Question

"What do you think is the ideal number of children for a family to have?" A poll asked this question of 1011 randomly chosen adults. Almost half (49%) thought two children was ideal. Suppose that p = 0.49 is exactly true for the population of all adults. The polling company announced a margin of error of ±3 percentage points for this poll. What is the probability that the sample proportion p̂ for an SRS of size n = 1011 falls between 0.46 and 0.52? You see that it is likely, but not certain, that polls like this give results that are correct within their margin of error. (Round your answer to four decimal places.)

Answer 1

Answer)

P = 0.49

N = 1011

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 495.39

N*(1-p) = 515.61

Both the conditions are met so we can use standard normal z table to estimate the probability

z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

P(0.46<p<0.52) = P(p<0.52) - P(p<0.46)

P(p<0.52)

Here observed P = 0.52, Claimed P = 0.49, N = 1011.

After substitution,

Z = 1.91

From z table, P(z<1.91) = 0.9719

P(p<0.46) = P(z<-1.91) = 0.0281

Required probability is 0.9719 - 0.0281 = 0.9438