(Please Show Work) Cholesterol levels are measured for 19 heart attack patients

Question

(Please Show Work)

Cholesterol levels are measured for 19 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 228.3 and standard deviation 44.6. The sample of other hospital patients had a mean cholesterol level of 213.7 and standard deviation 14.6. The degrees of freedom for the t-distribution in this case is df=21.

The doctors leading the study think cholesterol levels will be higher for heart attack patients. Test the claim at the 0.01 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?

fluffy-tailed two-tailed     left-tailed right-tailed

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

It is recommended that you submit your answers for parts a and b before submitting your answers for parts c and d below.

(c) What is the statistical decision?

Reject h0? fail to reject Ho? Reject Ha? Fail to reject Ha?

Again, you may want to submit your answer for part c before submitting your answer for part d.
(d) This means we...can? cannot? might? always? never?.... conclude that the population mean cholesterol level of heart attack patients is higher than the population mean cholesterol level of other hospital patients.

(e) Now create a 99% confidence interval for the difference between population mean cholesterol levels for heart attack patients and other hospital patients.

99% CI = ? to ?

Explanation / Answer

a)

The doctors leading the study think cholesterol levels will be higher for heart attack patients.

Hence, this is a RIGHT TAILED TEST. [ANSWER]

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0  
Ha:   u1 - u2   >   0  
At level of significance =    0.01          
As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    228.3          
X2 =    213.7          
              
Calculating the standard deviations of each group,              
              
s1 =    44.6          
s2 =    14.6          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    19          
n2 = sample size of group 2 =    28          

Here, df = 21.

Also, sD =    10.5974284          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    1.377692724   [ANSWER]

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c)      
              
where uD = hypothesized difference =    0          
              
Now, the critical value for t is              
              
tcrit = 2.517648016

As t < 2.517, we FAIL TO REJECT HO. [ANSWER]

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d)              

This means we CANNOT conclude that the population mean cholesterol level of heart attack patients is higher than the population mean cholesterol level of other hospital patients. [ANSWER]

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e)

For the   0.99   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.005          
df = 21

Hence,

t(alpha/2) =    2.831359558          

Thus,
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    -15.40513019          
upper bound = [X1 - X2] + t(alpha/2) * sD =    44.60513019          
              
Thus, the confidence interval is              
              
(   -15.40513019   ,   44.60513019   ) [ANSWER]

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