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 over the expected percentage. A sample of 200 patients found that 100 were uninsured. At the 0.10 level, is there enough evidence to support the director's claim?

(Please answer the following questions below)

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. P-value: ________________________________________

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? ________________________________________________

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p = 0.42

H_a : p > 0.42

= x / n = 100/200 = 0.50

P₀ = 0.42

1 - P₀ = 1-0.42 =0.58

b) Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.50-0.42/ [0.42*(1*0.58) / 200]

= 2.29

P(z < 2.92 ) = 0.0109

P-value = 0.0109

c) This is a one tailed test

= 0.10

p=0.0109 < 0.10, it is concluded that the null hypothesis is rejected.

Reject the null hypothesis .

There is sufficient evidence to suggest that claim that the population proportion pp is greater than _p_o​, at the α=0.10 significance level.