Question

A large collection of one-digit random numbers should have about 50% odd and 50% even digits because five of the ten digits are odd (1, 3, 5, 7, and 9) and five are even (0, 2, 4, 6, and 8) a. Find the proportion of odd-numbered digits in the following lines from a random number table. Count carefully. 9 0 49 0 5 0 7 8 1 7 8 3 4 5 0 7 1 5 7 5 1 2 3 4 5 9 6 2 1 b. Does the proportion found in part (a) represent p (the sample proportion) or p (the population proportion)? Find the error in this estimate, the difference between p and p (or p-p) C. a. The given random number table consists of% odd-numbered digits (Round to two decimal places as needed.)

Answer 1

a) Here table consist 30 random numbers. From which 18 numbers are odd. Thus the proportion of odd-numbered digits are = 18/30

= 0.6

Thus the given random number table consist 60% odd - numberd digits.

b) We have population proportion is 0.5

Thus the error of eatimate: p^ - P = 0.6-0.5 =0.1

Thus error of estimate is 0.1