Cholesterol levels are measured for 15 heart attack patients (two days after the

Question

Cholesterol levels are measured for 15 heart attack patients (two days after their attacks) and 27 other hospital patients who did not have a heart attack. The sample of heart attack patients had a mean cholesterol level of 229.9 and standard deviation 41. The sample of other hospital patients had a mean cholesterol level of 207.7 and standard deviation 15.6. The degrees of freedom for the t-distribution in this case is df=16.

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?

two-tailedfluffy-tailed    right-tailedleft-tailed

(b) What is the test statistic?  
(round your answer to three decimal places)

(c) What is the p-value?  
(round your answer to four decimal places)

It is recommended that you submit your answers for parts a, b, and c before submitting your answers for parts d and e below. That way, if you calculate the wrong p-value, you will have an opportunity to correct your answer before you use it to determine the statistical decision and its interpretation

(d) What is the statistical decision?  ---Select---  

Reject Ho

Fail to reject Ho

Reject Ha

Fail to reject Ha

Again, you may want to submit your answer for part d before submitting your answer for part e.
(e) This means we  ---Select---   conclude that the population mean cholesterol level of heart attack patients is higher than the population mean cholesterol level of other hospital patients.

can

cannot

might

always

never

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

here

hence degree of freedom =(S12/n1+S2/n2)2/((S12/n1)2/(n1-1)+ (S22/n2)2/(n2-1)) =16

also std error of difference =(S12/n1+S2/n2)1/2 =11.003

(a) What type of test is this?   right-tailed test

b)test stat =(X1-X2)/std error =2.0175

c)p value for above test stat and 16 degree of freedom =0.0304

d)as p value is lower then 0.05 level

hence Reject Ho

e)can

f) for 95% CI, t==2.1199

hence 95% CI=(x1-x2)+/- t*std error =-1.1267 ; 45.5267

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