Question

A hospital director is told that 68% of the treated patients are insured. The director wants to test the claim that the percentage of insured patients is above the expected percentage. A sample of 180 patients found that 135 were insured. At the 0.05 level, is there enough evidence to support the director's claim?

State the null and alternative hypotheses.

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

Specify if the test is one-tailed or two-tailed.

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

identify the value of the level of significance.

Make the decision to reject or fail to reject the null hypothesis.

State the conclusion of the hypothesis test.

Answer 1

Here we have: n= 180 x = 135 Sample proportion: Ô X === 135 = 0.75 180 n a = 0.05 Hypothesis: H; P = 0.68 H: P = 0.68 Test statistics: Z-test P-p 0.75-0.68 - ze 2.01 p(1-p) 0.68(1-0.68) 180 n

Two tailed test P-value = 2p (Z >|2.011) = 0.0444 Level of significance is 0.05 Reject the null hypothesis. Conclusion: Here we have sufficient evidence to reject the null hypothesis. P-value is less than alpha value. We cannot support the claim that the 68% of the treated patients are insured.