Question

A hospital director is told that 43% of the emergency room visitors are uninsured. The director wants to test the claim that the percentage of uninsured patients is above the expected percentage. A sample of 310 patients found that 155 were uninsured. At the 0.02 level, is there enough evidence to support the director's claim?

Step 2 of 6 :  

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

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ : p = 0.43

H_a : p > 0.43

n = 310

x = 155

= x / n = 155 / 310 = 0.5

P₀ = 0.43

1 - P₀ = 1 - 0.43 = 0.57

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.5 - 0.43 / [(0.43 * 0.57) / 310]

= 2.49

This is the right tailed test .

P(z > 2.49) = 1 - P(z < 2.49) = 1 - 0.9936 = 0.0064

P-value = 0.0064

= 0.02

P-value <

Reject the null hypothesis .

There is enough evidence to support the director's claim .