Cholesterol levels are measured for 16 heart attack patients (two days after the

Question

Cholesterol levels are measured for 16 heart attack patients (two days after their attacks) and 28 other hospital patients who did not have a heart attack. The sample of heart attack patients had a mean cholesterol level of 227.1 and standard deviation 44.3. The sample of other hospital patients had a mean cholesterol level of 210.7 and standard deviation 15.4. The degrees of freedom for the t-distribution in this case is df=l7. The doctors leading the study think cholesterol levels will be higher for heart attack patients. Test the claim at the 0.05 level of significance. Use heart attack patients as "Population 1" and non-heart attack patients as "Population 2." (a) What type of test is this? right-tailed left-tailed fluffy-tailed two-tailed (b) What is the test statistic? (c) What is the p-value? (d) What is the statistical decision? (e) This means we conclude that the population mean cholesterol level of heart attack patients is higher than the population mean cholesterol level of other hospital patients. Now create a 95% confidence interval for the difference between population mean cholesterol levels for heart attack patients and other hospital patients. 95% CI = to

Explanation / Answer

a) right tailed

b)std error of differnece=(s12/n1+s22/n2)1/2 =11.451

hence test stat =(X1-X2)/std error =1.4322

c) p value =0.0851

d)fail to reject null hypothesis

e)do not have sufficient evidence

95% CI, =-7.7595 ; 40.5595

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