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 38 randomly selected patients about the new fee, out of which 27 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

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p = 0.50

H_a : p > 0.50

= x / n = 27 / 38 = 0.711

P₀ = 0.50

1 - P₀ = 1 - 0.50 = 0.50

a) Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.711 - 0.50 / [(0.50 * 0.50) / 38 ]

Test statistic = z = 2.60

b) = 0.025

P(Z > z) = 0.025

= 1 - P(Z < z) = 0.025

= P(Z < z) = 1 - 0.025

= P(Z < 1.96) = 0.975

z critical value = 1.96

c) test statistic > critical value

Reject the null hypothesis .