and Student ID # 100 Given the result of experiments of two methods of weight lo

Question

and Student ID # 100 Given the result of experiments of two methods of weight loss procedures, in this case the data is the weight loss of participants so larger is better the summarized data is given as: 3. Procedure II Procedure I 45 48 10 48 52 N (sample size) Mean Standard deviation Standard error Find the 95% confidence interval of the population mean difference 2-. Can we conclude with =0.05 that the data show significant evidence that one procedure is better than the other. We assume no prior knowledge about which procedure is better. a. b.

Explanation / Answer

3.
a.
TRADITIONAL METHOD
given that,
mean(x)=48
standard deviation , 1 =10
population size(n1)=45
y(mean)=52
standard deviation, 2 =8
population size(n2)=48
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((100/45)+(64/48))
= 1.886
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
value of z table is 1.645
margin of error = 1.645 * 1.886
= 3.102
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (48-52) ± 3.102 ]
= [-7.102 , -0.898]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=48
standard deviation , 1 =10
number(n1)=45
y(mean)=52
standard deviation, 2 =8
number(n2)=48
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 48-52) ±Z a/2 * Sqrt( 100/45+64/48)]
= [ (-4) ± Z a/2 * Sqrt( 3.556) ]
= [ (-4) ± 1.645 * Sqrt( 3.556) ]
= [-7.102 , -0.898]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-7.102 , -0.898] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero

b.
Given that,
mean(x)=48
standard deviation , 1 =10
number(n1)=45
y(mean)=52
standard deviation, 2 =8
number(n2)=48
null, Ho: u1 = u2
alternate, H1: 1 < u2
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=48-52/sqrt((100/45)+(64/48))
zo =-2.121
| zo | =2.121
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =2.121 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -2.121 ) = 0.01695
hence value of p0.05 > 0.01695,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 < u2
test statistic: -2.121
critical value: -1.645
decision: reject Ho
p-value: 0.01695
we have enough evidence to support the claim that procedure 2 is better than the procedure 1

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