Question

A national study report indicated that 21% of Americans were identified as having medical bill financial issues. What if a news organization randomly sampled 370 Americans from 5 cities and found that 82 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

P-value for this test is 0.2946.

Step-by-step explanation:

We are given that a national study report indicated that 21% of Americans were identified as having medical bill financial issues.

A news organization randomly sampled 370 Americans from 5 cities and found that 82 reported having such difficulty.

Let p = % of Americans that were identified as having medical bill financial issues among the 5 cities

So, Null Hypothesis, $H_0$ : p $\leq$ 21%  {means that the % of Americans that were identified as having medical bill financial issues among the 5 cities is equal to 21% as reported by national study}

Alternate Hypothesis, $H_a$ : p > 21%  {means that the % of Americans that were identified as having medical bill financial issues among the 5 cities is more than 21%}

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

                T.S. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where,  $\hat p$ = % of Americans that were identified as having medical bill financial issues in the sample of 370 Americans = $\frac{82}{370} \times 100$ = 22.16%

            n = sample of Americans = 370

SO, for finding the p-value of the test, we have to find the test statistics first.

  Test statistics =  $\frac{\frac{82}{370} -0.21}{\sqrt{\frac{\frac{82}{370} (1-\frac{82}{370} )}{370} } }$  ~ N(0,1)

                           = 0.54

P-value tells us the exact percentage where our test statistics lie.

Now, P-value is given by = P(Z > 0.54) = 1 - P(Z $\leq$ 0.54)

                                          = 1 - 0.7054 = 0.2946

Therefore, the p-value for this test is 0.2946.