Question

The following table contains cholesterol levels collected from non-smoking males. The before cholesterol level is before they exercised, and the after cholesterol level was taken after the same subject regularly exercised for two months. It is claimed that the mean cholesterol level before exercise is greater than that after exercise. Assume that the population of cholesterol level difference (before-after) is normally distributed. Test the claim at the 0.01 level.

Cholesterol Before

176

163

169

169

166

182

Cholesterol After

181

165

160

156

148

168

Which statistical test should be used to test the claim?

2 sample T-test

2 Proportion Z-test

Z-test

Paired T-test

What is the test statistic? (Report answer accurate to 3 decimal places.)

Test statistic =

What is the p-value? (Report answer accurate to 4 decimal places.)

p-value =

What is the conclusion about the claim?

There is sufficient evidence to support the claim.

There is not sufficient evidence to support the claim.

There is sufficient evidence to reject the claim.

There is not sufficient evidence to reject the claim

1:06 1 . < The following table contains cholesterol levels collected from non-smoking males. The before cholesterol level is before they exercised, and the after cholesterol level was taken after the same subject regularly exercised for two months. It is claimed that the mean cholesterol level before exercise is greater than that after exercise. Assume that the population of cholesterol level difference (before-after) is normally distributed. Test the claim at the 0.01 level. Cholesterol Cholesterol Before After 176 181 163 165 169 160 169 156 166 148 182 168 Which statistical test should be used to test the claim? 2 sample T-test 2 Proportion Z-test Z-test Paired T-test What is the test statistir Rennrt answer

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =3.365
since our test is right-tailed
reject Ho, if to > 3.365
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 7.833
We have d = 7.833
pooled variance = calculate value of Sd= √S^2 = sqrt [ 799-(47^2/6 ] / 5 = 9.283
to = d/ (S/√n) = 2.067
critical Value
the value of |t α| with n-1 = 5 d.f is 3.365
we got |t o| = 2.067 & |t α| =3.365
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 2.067 ) = 0.0468
hence value of p0.01 < 0.0468,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud > 0
test statistic: 2.067
critical value: reject Ho, if to > 3.365
decision: Do not Reject Ho
p-value: 0.0468
we donot have enough evidence to support the claim that the mean cholesterol level before exercise is greater than that after exercise.