Question

An official of a large national union claims that the fraction of women in the union is not significantly different from 58%. To test this claim, a random sample of 400 individuals in the union is taken. Of those 400, 215 are women.

Use the p-value to conduct the hypothesis test and use α = 0.05 level of significance.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.58

H_a : p $\neq$ 0.58

n = 400

x = 215

$\hat p$ = x / n = 215 / 400 = 0.54

P₀ = 0.58

1 - P₀ = 1 - 0.58 = 0.42

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

=0.54 - 0.58 / [$\sqrt$(0.58*0.42) /400 ]

= -1.72

Test statistic = z = -1.72

P(z < -1.72 ) = 0.0427

P-value = 2 * 0.0427 = 0.0854

$\alpha$ = 0.05

P-value > $\alpha$

0.0854 > 0.05

Fail to reject the null hypothesis .

There is not sufficient evidence to suggest that