Need help with an analysis:


During 2003 my company implemented a number of policies aimed at
reducing the ages of our customers? accounts. In order to assess the
effectiveness of these measures, we randomly selected 10 customer
accounts. The average age of each account is determined for the years
2002 and 2003. This data is provided below.  Assuming that the
population of paired differences between the average ages in 2003 and
2002 is normally distributed:

		Average Age of Account (Days)
Account	2003		2002
1		27		35
2		19		24
3		40		47
4		30		28
5		33		41
6		25		33
7		31		35
8		29		51
9		15		18
10		21		28

a Please help us set up the null and alternative hypotheses to
establish that the mean average account age has been reduced by the
company?s new policies.


b Please help us determine rejection points to test these hypotheses
by setting ? equal to .10, .05, .01, and .001.

Excel Output of a paired difference analysis of the Account Age Data is below.

T=Test:  Paired Two Sample for Means
2003	2002
Mean			27			34
Variance		53.55556		104.2222
Observations		10			10
Pearson Correlation	0.804586		
Df			9
t-stat			-3.61211
P(T>=t) one-tall	0.00282
tCritical one-tail	1.833114
P(T<=t) Two-tail	0.005641
tCritical Two-tail	2.262159	

Additional Stats
-7.00 mean difference (2000 Average ? 1999 Average)
6.128 std. dev.
1.938 std. error
.0028 p-value (one-tailed, lower)	

c The excel output above gives the p-value for testing the hypotheses
of part ?. Please help us using the p-value to test these hypotheses
by setting ? equal to .10, .05, .01, and .001.

d And last, calculate a 95 percent confidence interval for the mean
difference in the average account ages between 2003 and 2002. We are
also trying this methode to estimate the minimum reduction in the mean
average account ages from 2003 to 2002.

Thank you in advance.

I am not sure about the hypotheses because confidence intervals will
give you not only that the hypothesis is true but also tell you how
loud you can say it

(a) The p-values at the confidence levels you requested are 
90% (? =0.1)	1.833113856
95% (? =0.05)	2.262158887
0.99(? =0.01)	3.249842848
0.999 (? = 0.001)	4.780886229

Take the difference of each column
 to yield the following table
2003    2002    Difference
---------------------------
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
------------------

Visual inspection alone tells you that 2003 data is less than 2002 but
nevertheless calculate the mean difference which is "-7". The variance
and standard deviation are 37.556 and 6.1829 respectively. The
standard error is 1.937. BTW I used MS-Excel for all these values.

Now you can calculate the configence intervals of the differences as 
----------------------------
90% ( ? =0.1)	(-10.55243823,	-3.447561766)
95% ( ? =0.05)	(-11.38389557,	-2.616104426)
0.99( ? =0.01)	(-13.29795359,	-0.702046412)
0.999 (? = 0.001)   (-16.26500172,	2.265001721)
-----------------

The above table tells you that except at the level of 99.9% (? =
0.001)  you can confidently say the mean average account age has been
reduced by the
company?s new policies. These confidence intervals tell you that if
you were to take such measurements 100 times 95 or 99 % of the time
the mean difference lies between those values.

The reason you cannot say this at 99.9% level is because the
confidence interval for the mean difference includes "0" in it.

I hope this is clear.Let me know if you have additional questions..

-Padmapani

Thank you so much, this was exactly what I was looking for.

Assuming that the paired differences should have an approximately
normal distribution is a "parametric" treatment.  A "nonparametric"
approach would avoid any assumption about the form of the distribution
and instead impose a "null hypothesis" of the form that a paired
difference is equally likely to be positive as negative (any zero
differences from the matched observations would be omitted for the
purpose of this analysis).

The nonparametric approach typically gives a somewhat less impressive
significance level, but one which is after all based on less stringent
assumptions.

Here we simply ask what the chance of getting 9 out of 10 Heads when
flipping a "fair and balanced" coin (as you got 9 out of 10 accounts
decreasing in aging of accounts).  For the sake of a "one-side" test,
excluding the equivalence of most accounts' ages increasing (because
the hypothesis concerns whether the policies cause a decrease), we
would count how many of the 2^10 = 1024 possible outcomes would
involve at most one increase.  The answer is:

  C(10,0) + C(10,1) = 11

So the results are significant in this "robust" version of the
(one-sided) test at the:

  11/1024 = 0.01 level

regards, mathtalk-ga

MathTalk,

Thats a cool way of looking at it.Thanks.I learnt something new.

-Paddy