hospital stay on the intervention ward is 6.1 days. Complete parts (a) through (c) below. a. For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 62 patients. The sampling distribution will be approximately normal with mean μ and standard deviation σ̄ₓ= days. (Round to four decimal places as needed.)
Transcript text: The standard deviation of the lengths of hospital stay on the intervention ward is 6.1 days. Complete parts (a) through (c) below. a. For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 62 patients. The sampling distribution will be approximately normal with mean $\mu$ and standard deviation $\sigma_{\bar{x}}=$ $\square$ days. (Round to four decimal places as needed.)
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Solution

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Solution Steps

Step 1: Calculate the Standard Error of the Mean

The standard deviation of the lengths of hospital stay on the intervention ward is given as \( \sigma = 6.1 \) days. For a sample size of \( n = 62 \) patients, the standard error of the mean (SEM) is calculated using the formula:

\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{6.1}{\sqrt{62}} \approx 0.7747 \text{ days} \]

Step 2: Describe the Sampling Distribution

The sampling distribution of the sample mean for samples of 62 patients will be approximately normal. The mean of this distribution is denoted as \( \mu \) (the population mean), and the standard deviation of the sampling distribution (standard error) is:

\[ \sigma_{\bar{x}} \approx 0.7747 \text{ days} \]

Final Answer

The sampling distribution will be approximately normal with mean \( \mu \) and standard deviation \( \sigma_{\bar{x}} = 0.7747 \) days.

\[ \boxed{\sigma_{\bar{x}} = 0.7747} \]

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