Question

A national study report indicated that​ 20.9% of Americans were identified as having medical bill financial issues. What if a news organization randomly sampled 400 Americans from 10 cities and found that 95 reported having such difficulty. A test was done to investigate whether the problem is more severe among these cities. What is the​ p-value for this​ test? (Round to four decimal places.)

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.209

H_a : p > 0.209

n = 400

x = 95

= x / n = 95 / 400 = 0.2375

P₀ = 0.209

1 - P₀ = 1 - 0.209 = 0.791

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.2375 - 0.209 / [(0.209 * 0.791) / 400]

= 1.40

Test statistic = 1.40

P(z > 1.40) = 1 - P(z < 1.40) = 1 - 0.9192 = 0.0808

P-value = 0.0808