Question

A certain drug is used to treat asthma. In a clinical trial of the​ drug, 17 of 271 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 10% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01significance level to complete parts​ (a) through​ (e) below. (1-PropZTest prop<0.1, z=-2.045104355, p=0.0204222855, p^=0.0627306273, n=271)  a. Is the test​ two-tailed, left-tailed, or​ right-tailed? b. What is the test​ statistic? c. What is the​ P-value? d. What is the null​ hypothesis, and what do you conclude about​ it? Identify the null hypothesis. Decide whether to reject the null hypothesis. Choose the correct answer below. A.Fail to reject the null hypothesis because the​ P-value is greater than the significance​ level, alphaα. B. Reject the null hypothesis because the​ P-value is greater than the significance​ level, alphaα. C.Fail to reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alphaα. D. Reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alphaα. e. What is the final​ conclusion A.There is not sufficient evidence to support the claim that less than 10% of treated subjects experienced headaches. B.There is sufficient evidence to support the claim that less than 10% of treated subjects experienced headaches. C.There is sufficient evidence to warrant rejection of the claim that less than 10​% of treated subjects experienced headaches. D.There is not sufficient evidence to warrant rejection of the claim that less than 10​% of treated subjects experienced headaches.

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p = 0.10

H_a : p < 0.10

n = 271

x = 17

= x / n = 17 / 271 = 0.0627

P₀ = 0.10

1 - P₀ = 1 - 0.10 = 0.90

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.0627 - 0.10 / [(0.10 * 0.90) / 271]

= -2.045

Test statistic = -2.045

P(z < -2.045) = 0.0204

P-value = 0.0204

= 0.01

P-value >

Fail to reject the null hypothesis .

B. Reject the null hypothesis because the​ P-value is greater than the significance​ level, α .

D. There is not sufficient evidence to warrant rejection of the claim that less than 10​% of

treated subjects experienced headaches.