Question

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

Answer 1

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)