During 2011 a company implemented a number of policies aimed at reducing the age
ID: 3364910 • Letter: D
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:
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.
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.)
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.)
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.)
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:
Average Account Ages in 2010 and 2011 for 10 Randomly Selected Accounts (for Exercise 10.20) AcctAge TABLE 10.4 Average Age of Account in 2010 (Days) 35 240 47 Average Age of Account in 2011 Account (Days) 27 19 40 30 2 4 5 6 28 41 25 31 29 15 21 35 51 18 28 8 10Explanation / Answer
1)
ho: there is no significant difference in the mean account age between 2011 and 2010. ud >= 0
h1: mean average account age has been reduced by the company’s new policies. ud < 0
2)
alpha
t(a,n-1)
1)
0.1
-1.38303
2)
0.05
-1.83311
3)
0.01
-2.82144
4)
0.001
-4.29681
T=-3.61211
REJECT Ho If -t < -t(a,df)
Reject H0 at = 0.1, 0.05, and 0.01
Very strong evidence
3)
P-value = 0.00282
Reject H0 at = 0.1, 0.05, and 0.01
Very strong evidence
4)
mean_diff
-7
sd_diff
6.12825877
n
10
t(A/2,DF)
2.262157163
lower
=-7+2.26*(6.128/SQRT(10))
-2.6205
upper
=-7-2.26*(6.128/SQRT(10))
-11.3795
ci = mean_diff +- t(a,df)*sd_diff/sqrt(n)
minimum reduction =2
2011
2010
diff
27
35
-8
19
24
-5
40
47
-7
30
28
2
33
41
-8
25
33
-8
31
35
-4
29
51
-22
15
18
-3
21
28
-7
alpha
t(a,n-1)
1)
0.1
-1.38303
2)
0.05
-1.83311
3)
0.01
-2.82144
4)
0.001
-4.29681
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