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 90 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:

Here, we have to use z test for population proportion.

H₀: p = 0.209 versus H_a: p > 0.209

This is an upper tailed test.

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 90

n = sample size = 400

p̂ = x/n = 90/400 = 0.225

p = 0.209

q = 1 - p = 1 – 0.209 = 0.791

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.225 – 0.209)/sqrt(0.209*0.791/400)

Z = (0.225 – 0.209)/ 0.0203

Z = 0.7870

P-value = 0.2156

(By using z-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the problem is more severe among these cities.