Question

Hi,

See the data set below.

Based on this, Excel comes with the following analysis (paired difference).

I need to calculate a 95% confidence interval. Based on this, I came to the following formula: [Mean of the difference (D-mean) +/ t(alpha/2) * Sd / sqrt(n)]. I can fill in the following information:

D-mean = 27-34 = -7.

t(alpha/2) = 2,262

sqrt(n) = sqrt(10)

Do you now how I can find the standard deviation of the sample? (Sd) I think it has something to do with both variances in the Excel output, but I don't know it exactly. Thanks for your help!

Best,

Maikel

TABLE 10.4 Average Account Ages in 2010 and 2011 for 10 Randomly Selected Accounts (for Exercise 10.20) AcctAge Average Age of Account in 2011 Average Age of Account in 2010 (Days) 35 240 47 28 41 Account (Days) 27 19 40 30 2 4 5 6 25 31 29 15 21 35 51 18 28 8 10 FIGURE 10.10 Excel Output of a Paired Difference Analysis of the Account Age Data (for Exercise 10.20) t-Test: Paired Two Sample for Means 2011 Age 27 53.55556 10 0.804586 0 2010 Age 34 104.2222 10 Mean Variance Observations Pearson Correlation Hypothesized Mean df t Stat P(T

Answer 1

Solution

Back-up Theory

Let X and Y be two random variables

E(aX + bY) = aE(X) + bE(Y) …………………………………………………………… (1a)

In particular,

E(X + Y) = E(X) + E(Y) ……………………………………..…………………..………(1b)

E(X - Y) = E(X) - E(Y) ……………………………………..…………………...………(1c)

Var(aX + bY) = a²Var(X) + b²Var(Y) + 2abCov(X,Y)………………………………..(3a)

In particular, Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y)……..……………..………(3b)

                       Var(X - Y) = Var(X) + Var(Y) - 2Cov(X,Y)……..……….……………(3c)

Cov(X, Y) = r_xy.s_x.s_y ………………………………………………………….……….. (4),

where r_xy = correlation between X and Y,s_x = standard deviation of X and s_y= standard deviation of Y

Now, to work out the solution,

Here, if X = Average age of accounts in 2011 and Y = Average age of accounts in 2010, then

D = X – Y.

So, vide (3c),

Var(D) = 53.55556 + 104.2222 - 2Cov(X,Y) and vide (4),

Cov(X,Y) = 0.804586 x √(53.55556 x 104.2222)

= 60.11113.

Hence,

Var(D) = 53.55556 + 104.2222 – (2 x 60.11113)

= 37.55551

Thus, standard deviation of D = √37.55551

= 6.1283 Answer