Asked by Michael
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
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
Answers
Answered by
PsyDAG
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.
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.
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