ExhibitI The following information was obtained from matched samples Individual

Question

ExhibitI The following information was obtained from matched samples Individual Method Method 2 1. Refer to Exhibit I. The point estimate for the difference between the means of the two populations (Method 1- Method 2) is b. 0 C -4 2. efer to Exhibit . The 9S% confidence interval for the difference between the two population i en b. c. -3.776 to 1.776 -2.776 to 2.776 -1.776 to 2.776 d. 0 to 3.776 3. Refer to Exhibit . Sup pose we test the hypotheses Hau-o versus H..0. The test statistic for the for this test is a. b. 0 4. Refer to Exhibit I. If the null hypothesis is tested at the 5% level of significance, the null hypothesis a. should be rejected should not be rejected should be revised None of these alternatives is correct. c. d. 5. In the ANOVA, treatment refers to experimental units different levels of a factor the dependent variable applying antibiotic to a wound c. d.

Explanation / Answer

Q2.

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 ) =-5/5=-1
Pooled Sd( Sd )= Sqrt [ 25- (-5^2/5 ] / 4 = 2.236
Confidence Interval = [ -1 ± t a/2 ( 1.291/ Sqrt ( 5) ) ]
= [ -1 - 2.776 * (1) , -1 + 2.776 * (1) ]
= [ -3.776 , 1.776 ]

Q3.

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -1
We have d = -1
pooled variance = calculate value of Sd= S^2 = sqrt [ 25-(-5^2/5 ] / 4 = 2.24
to = d/ (S/n) = -1
critical Value
the value of |t | with n-1 = 4 d.f is 2.776
we got |t o| = 1 & |t | =2.776
make Decision
hence Value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1 ) = 0.3739
hence value of p0.05 < 0.3739,here we do not reject Ho
ANSWERS
---------------
test statistic: -1

X Y X-Y (X-Y)^2 7 5 2 4 5 9 -4 16 6 8 -2 4 7 7 0 0 5 6 -1 1 -5 25

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