Question

In: Statistics and Probability

A pharmaceutical industry survey claims that 9 out of 10 doctors recommend aspirin for their patients...

A pharmaceutical industry survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test the claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend aspirin. What is the value of the test statistic used in evaluating this claim?  Include 2 decimal places in your answer.

Solutions

Expert Solution

Given that,
possibile chances (x)=83
sample size(n)=100
success rate ( p )= x/n = 0.83
success probability,( po )=0.9
failure probability,( qo) = 0.1
null, Ho:p=0.9
alternate, H1: p<0.9
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.83-0.9/(sqrt(0.09)/100)
zo =-2.333
| zo | =2.333
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =2.333 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -2.33333 ) = 0.00982
hence value of p0.05 > 0.00982,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.9
alternate, H1: p<0.9
test statistic: -2.333 =-2.33
critical value: -1.645
decision: reject Ho
p-value: 0.00982
we have enough evidence to support the claim that the actual proportion of doctors who recommend aspirin is less than 0.90


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