According to a study by the Centers for Disease Control, the national mean hospital stay after childbirth is 2.1 days. Reviewing records at her own hospital, a hospital administrator calculates that the mean hospital stay for a random sample of 81 women after childbirth is 2.3 days with a standard deviation of 1.2 days.
Test the claim that the mean hospital stay after childbirth for this hospital is significantly longer than the national mean of 2.1 days.
What is the p-value?
A) 0.9312
B) 0.0688
C) 0.1376
D) 0.164
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To test the claim that the mean hospital stay after childbirth for this hospital is significantly longer than the national mean of 2.1 days, we can use a hypothesis test.
Let's define our hypotheses:
Null hypothesis (H0): The mean hospital stay for this hospital is equal to the national mean of 2.1 days.
Alternative hypothesis (Ha): The mean hospital stay for this hospital is significantly longer than the national mean of 2.1 days.
Next, we need to determine the test statistic. In this case, we will use the t-test since we have the sample mean, sample standard deviation, and sample size.
The formula for the t-test is:
t = (x̄ - μ) / (s / √n)
where:
- x̄ is the sample mean (2.3 days),
- μ is the population mean (2.1 days),
- s is the sample standard deviation (1.2 days), and
- n is the sample size (81).
Let's substitute the values into the formula to calculate the t-test statistic:
t = (2.3 - 2.1) / (1.2 / √81)
t = 0.2 / (1.2 / 9)
t = 0.2 / 0.133
t = 1.504
Now that we have the t-test statistic, we need to find the p-value associated with this test statistic. The p-value is the probability of observing a sample mean as extreme or more extreme than the one obtained, assuming the null hypothesis is true.
To find the p-value, we will consult a t-distribution table or use a statistical calculator. The degrees of freedom for this test is (n - 1) = (81 - 1) = 80.
Looking up the t-score of 1.504 with 80 degrees of freedom (rounded to the nearest thousandth) in a t-distribution table or using a calculator, we find that the p-value is approximately 0.1376.
Therefore, the answer is C) 0.1376.