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?
A. 0.4312
B. 0.1078
C. 0.0539
D. 0.2156
Given : n=400 , X=90 , The estimate of the sample proportion is ,
The null and alternative hypothesis is ,
The test is one-tailed test.
The test statistic is ,
The p-value is ,
p-value=
; The Excel function is , =1-NORMDIST(0.787,0,1,TRUE)
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