Question

2.) I want to estimate the proportion of people who drink coffee in the morning. How many people should I poll to be 97% confident that I am within 1.5% of the actual proportion? 3.) Repeat #2, but assume that last year the polls showed that 40% drink coffee.

Answer 1

Now we have to calculate the sample size 'n' for 97% confidence interval with margin of error MOE = 1.5% = 0.015. Step (1): =1-97= 1 – 0.97 = 0.03 0.03 = 0.015 100 Step (2): 1 -0.015 = 0.985 Now we have to find the value of z such that P(Z < Za/2) = 0.985 From standard normal table Za/2 = 2.17 (Or) Use excel formula: =NORM.S.INV(0.985)

Step (3): MOE = Pa12 • 24- => 341-5) = (MOE) =n= (4.) * -p(1 – p) =n= (2017)** Þ(1 – ) an=20928.44* p*(1-0) n=20928.44* * (1-P)

Given that sample proportion p = 40% = 0.40. Now we have to calculate the sample size 'n'. n = 20928.44 * P * (1 - p) = 20928.44 * 0.4 * (1 – 0.4) = 20928.44 * 0.4 * (0.6) = 5022.83 5023 . The sample size n=5023

thank you.