Question

In: Statistics and Probability

Low-fat or low-carb? Are low-fat diets more effective for weight loss? A sample of 58 subjects...

Low-fat or low-carb? Are low-fat diets more effective for weight loss? A sample of

58

subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was

3.2

kilograms with a sample standard deviation of

5.42

kilograms. A second sample of

60

subjects went on a low-fat diet. Their sample mean weight loss was

2.4

kilograms with a standard deviation of

4.48

kilograms. Can you conclude that the mean weight loss differs between the two diets? Let

μ1

denote the mean weight lost on the low-carb diet and

μ2

denote the mean weight lost on the low-fat diet. Use the

=α0.05

level and the

P

-value method.

Part 1 of 6

Your Answer is correct

State the appropriate null and alternate hypotheses.

H0:=μ1μ2

H1:≠μ1μ2

This is a ▼two-tailed test.

Part 2 of 6

Your Answer is incorrect

Compute the test statistic. Round the answer to three decimal places.

=t

  

4.826

Correct Answer:

=t0.872

Part: 2 / 6

2 of 6 Parts Complete

Part 3 of 6

How many degrees of freedom are there, using the simple method?

The degrees of freedom using the simple method is .

Solutions

Expert Solution

Given that,
mean(x)=3.2
standard deviation , s.d1=5.42
number(n1)=58
y(mean)=2.4
standard deviation, s.d2 =4.48
number(n2)=60
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.002
since our test is two-tailed
reject Ho, if to < -2.002 OR if to > 2.002
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3.2-2.4/sqrt((29.3764/58)+(20.0704/60))
to =0.872
| to | =0.872
critical value
the value of |t α| with min (n1-1, n2-1) i.e 57 d.f is 2.002
we got |to| = 0.87235 & | t α | = 2.002
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.8724 ) = 0.387
hence value of p0.05 < 0.387,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.872
critical value: -2.002 , 2.002
decision: do not reject Ho
p-value: 0.387
we do not have enough evidence to support the claim that the mean weight loss differs between the two diets.


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