To combat antibiotic resistance, the Quality Improvement Consortium recommends a
ID: 3174675 • Letter: T
Question
To combat antibiotic resistance, the Quality Improvement Consortium recommends a throat s' confirm strep throat before a physician prescribes antibiotics to children under age 5. In a random of 97 children who received antibiotics for throat infections. 29 did not have a throat swab. At this a statistically significant reduction over last year's national rate of 40 percent? Choose the correct null and alternative hypotheses. H_0: pi greaterthanorequalto 0.40 vs. H_1: pi 0.40 Calculate the critical value. (A negative value should be indicated by a minus sign, answer to 3 decimal places.) Calculate the test statistic. (A negative value should be indicated by a minus sign. Round your answers to 2 decimal places.) Interpret the results. Reject H_0 Fail to reject H_0 Is it safe to assume normality of the sample proportion p? No, normality is not justified. Yes, normality is justified.Explanation / Answer
Given that,
possibile chances (x)=29
sample size(n)=97
success rate ( p )= x/n = 0.299
success probability,( po )=0.4
failure probability,( qo) = 0.6
null, Ho:p=0.4
alternate, H1: p!=0.4
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.29897-0.4/(sqrt(0.24)/97)
zo =-2.0311
| zo | =2.0311
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =2.031 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.03112 ) = 0.04224
hence value of p0.05 > 0.0422,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.4
alternate, H1: p!=0.4
test statistic: -2.0311
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.04224
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