A study involving the economic burden of congestive heart failure found that the lengths of hospital stays for patients had a mean of 7.8 days with a standard deviation of 9.1 days. For random samples of size 40, find the sample mean length of stay that is smaller than 10% of all sample means.
To find the sample mean length of stay that is smaller than 10% of all sample means, we need to find the z-score that corresponds to the 10th percentile of the standard normal distribution.
The formula to calculate the z-score is:
z = (x - μ) / (σ / √n)
Where:
- x is the value we want to find the percentile for (in this case, the sample mean length of stay)
- μ is the population mean (7.8 days)
- σ is the population standard deviation (9.1 days)
- n is the sample size (40)
To find the z-score corresponding to the 10th percentile, we can use a standard normal distribution table or a calculator.
Using a standard normal distribution table, we can look up the z-score that corresponds to a cumulative probability of 0.10 (10th percentile).
The z-score corresponding to a cumulative probability of 0.10 is approximately -1.28.
Now, we can rearrange the z-score formula to solve for the sample mean length of stay (x):
x = z * (σ / √n) + μ
Plugging in the values:
x = -1.28 * (9.1 / √40) + 7.8
Calculating this:
x = -1.28 * (9.1 / 6.324) + 7.8
x = -1.848 + 7.8
x ≈ 5.952
Therefore, the sample mean length of stay that is smaller than 10% of all sample means is approximately 5.952 days.
To find the sample mean length of stay that is smaller than 10% of all sample means, we need to calculate the z-score associated with a sample mean of this value.
The formula for calculating the z-score is:
z = (x - μ) / (σ / √n)
Where:
- z is the z-score
- x is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size
In this case, we are given the population mean (μ) as 7.8 days, the population standard deviation (σ) as 9.1 days, and the sample size (n) as 40.
First, we need to find the z-score corresponding to a percentile of 10% or 0.1. We can look up this z-score in the standard normal distribution table or use statistical software like R or Python.
Using a standard normal distribution table, we find that the z-score corresponding to a percentile of 10% is approximately -1.28.
Now we can substitute the given values into the z-score formula:
-1.28 = (x - 7.8) / (9.1 / √40)
Next, we can solve for x (the sample mean):
-1.28 = (x - 7.8) / (9.1 / 6.3246)
Multiply both sides of the equation by (9.1 / 6.3246):
-1.28 * (9.1 / 6.3246) = x - 7.8
Simplifying the equation:
-1.84852 = x - 7.8
Add 7.8 to both sides of the equation:
x = -1.84852 + 7.8
x ≈ 5.95148
Therefore, the sample mean length of stay that is smaller than 10% of all sample means is approximately 5.95148 days.