:8. Solution. Under the null hypothesis, the expected value of the percentage of defectives in the sample is 20%. The observed percentage of defectives in the sample is 23%. To figure out the SE of the sample percentage, we consider a 0-1 box with 10,000 tickets, 20% of them marked with 1. The number of defectives in the sample is like the sum of 100 draws from this box, and the SE of this sum is
10 x (.20x.80)^(1/2) = 10 x .16^(1/2)
= 10 x .4
This is the SE of the number of defectives in the sample. The SE of the percentage of defectives is
4/100 x 100% = 4% So
(observed - expected)/SE = (23-20)/4 = 3/4 = .75 .
The observed level of significance is the area under the normal curve to the right of .75. From the table, we compute this area to be about 22.5% percent. Of the choices, 25% is the closest.