During 2011 a company implemented a number of policies aimed at reducing the age

Question

During 2011 a company implemented a number of policies aimed at reducing the ages of its customers’ accounts. In order to assess the effectiveness of these measures, the company randomly selects 10 customer accounts. The average age of each account is determined for the years 2010 and 2011. These data are given in Table 10.4. Assuming that the population of paired differences between the average ages in 2011 and 2010 is normally distributed:

(a) Set up the null and alternative hypotheses needed to establish that the mean average account age has been reduced by the company’s new policies.

H0: µd ?  versus Ha: µd <

(b) Figure 10.10 gives the Excel output needed to test the hypotheses of part a. Use critical values to test these hypotheses by setting equal to .10, .05, .01, and .001. How much evidence is there that the mean average account age has been reduced? (Round your answer to 3 decimal places. Negative amount should be indicated by a minus sign.)

(c) Figure 10.10 gives the p-value for testing the hypotheses of part a. Use the p-value to test these hypotheses by setting equal to .10, .05, .01, and .001. How much evidence is there that the mean average account age has been reduced? (Round your answer to 4 decimal places.)

(d) Calculate a 95 percent confidence interval for the mean difference in the average account ages between 2010 and 2011. Estimate the minimum reduction in the mean average account ages from 2010 to 2011. (Do not round intermediate calculations. Round your final answers to 4 decimal places. Negative amounts should be indicated by a minus sign.)

TABLE 10. 4 Average Account Ages in 2010 and 2011 for 10 Randomly Selected Accounts (for Exercise 10.20) OS AcctAge Average Age of Average Age of Account in 2011 Account in 2010 Account (Days) (Days) 35 27 24 40 33 41 33 25 35 31 29 51 15 18 10 21

Explanation / Answer

Given that,
population mean(u)=40
sample mean, x =30
standard deviation, s =33
number (n)=25
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.833
since our test is two-tailed
reject Ho, if to < -1.833 OR 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) = -7
We have d = -7
Pooled variance = Calculate value of Sd= S^2 = Sqrt [ 828-(-70^2/10 ] / 9 = 6.13
to = d/ (S/n) = -3.61

AT 0.10
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |t o| = 3.61 & |t | =1.833
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.6121 ) = 0.0056
hence value of p0.1 > 0.0056,here we reject Ho
ANSWERS
---------------
null, Ho: d = 0
alternate, H1: d < 0
test statistic: -3.61
critical value: reject Ho, if to < -1.833 OR if to > 1.833
decision: Reject Ho
p-value: 0.0056

AT 0.05
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |t o| = 3.61 & |t | =1.833
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :left tail - Ha : ( p < -3.6121 ) = 0.00282
hence value of p0.05 > 0.00282,here we reject Ho
ANSWERS
---------------
null, Ho: d = 0
alternate, H1: d < 0
test statistic: -3.61
critical value: reject Ho, if to < -1.833
decision: Reject Ho
p-value: 0.00282

AT 0.01
Critical Value
The Value of |t | with n-1 = 9 d.f is 2.821
We got |t o| = 3.61 & |t | =2.821
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
p-value :left tail - Ha : ( p < -3.6121 ) = 0.00282
hence value of p0.01 > 0.00282,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: -3.61
critical value: reject Ho, if to < -2.821
decision: Reject Ho
p-value: 0.00282

AT 0.001
Critical Value
The Value of |t | with n-1 = 9 d.f is 4.297
We got |t o| = 3.61 & |t | =4.297
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
p-value :left tail - Ha : ( p < -3.6121 ) = 0.00282
hence value of p0.001 < 0.00282,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: -3.61
critical value: reject Ho, if to < -4.297
decision: Do not Reject Ho
p-value: 0.00282

d.

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 ) =-70/10=-7
Pooled Sd( Sd )= Sqrt [ 828- (-70^2/10 ] / 9 = 6.128
Confidence Interval = [ -7 ± t a/2 ( 3.538/ Sqrt ( 10) ) ]
= [ -7 - 2.262 * (1.938) , -7 + 2.262 * (1.938) ]
= [ -11.384 , -2.616 ]

X Y X-Y (X-Y)^2 27 35 -8 64 19 24 -5 25 40 47 -7 49 30 28 2 4 33 41 -8 64 25 33 -8 64 31 35 -4 16 29 51 -22 484 15 18 -3 9 21 28 -7 49 -70 828

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