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?

Answer 1

Answer:

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, $H_0$ : p $\leq$ 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, $H_A$ : 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.  = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of 400 Americans from 10 cities who were found having such difficulty =  $\frac{90}{400}$ = 0.225 or 22.5%

            n = sample of Americans = 400

So, test statistics  =  $\frac{0.225-0.209}{{\sqrt{\frac{0.225(1-0.225)}{400} } } } }$

                               =  0.77

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

         P-value = P(Z > 0.77) = 1 - P(Z $\leq$ 0.77)

                                            = 1 - 0.77935 = 0.22065