I\'m looking for the answer to e and f The water diet requires one to drink two

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I'm looking for the answer to e and f

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 Person 2 3 4 Weight before diet Weight after six weeks Difference 178120242 155 170124221 160 8 4 21 5 The goal is to test whether the water 'diet' leads to a mean weight loss.(Hint: the differences have the mean of 5 and the standard deviation 12.193) a) What is the null hypothesis? Let subscript 1 indicate the measurement before the diet, and subscript 2 the measurement after six weeks. Incorrect H0: x 1-x 2 = 5 Incorrect H0: 1-2=5 Incorrect H0: X 1-X 2-0 Correct: H0: 1-2=0 You are correct. revious Tries Your receipt no. is 154-9846 b) The alternative hypothesis for this test will be: Incorrect Ha: 1-x 2 > 5 Incorrect Ha : 1-2> 5 Incorrect Ha: 1-x 2 > 0 Correct. Ha: 1-2>0 Incorrect Ha: 1-2

Explanation / Answer

Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.353
since our test is right-tailed
reject Ho, if to > 2.353
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 5
We have d = 5
pooled variance = calculate value of Sd= S^2 = sqrt [ 546-(20^2/4 ] / 3 = 12.19
to = d/ (S/n) = 0.82
critical Value
the value of |t | with n-1 = 3 d.f is 2.353
we got |t o| = 0.82 & |t | =2.353
make Decision
hence Value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.8201 ) = 0.23611
hence value of p0.05 < 0.23611,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud > 0
test statistic: 0.82
critical value: reject Ho, if to > 2.353
decision: Do not Reject Ho
p-value: 0.23611 , P>0.1

Confidence Interval
CI = d ± t a/2 * (Sd/ Sqrt(n))
Where,
d = di/n
Sd = Sqrt( di^2 – ( di )^2 / n ] / ( n-1 ) )
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
d = ( di/n ) =98/9=10.889
Pooled Sd( Sd )= Sqrt [ 1802- (98^2/9 ] / 8 = 9.584
Confidence Interval = [ 10.889 ± t a/2 ( 5.534/ Sqrt ( 9) ) ]
= [ 10.889 - 3.355 * (3.195) , 10.889 + 3.355 * (3.195) ]
= [ 0.17 , 21.607 ]

X Y X-Y (X-Y)^2 178 170 8 64 120 124 -4 16 242 221 21 441 155 160 -5 25 20 546

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