Question

A hospital director is told that 44 % of the treated patients are uninsured. The director wants to test the claim that the percentage of uninsured patients is more than the expected percentage. A sample of 290 patients found that 145 were uninsured. At the 0.01 level, is there enough evidence to support the director's claim?

Step 1 of 7: State the null and alternative hypotheses.

Step 2 of 7: Find the value of the test statistic. Round your answer to two decimal places.

Step 3 of 7: Specify if the test is one-tailed or two-tailed.

Step 4 of 7: Determine the P-value of the test statistic. Round your answer to four decimal places.

Step 5 of 7: Identify the value of the level of significance.

Step 6 of 7: Make the decision to reject or fail to reject the null hypothesis.

Step 7 of 7: State the conclusion of the hypothesis test.

Answer 1

Ра; Solution:- Given that, n = 290 x = 145 d = 0.01 Ñ = x X = 145 = 0.5 290 Po. = 44 % =0.44 le The null and alternative hypothesis, Ho: P = 0.44 > 0.44 На • Р 2. This is right- tailed test, Test statistics Z P-po POLI-Pol n Z 0.5 -0.44 1.44 (-0.44 290 0. z = 2. 058 3. The test statistic Z=2.06

4. P-Value = 0.0197 5. L = 0.01 0.0197 P-Value > o.ol > x 6. The P-value greater than a fail to reject Ho. 7. Conclusion:- fail to reject the null by pothesis, There not sufficent evidence to claim that the percentage of uninsured patients is 44% the expected percentage. at d = 0.01 significance level. is more than