:4. Solution.  <p>

SE of sum of 1971 sample scores = 14 x 40 = 560  <p>

SE of average of 1971 sample scores = (560/1600) x 100%  <p>

= 14/40 = 7/20 = 35/100  <p>

= .35  <p>

SE of sum of 1975 sample scores = same as 1971's = .35  <p>

(The box SD is bootstraped by the sample SD in both cases) <p>

SE of the difference = the square root of (.35)^2 + (.35)^2  <p>

= .49  <p>

z = ( 1975 ave - 1971 ave - 0 )/ SE of difference  <p>

= (68.5 - 67.2)/.49 = 1.2/.49 = 2.6  <p>

The area under the normal curve from -2.6 to 2.6 is 99.07%, so
the area to the right of 2.6 is .93/2 = .465%. <p>

.465/100 = 4.65/1000  <p>

= 5/1000 (approx)  <p>

= 1/200