Let p = population proportion of adult Americans who would favor the drafting of women.
The hypotheses to be tested are
H0: p >= 0.5
H1: p < 0.5 .
(Note that H0 is always the hypothesis that includes the possibility of equality.)
The sample proportion is phat = 0.49, and the sample size is n = 1000.
The sample proportion has mean p = 0.5 and standard deviation
sqrt(p(1-p)/n) = sqrt(0.5(1-0.5)/1000).
So the z-score (or test statistic) associated with phat = 0.49 is
z = (0.49 - 0.5) / sqrt(0.5(1-0.5)/1000) = -0.63.
Since Ha uses the inequality symbol for "less than", the P-value is the probability that the z-score is less than -0.63.
P-value = P(Z < -0.63)
= P(Z > 0.63) by symmetry of the normal distribution
= 1 - P(Z <= 0.63)
= 1 - 0.7357
= 0.2643 .
Since the P-value, 0.2643, exceeds the significance level, 0.05, there is *not* enough evidence to reject H0. Therefore, there is *not* convincing evidence that fewer than half of adult Americans would favor the drafting of women.
Lord bless you today!