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?

The p-value for this test is 0.22065.

Step-by-step explanation:

We are given that a national study report indicated that 20.9% of Americans were identified as having medical bill financial issues.

A news organization randomly sampled 400 Americans from 10 cities and found that 90 reported having such difficulty.

Let p = proportion of Americans who were identified as having medical bill financial issues in 10 cities.

SO, Null Hypothesis, : p 20.9%   {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is less than or equal to 20.9%}

Alternate Hypothesis, : p > 20.9%   {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is more than 20.9% and is more severe}

The test statistics that will be used here is One-sample z proportion statistics;

                                  T.S.  =  ~ N(0,1)

where, = sample proportion of 400 Americans from 10 cities who were found having such difficulty =   = 0.225 or 22.5%

            n = sample of Americans = 400

So, test statistics  =  

                               =  0.77

Now, P-value of the test statistics is given by the following formula;

         P-value = P(Z > 0.77) = 1 - P(Z 0.77)

                                            = 1 - 0.77935 = 0.22065