Question

13.

Saving for college: In a survey of 909 U.S. adults with children conducted by the Financial Industry Regulatory Authority, 309 said that they had saved money for their children’s college education. Can you conclude that more than 30% of U.S. adults with children have saved money for college? Use the α = 0.05 level of significance.

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p = 0.30

H_a : p > 0.30

$\hat p$ = x / n = 309/909 = 0.3399

P₀ = 0.30

1 - P₀ = 1-0.30 = 0.70

Test statistic = z

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

=0.3399 - 0.30 / [$\sqrt$030*(0.70) /909 ]

= 2.627

P(z > 2.627) = 1 - P(z < 2.627 ) = 0.0043

P-value = 0.0043

$\alpha$ = 0.05    

p=0.0043<0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that the population proportion pp is greater than p₀​, at the α=0.05 significance level.