Question

Each student in a class of 40 was randomly assigned one line of a random number table. Each student then counted the​ odd-numbered digits in a 30​-digit line. The even digits are​ 0, 2,​ 4, 6 and 8.

a. On​ average, in the list of 30 digits, how many​ odd-numbered digits would each student​ find?

b. If each student found a 80​% confidence interval for the percentage of​ odd-numbered digits in the entire random number​ table, how many intervals​ (out of 40) would you expect not to capture the population percentage of​ 50%?

Answer 1

Answer:

a)

Here there are 5 odd numbers & 5 even

i.e., we have 50% of odd & 50% of even digits

so

p = 0.50

Given,

sample n = 30

p = 0.5

Number of odd numbered digits = np

= 30*0.50

= 15

b)

Total number of confidence interval = 40

Percentage = 80%

Expected number = Total number of confidence interval*Percentage

= 40*0.80

= 32

Now the expected number of intervals which doesn't capture = Total number of confidence intervals*(100-80)%

= 40*20%

= 40*0.20

= 8

So the expected number of intervals which doesn't capture the population percentage of​ 50% is 8.