Question

A hospital director is told that 55% of the treated patients are insured. The director wants to test the daim that the percentage of insured patients is less than the expected percentage. A sample of 380 patients found that 190 were insured. Find the value of the test statistic. Round your answer to two decimal places Answer How to enter your answer e Tables Keypad

Answer 1

Let p: proportion of insured patients

The proportion/percentage of insured patients is pess than the expected proportion/percentage

The following information is provided: The sample size is N = 380, the number of favorable cases is X = 190, and the sample proportion is p = X = 0.5, and the significance level is a = 0.05 190 380 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p > 0.55 Ha: p < 0.55 This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left-tailed test is ze = -1.64. The rejection region for this left-tailed test is R= {z:z < -1.64}

(3) Test Statistics The Z-statistic is computed as follows: z = p – po V po(1 – po)/n 0.5 – 0.55 70.55(1 – 0.55)/380 (4) Decision about the null hypothesis Since it is observed that z = -1.959 < 2 = -1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is 0.025, and since p= 0.025 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is less than po, at the a = 0.05 significance level.