Ien government employees working for the ministry of revenue verifying personal

Question

Ien government employees working for the ministry of revenue verifying personal income tax returns were randomly chosen to attend a one-day conference on how to increase returns verified per day for each of ten employees before and atter attending tne productivity. The following table contains the average number of income tax conference. (Assume that the table entries are normally distributed.) Before | 9 | 11 | 12 | 13 | 101 12 | 8 | .9 | 15 | 9 a) Test, at the 5% level of significance, the claim that an employee will be more p roductive after attending the conference productivity before and after attending the conference. b) Construct a 98% confidence interval for the true mean difference between an employee's

Explanation / Answer

PART A.

Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.833
since our test is left-tailed
reject Ho, if to < -1.833
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -0.2
We have d = -0.2
pooled variance = calculate value of Sd= S^2 = sqrt [ 10-(-2^2/10 ] / 9 = 1.03
to = d/ (S/n) = -0.61
critical Value
the value of |t | with n-1 = 9 d.f is 1.833
we got |t o| = 0.61 & |t | =1.833
make Decision
hence Value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.6124 ) = 0.27772
hence value of p0.05 < 0.27772,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: -0.61
critical value: reject Ho, if to < -1.833
decision: Do not Reject Ho
p-value: 0.27772

no evidence that employee will be more productive after the conference

PART B.

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 ) =-2/10=-0.2
Pooled Sd( Sd )= Sqrt [ 10- (-2^2/10 ] / 9 = 1.033
Confidence Interval = [ -0.2 ± t a/2 ( 0.596/ Sqrt ( 10) ) ]
= [ -0.2 - 2.821 * (0.327) , -0.2 + 2.821 * (0.327) ]
= [ -1.121 , 0.721 ]

X Y X-Y (X-Y)^2 9 10 -1 1 11 11 0 0 12 11 1 1 13 14 -1 1 10 10 0 0 12 11 1 1 8 9 -1 1 9 11 -2 4 15 14 1 1 9 9 0 0 -2 10

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