Question

A national study report indicated that​ 19.9% 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?

A) 0.096

B) 0.054

C) 0.216

D) 0.108

Answer 1

Solution:

Given:

A national study report indicated that​ 19.9% of Americans were identified as having medical bill financial issues.

that is: p = 0.199

Sample size = n = 400

x = Number of Americans  reported having such difficulty.= 90

We have to test whether the problem is MORE SEVERE among these cities.

thus hypothesis are:

H0: p = 0.199 Vs H1: p > 0.199 ( this is right tailed ( > ) test)

We have to find P-value:

For right tailed test , p-value is:

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

where

where

thus

thus we get:

P-value = P( Z > z )

P-value = P( Z > 1.30245)

P-value = 1 - P( Z < 1.30245)

Use following Excel command to find P-value:

=1-NORM.S.DIST(z,cumulative)

=1-NORM.S.DIST(1.30245,TRUE)

=0.0963813

=0.096

P-value = 0.096

Thus correct answer is:

A) 0.096