In a clinical trial, 26 out of 890 patients taking a prescription drug daily com
ID: 3046330 • Letter: I
Question
In a clinical trial, 26 out of 890 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 2.5% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.5% of this drug's users experience flulike symptoms as a side effect at the alpha equals 0.05 level of significance? Because np 0 left parenthesis 1 minus p 0 right parenthesisequals nothing less than equals not equals greater than 10, the sample size is 5% of the population size, and the sample the requirements for testing the hypothesis are not are satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Upper H 0: p mu sigma greater than less than not equals equals nothing versus Upper H 1: sigma p mu less than equals not equals greater than nothing (Type integers or decimals. Do not round.) Find the test statistic, z 0. z 0equals nothing (Round to two decimal places as needed.) Find the P-value. P-valueequals nothing (Round to three decimal places as needed.) Choose the correct conclusion below. A. Since P-valueless thanalpha, do not reject the null hypothesis and conclude that there is sufficient evidence that more than 2.5% of the users experience flulike symptoms. B. Since P-valuegreater thanalpha, do not reject the null hypothesis and conclude that there is not sufficient evidence that more than 2.5% of the users experience flulike symptoms. C. Since P-valuegreater thanalpha, reject the null hypothesis and conclude that there is not sufficient evidence that more than 2.5% of the users experience flulike symptoms. D. Since P-valueless thanalpha, reject the null hypothesis and conclude that there is sufficient evidence that more than 2.5% of the users experience flulike symptoms.
Explanation / Answer
Given that,
possibile chances (x)=26
sample size(n)=890
success rate ( p )= x/n = 0.0292
success probability,( po )=0.025
failure probability,( qo) = 0.975
null, Ho:p=0.025
alternate, H1: p>0.025
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.02921-0.025/(sqrt(0.024375)/890)
zo =0.8051
| zo | =0.8051
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =0.805 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 0.80513 ) = 0.21037
hence value of p0.05 < 0.21037,here we do not reject Ho
ANSWERS
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null, Ho:p=0.025
alternate, H1: p>0.025
test statistic: 0.8051
critical value: 1.64
decision: do not reject Ho
p-value: 0.21037
B. Since P-valuegreater thanalpha, do not reject the null hypothesis and conclude that there is not sufficient evidence that more than 2.5% of the users experience flulike symptoms