The water diet requires one to drink two cups of water every

The water diet requires one to drink two cups of water

every half hour from when one gets up until one goes to bed,

but otherwise allows one to eat whatever one likes. Four adult

volunteers agree to test the diet. They are weighed prior to

beginning the diet and after six weeks on the diet. The weights

(in pounds) are

The goal is to test whether the water \'diet\' leads to a mean weight loss.

i) What is the null hypothesis? Let subscript 1 indicate the measurement before the diet, and subscript 2 the measurement after six weeks.

(Pick letter corresponding to the best answer.)

a - H₀: x?₁-x?₂ = 7

b - H₀: ?₁-?₂ = 7

c - H₀: x?₁-x?₂ = 0

d - H₀: ?₁-?₂ = 0

ii) The alternative hypothesis for this test will be:

(Pick letter corresponding to the best answer.)

a - H_a: x?₁-x?₂ > 7

b - H_a: ?₁-?₂ > 7

c - H_a: x?₁-x?₂ > 0

d - H_a: ?₁-?₂ > 0  

e - Ha: ?1-?2 < 0

iii) Give a value for the test statistic, to 2 places past the decimal.

iv) How many degrees of freedom are there?

v) What are bounds on the P value?

Choose answer from the following and report the letter.

     a P < .001

     b .01 < P < .05

     c .05 < P < .1

     d P > .1

vi) Construct a 99% confidence interval for the mean weight loss. Give answer to TWO places past the decimal place.

Lower Limit

Upper Limit

Person	1	2	3	4	Mean	SD
Weight before diet	180	125	240	150	173.75	49.56
Weight after six weeks	170	130	215	152	166.75	36.09
Difference	10	-5	25	-2	7	13.64

Solution

(i) d

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(ii) d

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(iii) test statistic:

t=mean difference/(s/vn)

=7/(13.64/sqrt(4))

=1.03

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(iv.) degrees of freedom = n-1=4-1=3

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(v) It is a right-tailed test.

So the p-value= P(t with df=3 >1.03) =0.1894 (from student t table)

d P > .1

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(vi) Given a=1-0.99=0.01, t(0.005, df=3) =5.84 (from student t table)

Lower Limit:

mean differenece - t*s/vn

= 7- 5.84*13.64/sqrt(4)

=-32.83

Upper Limit:

mean differenece + t*s/vn

= 7+ 5.84*13.64/sqrt(4)

=46.83