A national study report released by the Center for Studying Health System Change (HSC) in 2010, indicated that 20.9% of Americans were identified as having medical bill financial issues. Many people in families with problems paying medical bills in 2010 experienced severe financial consequences from their medical debt, with about two-thirds reporting problems paying for other necessities and a quarter considering bankruptcy, the study found. 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. The result of this test is ________.

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nuhulawal20 nuhulawal20 · Answer 1 · 2021-04-21T07:48:55+02:00

Answer:

The result of this test is 0.2156

Step-by-step explanation:

Given the data in the question;

The proportion of Americans having medical bill financial issues is ;

p = 20.9% = 0.209

CLAIM;

The problem is more severe among the cities

so

H₀ : p = 0.209

H₁ : p > 0.209 { this is our claim }

population n = 400

number of people that reported difficulty x = 90

so

sample proportion p" = x / n = 90 / 40 = 0.225

Test Statistics;

z = [ (p"-p) / √( (p(1-p))/n ) ]

we substitute

z = [ (0.225 - 0.209) / √( (0.209(1-0.209)) / 400 ) ]

z = [ 0.016 / √0.0004132976 ]

z = [ 0.016 / 0.0203297 ]

z = 0.787

P-VALUE

p-value = P( Z ≥ z )

p-value = P( Z ≥ 0.787 )

p-value = P( Z ≤ -0.787 )

from normal distribution table;

p-value = P( Z ≤ -0.787 ) = 0.2156

p-value = 0.2156

Therefore, The result of this test is 0.2156