A certain drug is used to treat asthma, In a clinical trial of the drug, 26 of 2

Question

A certain drug is used to treat asthma, In a clinical trial of the drug, 26 of 299 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 7518094 distribution and assume a 0.05 significance level to complete parts (a) through (e) below. a. Is the best two-tailed, left-tailed, or right-tailed? Two-tailed test Right tailed test Left-tailed test b. What is the test statistic? (Round to two decimal places as needed) c, What is the P-value? (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis. A. H_0: P = 0.1 B. p 0.1 Decide whether to reject the null hypothesis. Choose the correct answer below. A. Reject the null hypothesis because the P-value is less than or equal to the significance level, alpha B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, alpha. C. Reject the null hypothesis because the P-value is greater than the significance level, alpha. D. Fail to reject the null hypothesis because the P - value is greater than the significance level, alpha. e. What is the final conclusion? A. There is sufficient evidence to support the claim that less than 10% of treated subjects experienced headaches. B. There is not sufficient evidence to warrant rejection of the claim that less than 10% of treated subjects experienced headaches. C. There is not sufficient evidence to support to support the claim that less than 10% for treated subject experienced headaches. D. There is sufficient evidence to warrant rejection of the claim that less than 10% of treated subject experienced headaches.

Explanation / Answer

Given that,
possibile chances (x)=26
sample size(n)=299
success rate ( p )= x/n = 0.087
success probability,( po )=0.1
failure probability,( qo) = 0.9
null, Ho:p=0.1
alternate, H1: p<0.1
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.08696-0.1/(sqrt(0.09)/299)
zo =-0.7518
| zo | =0.7518
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =0.752 & | z | =1.64
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.75181 ) = 0.22608
hence value of p0.05 < 0.22608,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.1
left tailed test, alternate, H1: p<0.1
test statistic: -0.7518
critical value: -1.64
decision: do not reject Ho
p-value: 0.22608

failed to reject the null hypothesis becuase the null value greater than
the significance level 0.05

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