Question 1118308: Statistics help--please show me how to step by step set up and solution for this problem Hospital Stays for Maternity Patients Health Care Knowledge Systems reported that an insured woman spends on the average 2.3 days in the hospital for a routine childbirth, while an uninsured woman spends and average of 1.9 days. Assume two random samples of 16 woman each were used in both samples. The standard deviation of the first sample is 0.3 day. At a=0.05, test the claim that the means are equal. Find the 99% confidence interval for the difference of the means. Use the P-value method.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Hospital Stays for Maternity Patients Health Care Knowledge Systems reported that an insured woman spends on the average 2.3 days in the hospital for a routine childbirth, while an uninsured woman spends and average of 1.9 days.
Assume two random samples of 16 woman each were used in both samples.
The standard deviation of the first sample is 0.3 day.
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Question:: What is the std for the second sample ??????
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At a=0.05, test the claim that the means are equal.
Ho: i-u = 0
Ha: i-u # 0
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Sample difference:: 2.3-1.9 = 0.4
z(0.4) = 0.4/sqrt[(0.3^2/16)+(?^2/16)] = ?????
Comment:: Cannot complete without know the std of the second sample.
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Cheers,
Stan H.
Find the 99% confidence interval for the difference of the means.
Use the P-value method.
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