Question

An office administrator for a physician is piloting a new "no-show" fee to attempt to deter some of the numerous patients each month that do not show up for their scheduled appointments. However, the administrator wants the majority of patients to feel that the fee is both reasonable and fair. She administers a survey to 36 randomly selected patients about the new fee, out of which 25 respond saying they believe the new fee is both reasonable and fair. Test the claim that more than 50% of the patients feel the fee is reasonable and fair at a 2.5% level of significance.

Standard Normal Distribution Table

a. Calculate the test statistic.

z=z=

Round to two decimal places if necessary

Enter 0 if normal approximation to the binomial cannot be used

b. Determine the critical value(s) for the hypothesis test.

• +

Round to two decimal places if necessary

Enter 0 if normal approximation to the binomial cannot be used

c. Conclude whether to reject the null hypothesis or not based on the test statistic.

Reject

Fail to Reject

Cannot Use Normal Approximation to Binomial

h(i,x)=

{10if  program i halts on input x,otherwise.{1if  program i halts on input x,0otherwise.

Answer 1

a) The sample proportion here is computed as:
p = x/n = 25/36 = 0.6944

The test statistic here is computed as:

Therefore 2.33 is the test statistic value here.

b) As this is a one tailed test, as we are testing whether the proportion is more than 0.5, therefore we have from the standard normal tables:
P( Z > 1.96) = 0.025

Therefore 1.96 is the required critical value here for 2.5% level of significance.

c) As the test statistic value is 2.33 > 1.96, therefore it lies in the rejection region. Therefore we Reject H0 here.