Question

A hospital director is told that 42% of the treated patients are uninsured. The director wants to test the claim that the percentage of uninsured patients is above the expected percentage. A sample of 190 patients found that 95 were uninsured. At the 0.01 level, is there enough evidence to support the director's claim?

a. State the null and alternative hypotheses.

b. Find the value of the test statistic. Round your answer to two decimal places.

c. Is this test one-tailed or two-tailed?

d. Find the P-value of the test statistic. Round your answer to four decimal places.

e. Identify the level of significance for the hypothesis test.

f. Based on your results, do we reject or fail to reject the null hypothesis?

Please show steps. :)

Answer 1

Answer:

Given,

x = 95

sample n = 190

sample proportion p^ = x/n

= 95/190

= 0.5

Null hypothesis Ho : p = 0.42

Alternative hypothesis Ha : p > 0.42

consider,

test statistic z = (p^ - p)/sqrt(p(1-p)/n)

substitute values

= (0.5 - 0.42)/sqrt(0.42(1-0.42)/190)

z = 2.234

Corresponding p value = P(z > 2.234) [right tailed]

= 0.0127415 [since from z table]

= 0.0127

P value = 0.0127

Here we observe that p value > 0.01, so we fail to reject null hypothesis.