Cholesterol levels are measured for 20 heart attack patients (two days after the

Question

Cholesterol levels are measured for 20 heart attack patients (two days after their attacks) and 26 other hospital patients who did not have a heart attack. The sample of heart attack patients had a mean cholesterol level of 225 and standard deviation 43.8. The sample of other hospital patients had a mean cholesterol level of 206.9 and standard deviation 15.3. The degrees of freedom for the t-distribution in this case is df=23.

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     right-tailed left-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?

Again, you may want to submit your answer for part d before submitting your answer for part e.
(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 99% confidence interval for the difference between population mean cholesterol levels for heart attack patients and other hospital patients.

99% CI = ? to ?

Explanation / Answer

Given that,
mean(x)=225
standard deviation , s.d1=43.8
number(n1)=20
y(mean)=206.9
standard deviation, s.d2 =15.3
number(n2)=26
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.539
since our test is right-tailed
reject Ho, if to > 2.539
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =225-206.9/sqrt((1918.44/20)+(234.09/26))
to =1.767
| to | =1.767
critical value
the value of |t | with min (n1-1, n2-1) i.e 19 d.f is 2.539
we got |to| = 1.76701 & | t | = 2.539
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.767 ) = 0.04664
hence value of p0.01 < 0.04664,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
critical value: 2.539
a. right-tailed
b. test statistic: 1.767
c. p-value: 0.04664
d. decision: do not reject Ho

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