A. The mean of the sampling distribution of P^ is 0.7.
B. The SE = 0.08366
C. The probability that 80% or more of this sample will prefer Candidate A, using the normal approximation, is approximately 0.116.
D. The probability that 45% or more of this sample will prefer some other candidate is approximately 0.9635.
How do we find the mean, standard error and
p = 0.7 (proportion preferring Candidate A)
n = 30 (sample size)
B. SE( p^) = √((p(1-p))/n)
SE = 0.7(0.3)/30
SE = √(0.21/30)
SE = 0.0837
z = (p^ - p)/SE( p^)
where p^ is the sample proportion we're interested in (in this case, 0.8 for 80%), p is the population proportion (0.7), and SE(p^) is the standard error of the sample proportion
(0.8 -0.7)/0.0837 = 1.195
SE = 0.0837 to find the z-score for p^ = 0.8 = 1.195
This probability is approximately 0.116.
D. 45% is equivalent to the probability of less than 55% (or 0.55) preferring Candidate A. We standardize 0.55 and find the probability for that z-score.
z = (0.55−0.70)/0.0837 = -1.792
The probability of a z-score being less than -1.79 in a standard normal distribution is approximately 0.0365.
1 - 0.0365 = 0.9635