Question

A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A.

Answer 1

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 = $(p' - p)/\sqrt{p(1 - p)/n}$

Substituting the values, we have

Z = $(0.8 - 0.75)/\sqrt{0.75(1 - 0.75)/100}$

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%