Answer:
C
Step-by-step explanation:
Given:
Number of samples, n = 100
let number of sample that favoured candidate A be x = 80
probability, p' = x/n = 80/100 = 0.8
probability, p for significance = 75% = 0.75
Using Z-proportion test statistics, we have
Z = ![(p' - p)/\sqrt{p(1 - p)/n}](https://tex.z-dn.net/?f=%20%28p%27%20-%20p%29%2F%5Csqrt%7Bp%281%20-%20p%29%2Fn%7D%20)
Substituting the values, we have
Z = ![(0.8 - 0.75)/\sqrt{0.75(1 - 0.75)/100}](https://tex.z-dn.net/?f=%20%280.8%20-%200.75%29%2F%5Csqrt%7B0.75%281%20-%200.75%29%2F100%7D%20)
Z = 0.05 / 0.0433 = 1.1547
The p-value is
P(Z>1.1547) = 1 - P(Z≤1.1547)
Checking the Z-score table, P(Z≤1.1547) = 0.8789
Hence, we have 1 - 0.8789
= 0.1241
Hence, the p-value is greater than 0.05, so we cannot say it is very significant.
Hence, the population in favour of candidate A is not significantly greater than 75%