Answer:
Null hypothesis: H0 = 0.50
Alternative hypothesis: Ha < 0.50
z = -3.16
P value = P(Z<-3.16) = 0.0008
Decision we reject the null hypothesis and accept the alternative hypothesis. That is, there is convincing evidence that fewer than half of adult Americans would favor the drafting of women.
Rule
If;
P-value > significance level --- accept Null hypothesis
P-value < significance level --- reject Null hypothesis
Z score > Z(at 95% confidence interval) ---- reject Null hypothesis
Z score < Z(at 95% confidence interval) ------ accept Null hypothesis
Explanation:
Given;
n=1000 represent the random sample taken
Null hypothesis: H0 = 0.50
Alternative hypothesis: Ha < 0.50
Test statistic z score can be calculated with the formula below;
z = (p^−po)/√{po(1−po)/n}
Where,
z= Test statistics
n = Sample size = 1000
po = Null hypothesized value = 0.50
p^ = Observed proportion = 0.45
Substituting the values we have
z = (0.45-0.50)/√{0.50(1-0.50)/1000}
z = -3.16
z = -3.16
To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.
P value = P(Z<-3.16) = 0.0008
Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = -3.16) which doesn't falls with the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.0008 which is lower than 0.05. Then we can conclude that we have enough evidence to FAIL or reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is invalid, therefore we accept the alternative hypothesis.