Question

In a clinical​ trial, 16 out of 874 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 1.6​% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 1.6​% of this​ drug's users experience flulike symptoms as a side effect at the alpha equals 0.1 level of​ significance? Because np 0 left parenthesis 1 minus p 0 right parenthesisequals nothing ▼ less than equals greater than not equals ​10, the sample size is ▼ greater than less than ​5% of the population​ size, and the sample ▼ is given to not be random, is given to be random, cannot be reasonably assumed to be random, can be reasonably assumed to be random, the requirements for testing the hypothesis ▼ are are not satisfied. ​(Round to one decimal place as​ needed.) Find P value Find null and alternative hypothesis

Answer 1

given that

n = 874 and population proportion po = 1.6/100 = 0.016

n*po*(1-po)

= 874*0.016*(1-0.016)

= 13.8 > 10 (satisfied)

Because np 0 left parenthesis 1 minus p 0 right parenthesisequals 13.8 equals or greater than ​10, the sample size is less than ​5% of the population​ size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied

We have to test whether the proportion is 1.6% or higher. So, it is a right tailed hypothesis test

Ho: p = 0.016

H1: p >0.016

Using TI 84 calculator

press stat then tests then 1-PropZtest

po = 0.016

x = 16

n = 874

prop>po

press ENTER

p value = 0.295 (3 decimals)