Question

QUESTION 22 Problem 22) For a project that investigates the average weight difference by gender of patients from the Mayo clinic the researcher collected a sample of 64 male patients and another sample of 36 female patients. The researcher then calculated the sample mean of male patients as 160 lbs with a sample standard deviation of 20 lbs, and the sample mean of female patients as 140 lbs with a sample standard deviation of 10 lbs. Test the claim that the average weight of male patients is more than that of female patients. Assume unequal population standard deviation. Given the significance level equals 5%. (Question). Step 2: Find the Critical Value. QUESTION 23 Problem 23) For a project that investigates the average weight difference by gender of patients from the Mayo clinic, the researcher collected a sample of 61 male patients and another sample of 36 female patients. The researcher the calculated the sample mean of male patients as 160 lbs with a sample standard deviation of 20 lbs, and the sample mean of female patients 140 lbs with a sample standard deviation of 10 lbs. Test the claim that the average weight of male patients is more than that of female patients. Assume unequal population Standard deviation. Given the significance level equals 5. (Question), Step 3: Find the test statistics ve

Answer 1

Given that,
mean(x)=160
standard deviation , s.d1=20
number(n1)=64
y(mean)=140
standard deviation, s.d2 =10
number(n2)=36
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.69
since our test is right-tailed
reject Ho, if to > 1.69
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =160-140/sqrt((400/64)+(100/36))
to =6.656
| to | =6.656
critical value
the value of |t α| with min (n1-1, n2-1) i.e 35 d.f is 1.69
we got |to| = 6.6564 & | t α | = 1.69
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 6.6564 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 > u2
23.
test statistic: 6.656
22.
critical value: 1.69
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that average weight ofmale ismore than females.