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:

Given: A national study report indicated that​ 20.9% of Americans were identified as having medical bill financial issues.

That is : p = proportion Americans were identified as having medical bill financial issues = 0.209

n = Sample size = 400

x = Number of Americans were identified as having medical bill financial issues = 95

Thus sample proportion is:

A test was done to investigate whether the problem is more severe among these cities.

Thus hypothesis of the study are:

H0: p = 0.209 Vs H1: p > 0.209

We have to find p-value.

p-value = P( Z > z test statistic value)

( it is Z > z test statistic value , since this is right tailed test)

where z test statistic value is given by:

Thus p-value is:

p-value = P( Z > 1.40)

p-value = 1 - P( Z < 1.40)

Look in z table for z = 1.4 and 0.00 and find area.

P( Z < 1.40) = 0.9192

Thus

p-value = 1 - P( Z < 1.40)

p-value = 1 - 0.9192

p-value = 0.0808