Figure this out for me.
1 answer:
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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.
<u><em>Let p = proportion of Americans who were identified as having medical bill financial issues in 10 cities.</em></u>
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 <u>One-sample z proportion</u> <u>statistics</u>;
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, <u><em>test statistics</em></u> =
= 0.77
<u></u>
<u>Now, P-value of the test statistics is given by the following formula;</u>
P-value = P(Z > 0.77) = 1 - P(Z 0.77)
= 1 - 0.77935 = 0.22065