Respuesta :
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, [tex]H_0[/tex] : p [tex]\leq[/tex] 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, [tex]H_A[/tex] : 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. = [tex]\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of 400 Americans from 10 cities who were found having such difficulty = [tex]\frac{90}{400}[/tex] = 0.225 or 22.5%
n = sample of Americans = 400
So, test statistics = [tex]\frac{0.225-0.209}{{\sqrt{\frac{0.225(1-0.225)}{400} } } } }[/tex]
= 0.77
Now, P-value of the test statistics is given by the following formula;
P-value = P(Z > 0.77) = 1 - P(Z [tex]\leq[/tex] 0.77)
= 1 - 0.77935 = 0.22065
Testing the hypothesis, using the information given, it is found that the p-value is of 0.2148.
At the null hypothesis, it is tested if the proportion for these cities is of 20.9% = 0.209, hence:
[tex]H_0: p = 0.209[/tex]
At the alternative hypothesis, it is tested if the proportion for these cities is greater than 0.209, hence:
[tex]H_1: p > 0.209[/tex].
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
For this problem, the parameters are: [tex]p = 0.209, n = 400, \overline{p} = \frac{90}{400} = 0.225[/tex].
Then, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.225 - 0.209}{\sqrt{\frac{0.209(0.791)}{400}}}[/tex]
[tex]z = 0.79[/tex]
The p-value for this test is the probability of finding a sample proportion above 0.225, which is 1 subtracted by the p-value of z = 0.79.
Looking at the z-table, z = 0.79 has a p-value of 0.7852.
1 - 0.7852 = 0.2148.
The p-value for this test is of 0.2148.
A similar problem is given at https://brainly.com/question/24166849
