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, the clerk used a method favoring Party A.
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
Step-by-step explanation:
Given;
n = 40 represent the number of samples 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 = 40
po = Null hypothesized value = 0.50
p^ = Observed proportion = 30/40 = 0.75
Substituting the values we have
z = (0.75-0.50)/√(0.50(1-0.50)/40)
z = 3.16227766
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.
